We are asked: "Alex divides a disk into four quadrants with two perpendicular diameters intersecting at the center of the disk. He draws $25$ more lines segments through the disk, drawing each segment by selecting two points at random on the perimeter of the disk in different quadrants and connecting these two points. Find the expected number of regions into which these $27$ line segments divide the disk."

Interpretation: Initially, the disk is divided by two perpendicular diameters (presumably they are lines through the center dividing the disk into 4 equal quadrants). So initially, there are 2 lines (the diameters) dividing the disk into 4 regions (the quadrants). Then he draws 25 additional line segments, each segment is drawn by selecting two points at random on the perimeter (the circumference) of the disk, but with the condition that the two points are in different quadrants (presumably the quadrants defined by the initial two diameters). And then connecting these two points with a line segment (a chord). So we have total 2 diameters + 25 chords = 27 line segments (some are diameters, some are chords). The question: Expected number of regions into which these 27 line segments divide the disk.

We need to compute expected number of regions formed by these lines (chords and diameters). The lines are placed randomly subject to the constraint that each chord's endpoints lie in different quadrants.

We need expected number of regions. Typically, for a set of lines (chords) in a disk, the number of regions formed can be analyzed using Euler's formula: R = 1 + E - V + C? Actually for planar graph: regions = 1 + edges - vertices + (number of connected components). But careful: The disk is a bounded region, lines are chords inside the disk, intersections occur at points where chords cross, or at endpoints on boundary. The boundary (circle) is also part of the graph. So we have a planar graph embedded in the disk. The number of regions (faces) is given by F = 1 + E - V + C, where C is number of connected components? Actually for a planar graph drawn on a sphere, F = E - V + 2. For a planar graph drawn on a disk with boundary, we have to treat the outer boundary as part of the graph. If we consider the graph including the boundary cycle, then Euler formula: V - E + F = 1 + C? Actually standard Euler formula for a connected planar graph drawn on a disk: V - E + F = 1 + (number of connected components)? Let's recall: For a planar graph with possibly multiple components, embedded in the plane, the formula is V - E + F = 1 + C, where C is number of components, and F includes the unbounded face. If we restrict to a disk, the boundary is a cycle that is part of the graph. So we treat the disk as a closed disc, the boundary is a cycle. Then the graph is drawn inside, with vertices on boundary. The outer face is the "outside" of the disk? Actually we only consider the disk, so the region outside the disk is not considered; the boundary is part of the graph, so the disk's interior is the whole drawing. So Euler's formula for a planar graph embedded in a closed disk (with boundary as part of graph) is: V - E + F = 1 + C, where C is number of connected components of the graph (including the boundary? Usually the boundary is part of the graph, so it might connect components). But careful: In a planar graph with boundary, the outer region is not counted because the boundary encloses the entire drawing; the graph is contained within a closed disc. The Euler formula for a planar graph drawn on a sphere is V - E + F = 2. If we cut out a disc, we get a planar graph with boundary. For a planar graph with boundary (like a planar map), Euler formula: V - E + F = 1 + B, where B is number of boundary components? Actually there is a formula for planar graphs with holes: V - E + F = 1 + C, where C is number of connected components of the graph? Not exactly. Let's derive: Consider the entire disk including its boundary as a topological sphere? Actually the disk is homeomorphic to a sphere with one puncture. If we take the disk and consider its boundary as a cycle, we can cap it with a hemisphere to close into a sphere. The graph on the disk plus the boundary cycle (which is a circle) forms a graph on the sphere if we add a vertex for each intersection? Wait, easier: The number of regions inside the disk is exactly the number of faces in the planar graph formed by the chords and the boundary circle, where the boundary circle is considered as an edge. The graph includes the circle as a cycle. The disk interior is the domain. So we can apply Euler's formula: V - E + F = 1 + C, where C is number of connected components of the graph? Actually standard Euler formula for a planar graph (with possible multiple components) is V - E + F = 1 + C, where F includes the unbounded face. But if the graph includes the outer boundary as part of the edges, then the "unbounded face" is outside the disk, which we don't consider. So we need a version that accounts for the fact that the disk boundary is part of the graph, so there is no external face. Alternatively, we can think of the graph as lying on the sphere by adding a point at infinity; but then the boundary circle becomes a cycle that is not necessarily the outer face. Actually we can treat the disk as a closed disk, and consider the graph formed by the chords and the boundary circle. The interior of the disk is a planar region. The Euler formula for a planar graph drawn on a disk (with the boundary as part of the graph) is V - E + F = 1, where F includes all faces inside the disk. This holds if the graph is connected (including the boundary). But if the graph is disconnected, there may be isolated vertices? However, all vertices lie on the boundary circle? Actually endpoints of chords lie on the boundary, and the diameters also have endpoints on the boundary (since they are diameters, they intersect the boundary at two opposite points). The initial two diameters intersect at center, so center is a vertex interior to the disk. Also chord intersections can occur inside. So there may be multiple components? Typically, the entire graph (including boundary circle) is connected because all vertices lie on the boundary circle, and the boundary circle is a cycle that connects all boundary points (if we consider the boundary circle as an edge connecting points along the circumference). However, the boundary circle is not drawn as a separate edge? Actually the boundary of the disk is the circumference; it is a curve that is part of the region's boundary. In the problem, we are dividing the disk by line segments; the boundary is already present as the perimeter. So we consider the disk's boundary as a closed curve. It is not drawn as a "line segment", but it is part of the division. The line segments are drawn inside. So the division of the disk into regions is determined by the arrangement of the line segments and the boundary. So the graph consists of: the boundary circle (which is a closed curve), and the line segments (chords and diameters). The boundary circle is a single edge? Actually we can think of it as a cycle composed of arcs between consecutive intersection points on the boundary. But those intersection points are the endpoints of chords/diameters on the boundary. So the boundary circle is broken into arcs between these points. So the graph is a planar graph with vertices: all intersection points (on boundary where chords/diameters meet the boundary, and interior intersection points where chords/diameters cross), and edges: arcs along the boundary between consecutive boundary vertices, and the chord/diameter segments between vertices (including interior intersection points). This graph is connected because the boundary cycle connects all boundary vertices, and all chords/diameters connect to boundary vertices (except possibly the center is interior but connected to boundary via diameters). So the graph is connected (assuming at least one chord? The initial two diameters connect center to boundary, so the graph is connected). So we can use Euler's formula: V - E + F = 1 + C, but if graph is connected, C=1, then V - E + F = 2? Wait: For a connected planar graph drawn in the plane (including the unbounded face), V - E + F = 2. But here, if we consider the entire disk as the drawing, the "unbounded face" is outside the disk, which is not considered. However, the graph includes the boundary as edges; the region outside the disk is not part of the domain. So the faces we count are only those inside the disk. So the Euler formula for a planar graph drawn on a disc (with the boundary being part of the graph) is V - E + F = 1, because the "outer face" is not counted; the boundary encloses the drawing. Actually think: If we have a connected planar graph that includes the boundary cycle, then the entire graph is inside a disc; the region outside the disc is not part of the graph's faces. The Euler characteristic of a disc is 1 (since a disc is homeomorphic to a sphere with one hole; the Euler characteristic of a disc is 1). For a graph embedded in a disc with the boundary as part of the graph, the formula is V - E + F = 1, where F counts the faces (regions) inside the disc. This is standard: For a planar graph drawn on a disc with boundary, V - E + F = 1 (if the graph includes the boundary as edges). For example, consider a simple cycle (the boundary alone) with no vertices (or one vertex? Actually boundary cycle: if we have just the circle, no chords. Then there is one face (the interior). V = number of vertices on the circle? If we treat the circle as a closed curve, but no vertices, then we need to consider vertices? Actually if we have a circle with no distinguished points, it's not a graph. But we can consider the circle as an edge forming a cycle. To apply graph theory, we need vertices. For a simple closed curve dividing the disc into interior and exterior? But here we only consider interior. If we treat the circle as a single edge that loops back to itself? That's not a graph in the usual sense (edge must have two endpoints). So we need to place vertices at points where chords intersect the boundary. Without chords, the boundary is just a smooth curve, but we can still consider the disk as a region with no division. The division is by the line segments; the boundary is always present as the outer edge. So initially, with only the two diameters, we have 4 regions (quadrants). That matches: V = ? The two diameters intersect at center, and each diameter has two endpoints on the boundary, so total endpoints: 4 distinct boundary points (since diameters are perpendicular, they intersect at center, and the endpoints are at 0°, 90°, 180°, 270°). So vertices: center (1), plus 4 boundary points. Edges: each diameter is divided into two radii from center to boundary, so 4 radial edges. Also the boundary circle is divided into 4 arcs between consecutive boundary points. So total edges: 4 radial edges + 4 boundary arcs = 8 edges. Then faces: the 4 quadrants. Euler: V=5, E=8, F=4 => 5-8+4=1 => indeed V-E+F=1. So formula holds with F = number of regions. Good.

Thus we can compute expected number of regions = 1 + E - V, where E = total number of edges (segments of lines between vertices, including arcs on boundary) and V = total number of vertices (intersection points, including boundary points and interior crossings). Actually F = 1 + E - V (since graph is connected). So to compute expected number of regions, we need expected E and V.

Alternatively, we can use known formulas for arrangements of lines in a disk: number of regions = 1 + (number of lines) + (number of intersection points) + (number of intersections of lines with boundary?) Wait, there is a classic formula for arrangements of chords in a circle: For a set of chords (with endpoints on the circle), the number of regions formed inside the circle is 1 + (number of chords) + (number of interior intersections of chords). Actually, consider chords drawn in a circle, with endpoints on the circle. If we have n chords, and they intersect in interior points (no three chords concurrent except at endpoints?), then the number of regions is 1 + n + I, where I is the number of interior intersections. But is that correct? Let's test: For n=1 chord, regions = 2. 1 + 1 + 0 = 2, good. For n=2 chords that intersect (i.e., crossing), regions = 4. 1+2+1=4, good. For n=2 chords that do not intersect (non-crossing), regions = 3. 1+2+0=3, good. For n=3 chords all intersecting pairwise? Maximum? If three chords all intersect at one point (concurrent), regions = 6? Actually 3 chords through a common point inside: they divide circle into 6 regions? Let's think: Three chords passing through a common interior point: they emanate from that point to 6 endpoints on the boundary (since each chord has two endpoints). But if they all share a common interior intersection point, then there are 6 boundary points, and the interior point. The edges: from interior point to boundary: 6 edges; boundary arcs: 6 arcs; total edges 12; vertices: 7; then faces = 1+12-7=6. So formula: regions = 1 + n + I? Here n=3, I=1 (one intersection point) gives 1+3+1=5, not 6. So that formula assumes no three chords concurrent? Actually if three chords intersect at a single point, they create 6 regions, but the formula 1 + n + I gives 5. So need to adjust for concurrency: In planar arrangements of lines (or chords), the number of regions = 1 + number of lines + sum over intersections of (multiplicity) where each intersection point contributes (number of lines crossing at that point) - 1? Actually typical formula for arrangement of lines in projective plane: For lines, each intersection point adds 1 to number of regions beyond the base count. But if multiple lines intersect at a point, they add 1 per crossing pair? However, in an arrangement of chords in a circle, if we have chords that intersect only at endpoints (on boundary) and interior intersections, the formula R = 1 + C + I, where C is number of chords, I is number of interior intersection points, holds only when no three chords intersect at a common interior point (i.e., all intersections are simple crossings). But if there is concurrency, then the number of regions is less than that formula because intersections merge. So we need to consider expected number of intersections and also possibly multiple intersections. But in this problem, chords are drawn by selecting two points at random on the perimeter in different quadrants. Since the points are chosen uniformly at random on the circumference (presumably continuous), the probability that three chords intersect at a common point is zero (since that would require specific algebraic conditions). Also the probability that two chords share an endpoint is zero because endpoints are chosen randomly and independently? However, endpoints are chosen randomly, but there is a possibility that two chords might share the same endpoint if the random selection picks the same point twice? But points are chosen at random on the perimeter; the probability of exact coincidence is zero. So with probability 1, all chords have distinct endpoints and no three chords intersect at a single interior point (since that would require specific alignment). Also the diameters: they intersect at center, and they are fixed perpendicular diameters. So the center is an intersection point of the two diameters. There could be intersections between chords and diameters, and chords with chords. But note: chords are drawn by selecting two points from different quadrants. So endpoints are on the circumference, each in a specific quadrant (the quadrants defined by the two diameters). The quadrants are four arcs: between 0° and 90°, 90° and 180°, 180° and 270°, 270° and 360° (or whatever). The condition: the two endpoints of each chord must lie in different quadrants. So chords will always cross the interior; they might intersect the diameters, or intersect each other. Since the endpoints are random, intersections will be generic (no three concurrent, no two chords share endpoint). So we can treat intersections as simple crossings. So the formula R = 1 + N + I, where N is total number of line segments (including the two diameters?) and I is number of interior intersection points (including intersections between chords, and chords with diameters? But careful: The diameters are also lines; they intersect at the center. So that is an interior intersection point (center). Also intersections between chords and diameters: those are interior points (unless a chord passes through the center? Possibly, but probability zero). So we need to count all interior intersection points where two line segments (including diameters) cross. But note: The diameters themselves intersect at the center, but that is a single intersection point. Also chords may intersect diameters at interior points. And chords may intersect each other. Also note: The diameters are drawn initially, dividing the disk into quadrants. They are two lines through the center. They have endpoints on the boundary. So they are part of the arrangement. So total line segments N = 2 (diameters) + 25 (chords) = 27.

But careful: The formula R = 1 + N + I is valid for an arrangement of lines (or segments) in a convex region where each line is drawn from boundary to boundary? Actually chords are segments that connect two boundary points, but diameters are also such segments (connecting opposite boundary points). So each line segment (chord or diameter) is a chord of the circle. For a set of chords (including diameters) in a circle, if no three chords are concurrent at an interior point and no two chords share an endpoint (except possibly endpoints on the boundary that are shared? But here endpoints are distinct with probability 1), then the number of regions is: 1 + (number of chords) + (number of interior intersections). This is a known result: For an arrangement of n chords (with endpoints on the circle) in general position (no three chords intersect at a point, no two chords share endpoints), the number of regions formed inside the circle is 1 + n + I, where I is the number of intersection points (interior crossings). Because each chord adds 1 region when added, and each crossing adds an extra region (since crossing two chords increases regions by 1 compared to if they didn't cross). This can be derived from Euler: V = boundary vertices + interior vertices; boundary vertices = number of distinct endpoints on the circle (which is 2n if no sharing) plus maybe the initial endpoints from diameters? Actually diameters are chords too, so total chords = 27, each chord contributes 2 endpoints, but some endpoints may coincide? With random selection, distinct endpoints with probability 1. So total distinct endpoints = 2*27 = 54 points on the circle. But note: the two diameters have endpoints at fixed positions: they are the four cardinal points (0°, 90°, 180°, 270°). However, the 25 chords are drawn with random endpoints; these endpoints could fall anywhere on the circumference, including possibly at the same positions as the diameter endpoints? Probability of exactly hitting those points is zero, so we can assume all endpoints are distinct. So total boundary vertices = 54 distinct points? But wait, the diameters are already there: they have endpoints at 0°,90°,180°,270°. Those are 4 points. The chords have 25*2 = 50 endpoints, but these could fall anywhere, possibly near those points but not exactly. So total distinct endpoints = 4 (from diameters) + 50 (from chords) = 54. So boundary vertices count = 54.

Interior vertices: intersections between chords (including chords with diameters, and diameters with each other). The diameters intersect at the center (1 point). Also each chord may intersect diameters at most once per diameter? Since chords are straight lines connecting two points on boundary, and diameters are lines through the center. A chord could intersect a diameter if its endpoints are on opposite sides of the diameter line. Since endpoints are in different quadrants, it's possible that a chord crosses a diameter. Because the quadrants are defined by the two diameters; the chord's endpoints are in different quadrants, so the chord necessarily crosses at least one of the diameters? Actually consider the two perpendicular diameters: they divide the circle into four quadrants. If two points are in different quadrants, the chord connecting them may cross either the horizontal or vertical diameter (or both) depending on which quadrants. For example, if endpoints are in quadrants I and III (opposite), the chord will pass through the center? Actually a chord connecting opposite quadrants (I and III) will pass through the center if the points are diametrically opposite? But in general, chord connecting points in opposite quadrants will cross the center? Not necessarily: The line through two points on a circle; if they are antipodal (180° apart), then the chord is a diameter and passes through center. But if they are not exactly antipodal, the chord will not pass through the center. However, if one endpoint is in quadrant I (0-90) and the other in quadrant III (180-270), the chord will cross the horizontal diameter? Let's think: The horizontal diameter is the line from 0° to 180° (the right-left line). Quadrant I is top-right? Actually typical: angles: 0° at right? Let's set coordinates: Let the diameters be vertical and horizontal through center. So quadrants: Q1: top-right (0-90°), Q2: top-left (90-180°), Q3: bottom-left (180-270°), Q4: bottom-right (270-360°). The horizontal diameter goes from 0° (right) to 180° (left). The vertical diameter goes from 90° (top) to 270° (bottom). A chord connecting a point in Q1 (top-right) and Q3 (bottom-left) will cross both diameters? Actually it will cross the vertical diameter if the points are on opposite sides of the vertical line? Since Q1 is to the right of vertical axis (x>0) and Q3 is to the left (x<0), so the chord must cross the vertical line x=0 somewhere. Similarly, Q1 has y>0, Q3 has y<0, so it must cross the horizontal line y=0. So it will cross both diameters? Possibly at the same point if the chord passes through the center? That would require points antipodal, but general chord will cross vertical and horizontal lines at different points, so it will intersect each diameter once. So a chord connecting opposite quadrants will intersect both diameters (twice). For chords connecting adjacent quadrants (e.g., Q1 and Q2), they will cross only one diameter? Q1 (right, top) and Q2 (left, top): both have y>0, so they are both above horizontal axis, so they do not cross horizontal diameter; but Q1 has x>0, Q2 has x<0, so they cross vertical diameter. So chord will intersect vertical diameter. Similarly, Q1 and Q4: both x>0, so cross horizontal? Actually Q4 is bottom-right: x>0, y<0. So Q1 (y>0) and Q4 (y<0) cross horizontal diameter; but x both positive, so no crossing vertical. Q2 and Q3: both x<0, cross horizontal; Q2 and Q4: Q2 (x<0,y>0) and Q4 (x>0,y<0) cross both? Actually Q2 (left,top) and Q4 (right,bottom): x opposite, y opposite, so cross both diameters. Q3 and Q4: both y<0, cross vertical? Q3 (x<0,y<0) and Q4 (x>0,y<0): cross vertical. So chords always cross at least one diameter, and sometimes two. So chords will intersect diameters.

Thus interior vertices include:
- Center (intersection of the two diameters) — 1 point.
- Intersections between each chord and each diameter it crosses. Since endpoints are in different quadrants, each chord crosses exactly 1 or 2 diameters. The number depends on which pair of quadrants the endpoints lie in. Let's denote quadrants: 1,2,3,4. Condition: endpoints in different quadrants. So there are 6 possible unordered pairs: {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}. For each pair, the chord will cross:
- Adjacent quadrants (sharing a side): {1,2} share vertical line? Actually Q1 and Q2 share the vertical diameter? They share the boundary at the vertical diameter? Q1 and Q2 are separated by the vertical diameter (the line from top to bottom). So chord connecting a point in Q1 and a point in Q2 will cross the vertical diameter. Similarly {2,3} share horizontal? Actually Q2 and Q3 share horizontal diameter? Q2 top-left, Q3 bottom-left: they are separated by the horizontal diameter? Yes, the horizontal diameter runs left-right; Q2 has y>0, Q3 y<0, so they are separated by horizontal line. So {2,3} cross horizontal. {3,4} share vertical? Q3 bottom-left, Q4 bottom-right: separated by vertical? Yes, vertical line. {4,1} share horizontal? Q4 bottom-right, Q1 top-right: separated by horizontal. So adjacent quadrants (sharing a diameter) produce chords crossing exactly that one diameter (the one that separates them).
- Opposite quadrants: {1,3} and {2,4} are opposite pairs (diagonally opposite). Those chords cross both diameters (since they cross both horizontal and vertical). So chords from opposite quadrants intersect both diameters.

Thus each chord will intersect either 1 or 2 diameters, depending on the quadrant pair. So we need to count expected number of intersections between chords and diameters.

Additionally, chords may intersect each other. Two chords intersect if and only if their endpoints are interleaved around the circle (assuming chords are drawn as straight lines). For chords with endpoints on circle, two chords intersect (in interior) if and only if the endpoints are in alternating order around the circle. This is a known property: For four distinct points A,B,C,D on circle in order, chords AC and BD intersect if A and C are separated by B and D. More precisely, if we label endpoints in order around the circle, chords (a,b) and (c,d) intersect if and only if the endpoints are interleaved: one chord's endpoints are separated by the other chord's endpoints. So we need to compute expected number of intersections between chords, given that endpoints are chosen randomly with the constraint that each chord's endpoints lie in different quadrants. Also chords may intersect with each other and with diameters. But diameters are fixed lines, so we can compute expected number of intersections between chords and diameters based on quadrant assignments. Also chords may intersect each other; we need expected number of such intersections.

Also note: The diameters themselves intersect at center. That's one interior intersection. Also chords may intersect at center? Possibly if a chord passes through center, but that would require endpoints antipodal, probability zero. So ignore.

Thus we can compute total expected number of interior intersection points I = intersections among chords + intersections between chords and diameters + intersection of diameters (1). However, note that if a chord intersects both diameters, those are two distinct intersection points (unless the chord passes through center, which is unlikely). So each chord-diameter intersection is a point. Also, if two chords intersect each other, that's a point. All these points are distinct with probability 1 (since no three chords concurrent etc.). So I_total = I_diameters (1) + I_chord-diameter + I_chord-chord.

Then expected number of regions = 1 + N + I_total, where N = number of line segments (27). So we need E[I_total].

We need to compute expected number of intersections among 25 chords, and expected number of intersections between chords and diameters.

We must consider randomness: The 25 chords are drawn sequentially? But each chord is drawn by selecting two points at random on the perimeter of the disk in different quadrants. The selection is presumably independent for each chord, with uniform distribution over the circumference within each quadrant? The wording: "selecting two points at random on the perimeter of the disk in different quadrants". It doesn't specify distribution, but presumably each point is chosen uniformly at random from the circumference, and they are independent, conditioned on being in different quadrants. Since the quadrants are four arcs of equal length (π/2 each), the probability that a random point lies in a given quadrant is 1/4. However, the condition that the two points are in different quadrants means we choose two points independently uniformly from the circumference, then reject if they are in the same quadrant? Or we could directly choose one point uniformly from each quadrant? Usually, "selecting two points at random on the perimeter ... in different quadrants" could mean that we choose two points uniformly at random from the entire circumference, but we only consider pairs that lie in different quadrants (or we condition on that). Since the problem asks for expected number, it's likely that the selection process is uniform among all pairs of points on the circumference that belong to different quadrants. Equivalently, we can think of choosing two distinct quadrants uniformly among the four, then choose a point uniformly in each chosen quadrant. However, there is also possibility that both points could be in different quadrants but the quadrants are not necessarily uniformly chosen? Since the points are uniformly random on the circumference, the distribution of which quadrants they fall in is uniform over the four quadrants (each with probability 1/4), independent for the two points. Then conditioned on being in different quadrants, the pair of quadrants is uniformly distributed over the 4*3=12 ordered pairs? But since unordered, each unordered pair of distinct quadrants has probability 1/6? Actually for two independent uniform points on circle, probability that they are in same quadrant is 4*(1/4)^2 = 4*(1/16)=1/4? Actually each quadrant probability 1/4, so probability both in same quadrant = 4*(1/4)*(1/4)=1/4. So probability different quadrants = 3/4. Among different quadrants, the conditional probability for a specific unordered pair {i,j} is (2/4)*(1/4)*? Let's compute: For independent uniform points, the probability that point1 in quadrant i and point2 in quadrant j, i≠j, is (1/4)*(1/4)=1/16 each ordered pair. There are 4*3=12 ordered pairs, total probability 12/16=3/4. So conditional on different quadrants, each ordered pair has probability (1/16)/(3/4)=1/12. So unordered pair {i,j} has probability 2*(1/12)=1/6. So indeed each of the 6 unordered pairs equally likely. Then given quadrants i and j, each point is uniformly distributed within its quadrant (since conditional on being in that quadrant, the distribution is uniform on that arc). So we can treat each chord as: choose two distinct quadrants uniformly among the 6 pairs, then pick a point uniformly in each of those two quadrants. That is the distribution.

Thus for each chord, we have:
- The quadrant pair (i,j) is uniformly random among 6 possibilities.
- Given quadrants, the points are independent uniform on those arcs.

We need expected number of intersections among the 25 chords (chord-chord intersections). Also expected number of intersections between chords and the two diameters.

We have two diameters: vertical (from 90° to 270°) and horizontal (from 0° to 180°). We'll define coordinates: Let the circle be unit circle centered at origin. Horizontal diameter: line y=0 from x=-1 to 1, endpoints at (1,0) and (-1,0) (angles 0° and 180°). Vertical diameter: line x=0 from y=-1 to 1, endpoints at (0,1) and (0,-1) (angles 90° and 270°). Quadrants: Q1: angles (0°,90°) = right-top; Q2: (90°,180°) = left-top; Q3: (180°,270°) = left-bottom; Q4: (270°,360°) = right-bottom.

Given a chord connecting points at angles θ1 and θ2 (with θ1,θ2 in degrees or radians), the chord will intersect the horizontal diameter if the points are on opposite sides of the horizontal line (i.e., one with y>0, one with y<0). That is, if one angle is in (0°,180°) and the other in (180°,360°) but not exactly 0,180 etc. Since quadrants: Q1 and Q4 have y>0? Actually Q1: angles (0,90) => y>0, x>0; Q2: (90,180) => y>0, x<0; Q3: (180,270) => y<0, x<0; Q4: (270,360) => y<0, x>0. So horizontal diameter crossing condition: one endpoint in Q1 or Q2 (y>0) and the other in Q3 or Q4 (y<0). Similarly, vertical diameter crossing: one endpoint in Q1 or Q4 (x>0) and the other in Q2 or Q3 (x<0). So for a given chord, we can determine which diameters it crosses based on the quadrants of its endpoints. As we noted:
- Adjacent quadrants sharing a diameter: {1,2}: both y>0, so cross vertical (since x opposite); {2,3}: both x<0, cross horizontal; {3,4}: both y<0, cross vertical; {4,1}: both x>0, cross horizontal.
- Opposite quadrants: {1,3}: one y>0, one y<0 => cross horizontal; also x>0 and x<0 => cross vertical. So both. {2,4}: similarly both.

Thus each chord crosses exactly one diameter if the quadrants are adjacent, and both diameters if opposite.

Now, we need expected number of intersections between chords and diameters. For each chord, we count number of intersections with diameters: 1 or 2. Since diameters are fixed lines, each chord will intersect each diameter it crosses at exactly one point (since line is straight and diameters are lines through center, crossing condition ensures they intersect). The intersection point is interior (except possibly if chord endpoint lies exactly on diameter, probability zero). So for each chord, we have either 1 or 2 intersections with diameters. So expected number of chord-diameter intersections = 25 * E[number of diameter intersections per chord]. Since chords are independent? They are drawn independently, so total expected intersections = 25 * p, where p = expected number of intersections per chord with diameters. For a random chord with endpoints in different quadrants uniformly, we compute expected number of diameter intersections.

We can compute: For a random chord with endpoints in different quadrants, the distribution over quadrant pairs is uniform over 6 pairs. Among those, pairs:
- Adjacent pairs: {1,2}, {2,3}, {3,4}, {4,1}: these are 4 pairs, each yields 1 intersection.
- Opposite pairs: {1,3}, {2,4}: these are 2 pairs, each yields 2 intersections.
So expected intersections per chord = (4*1 + 2*2)/6 = (4+4)/6 = 8/6 = 4/3 ≈ 1.3333. So expected chord-diameter intersections total = 25 * 4/3 = 100/3 ≈ 33.333... But careful: This is expectation of number of intersections. However, we also have the diameters intersecting each other at center, which is 1 intersection. So I_diameters = 1.

Now chord-chord intersections: We have 25 chords. For any two distinct chords, they may intersect if their endpoints are interleaved around the circle. Given the chords are drawn independently with the quadrant constraint, we need to compute the probability that two chords intersect. Then expected number of chord-chord intersections = number of unordered pairs of chords (C(25,2) = 300) times probability that two random chords (with the given distribution) intersect. Since chords are independent and identically distributed, we can compute the probability that two chords intersect.

So we need to compute: For two random chords, each formed by picking two points uniformly from the circumference conditioned on being in different quadrants (with distribution as described), what is the probability that the chords intersect? This depends on the endpoints' positions and the quadrant assignments. The intersection condition for two chords (a,b) and (c,d) (with endpoints on circle) is that the endpoints are in alternating order around the circle. More precisely, if we list the four points in circular order, the chords intersect if and only if the endpoints of the two chords are interleaved: e.g., a, c, b, d (in order) yields intersection. Equivalent to: one chord's endpoints separate the other chord's endpoints. So we need the probability that for two random chords, the four endpoints (all distinct with probability 1) are in alternating order.

Given that each chord's endpoints are chosen from different quadrants, we need to compute the probability that two chords intersect. Since chords are independent, we can consider the joint distribution of endpoints.

We can compute this probability by integrating over the positions. Alternatively, we can use symmetry: Since the quadrants are symmetric and the diameters are fixed, we might compute the probability that two chords intersect given the quadrant configuration of the endpoints. But maybe there is a simpler approach: The probability that two random chords intersect, when endpoints are chosen uniformly on the circle without any restrictions, is 1/3? Actually known result: For two random chords in a circle, with endpoints chosen uniformly and independently on the circle (four points independent uniform), the probability that the chords intersect is 1/3. Because there are three ways to pair the four points into two chords: the actual pairing is given; given four random points on a circle, the probability that the two chords (determined by pairing) intersect is 1/3. But here chords are not arbitrary pairings; they are specific pairs (each chord connects two points that are chosen as a pair). The chords are drawn by selecting two points per chord. So for two independent chords, we have four points chosen: two for chord1 (A,B) and two for chord2 (C,D). The chords intersect if and only if the order is interleaved: A, C, B, D (or A, D, B, C) etc. Since the points are random, the probability that a given pairing (A-B and C-D) results in crossing is 1/3 if the points are i.i.d. uniform on circle. Indeed, for four i.i.d. uniform points on circle, the probability that the chords (A,B) and (C,D) (with given pairing) intersect is 1/3, because among the three possible perfect matchings on the four points (pairings), exactly one yields crossing chords? Actually for four points on a circle, there are three ways to partition into two pairs (up to labeling). The chords intersect if the pairing is the "crossing" pairing (i.e., the pairs are interleaved). When the points are in random order, each pairing is equally likely? But here the pairing is fixed: we have chords (A,B) and (C,D). Given four random points, the order around the circle is random. The probability that A and B are separated by C and D (i.e., A,C,B,D in order) is 1/3. Because there are 3 equally likely ways to pair the points (if we consider all possible matchings), but given the labeling, the probability that the specific pairing (A-B, C-D) is the crossing one is 1/3. This is a known fact: For four random points on a circle, the probability that two chords (determined by pairing the points) intersect is 1/3. This can be derived: There are 3 ways to pair 4 points into 2 pairs; exactly one pairing yields crossing chords. Since the points are i.i.d., all pairings are equally likely? However, the pairing is not uniformly random among the three; it's predetermined by the chord assignment. But if the points are i.i.d., the order of points is uniform among all permutations. Given the labeling A,B,C,D, the probability that the order is A, C, B, D (up to cyclic rotation) is 1/3? Let's compute: For four random points on a circle, after ordering them (say by angle), the relative order of the four labels is uniformly random among the 4! = 24 possible orders (up to cyclic symmetry? Actually because circle, we can fix one point at angle 0, but still the order of the other three relative to it is uniform). More concretely: The probability that the two chords intersect is 1/3. This is a standard result: For two random chords, the probability of intersection is 1/3. Many sources: e.g., "If two chords are chosen at random from a circle, the probability that they intersect is 1/3". Indeed, because there are three ways to connect four points, and one yields crossing. But careful: That result often assumes that the chords are formed by selecting two random points on the circle and connecting them, and then independently selecting another two random points and connecting them. Then the four points are i.i.d. uniform. Then the probability that the chords intersect is 1/3. So if there were no quadrant restrictions, the expected number of intersections among 25 chords would be C(25,2)*(1/3). But here there is a restriction: each chord's endpoints must be in different quadrants. Does that change the probability? Possibly because the distribution of endpoints is not uniform over the whole circle; it's conditioned on being in different quadrants. Also the quadrants are defined by the diameters, which are fixed lines. This restriction could affect the relative likelihood of crossing. We need to compute the probability that two chords intersect given that each chord's endpoints are in different quadrants. The quadrants are arcs of length π/2. The condition that each chord's endpoints are in different quadrants might affect the distribution of angles; it may make chords more likely to be "long" chords (connecting different quadrants) and thus possibly increase crossing probability. Let's analyze.

We have four quadrants: Q1, Q2, Q3, Q4. For each chord, the two endpoints lie in two distinct quadrants, uniformly among the 6 pairs. And given quadrants, each endpoint is uniform on its quadrant arc. So we need to compute, for two independent chords, the probability that they intersect. Since chords are independent, the probability is the expectation over the random quadrant pairs and random positions within quadrants.

We can compute the probability by integrating over angles. Alternatively, we can use symmetry to reduce to cases based on the quadrant configurations. There are many possibilities: The two chords each have a pair of quadrants. The four endpoints (two per chord) lie in quadrants. The quadrants may be the same or different across chords. We need to compute the probability that the chords intersect given the quadrant configuration.

Because the quadrants are arcs of length π/2, and the condition of endpoints being in different quadrants ensures that each chord connects two arcs that are either adjacent or opposite. The crossing condition for two chords depends on the angular positions.

Let's denote the circle parameterized by angle θ in [0, 2π). Quadrant boundaries: 0, π/2, π, 3π/2, 2π. Q1: [0, π/2), Q2: [π/2, π), Q3: [π, 3π/2), Q4: [3π/2, 2π). For each chord, we have two angles θ1,θ2 in distinct quadrants. The chord's endpoints are (θ1,θ2). We'll treat unordered pairs.

We need probability that two chords intersect. For four points on a circle, the intersection condition is that the chords are crossing: if we order the angles cyclically, the endpoints of one chord are separated by the endpoints of the other chord. Equivalent to: the two chords intersect if and only if exactly one endpoint of chord1 lies in the arc between the endpoints of chord2, and the other endpoint lies outside that arc. More precisely, given chord2 with endpoints c and d (angles), the chord1 endpoints a,b intersect chord2 if and only if a and b are on opposite sides of the chord (i.e., one lies in the minor arc between c and d, the other in the complementary arc). This is equivalent to the alternating condition.

Given the quadrant constraints, we need to compute the probability that for two independent chords, the four angles satisfy the crossing condition.

We can compute by considering the distribution of the angles. Since the quadrants are intervals, we can use symmetry: Because the quadrants are symmetric, and the diameters are fixed, we can perhaps compute the probability that two chords intersect by integrating over the angles. Or we can compute using geometric probability: For two random chords with endpoints uniformly on the circle (without restrictions), the probability is 1/3. With the restriction that each chord's endpoints are in different quadrants, we need to compute the conditional probability given that each chord's endpoints are in different quadrants. Since the event "endpoints in different quadrants" has probability 3/4 for each chord, and chords are independent, the overall probability for two chords to have endpoints in different quadrants is (3/4)^2. But we need the conditional probability of intersection given that condition. Alternatively, we can compute directly the probability that two chords (each with endpoints in different quadrants) intersect by enumerating quadrant configurations and using integration.

We can attempt to compute the probability that two chords intersect, using the law of total probability over quadrant assignments. Let chord1 have quadrants (i1,j1) with i1≠j1, chord2 have quadrants (i2,j2) with i2≠j2, each uniformly over 6 pairs. Then, given quadrants, the endpoints are uniform on those arcs. So we need P(intersection | quadrants). Then average over quadrant pairs.

Because the quadrants are intervals of length π/2, and the chords may cross depending on the actual angular positions within arcs.

We need to compute for each combination of quadrant pairs, the probability that two chords intersect, given independent uniform points in those quadrants.

We have 6 possible quadrant pairs for each chord. For two chords, there are 6*6=36 combinations. But many are symmetric by rotations and reflections (since the quadrants are symmetric). Because the quadrants are arranged in order, we can reduce to equivalence classes.

Let's denote quadrants as 1,2,3,4 in order around the circle. The quadrants are adjacent: 1-2, 2-3, 3-4, 4-1 are adjacent; 1-3 and 2-4 are opposite.

Because of symmetry, we can fix chord1's quadrants to a canonical form, and then consider possibilities for chord2 relative to chord1. The overall probability will be symmetric.

We can compute the probability that two chords intersect by integrating over angles. Alternatively, we could compute using a combinatorial approach: For four random points chosen uniformly from the circle subject to each chord's endpoints being in different quadrants, we want probability that the pairing (A-B, C-D) yields crossing. Since the chords are independent, we can think of drawing four points: first chord's two points (A,B) with constraint different quadrants, then chord2's two points (C,D) with same constraint. But note: The quadrants are defined relative to the fixed diameters. So the distribution is not uniform over the circle; it's conditional on the quadrants. However, we might compute by direct integration over angles.

Let's set up coordinate: Let angles be in radians. For chord1, pick two angles θ1,θ2 uniformly from intervals I1 and I2, where I1 and I2 are two distinct quadrants (arcs of length π/2). Similarly for chord2, pick φ1,φ2 uniformly from two quadrants (possibly same as chord1's quadrants, but independent). We need probability that the four angles (θ1,θ2,φ1,φ2) satisfy crossing condition: when sorted, they alternate between the two chords.

We can compute by considering the order of the four angles. Since all angles are independent uniform over their respective arcs (which are subsets of [0,2π)), we can compute the probability that the order is alternating. Because the arcs are intervals, we might compute using symmetry: The relative order of points from different arcs depends on the arcs' positions. For example, if chord1's quadrants are adjacent (say 1 and 2), its endpoints lie in Q1=[0,π/2) and Q2=[π/2,π). So one endpoint is in [0,π/2), the other in [π/2,π). Note that Q1 and Q2 are adjacent, with boundary at π/2. Similarly, chord2 could have quadrants that are adjacent or opposite. The arcs are disjoint or overlapping? Quadrants are disjoint intervals covering the circle, but they are adjacent. So the endpoints from different quadrants are in specific intervals. The crossing probability depends on the relative positions of these intervals.

We can approach by enumerating all possible assignments of quadrants for the four endpoints. Since each chord's endpoints come from two distinct quadrants, there are 4 endpoints, each belonging to some quadrant. The quadrants may be all distinct, or some may coincide (e.g., chord1 and chord2 could share a quadrant). The condition that each chord's endpoints are in different quadrants imposes restrictions: For chord1, its two endpoints are in two different quadrants; for chord2 similarly. So overall, the four endpoints are in quadrants that could be: all four distinct (each chord uses two distinct quadrants, possibly overlapping). For example, chord1 uses Q1 and Q2; chord2 uses Q2 and Q3; then endpoints: one from Q1, two from Q2? Actually chord2 uses Q2 and Q3, so chord2 has one endpoint in Q2 and one in Q3. So total endpoints: chord1 gives one in Q1, one in Q2; chord2 gives one in Q2, one in Q3. So we have two endpoints in Q2? But each endpoint is a distinct point; they are independent, so two points in Q2 are possible, and they are independent uniform on Q2. So quadrants can be reused across chords.

Thus we need to compute for each combination of quadrant pairs (i1,j1) and (i2,j2), the probability that the chords intersect given uniform points in those quadrants. Then average over pairs.

Because of symmetry, we can fix chord1's quadrants to a specific pair, and then consider chord2's quadrants. Since the quadrants are arranged cyclically, we can consider cases based on the relationship between the two pairs: they could be the same pair, or adjacent pairs, opposite pairs, etc. But careful: The quadrants are labeled, but because of rotational symmetry? The diameters are fixed, so the quadrants are fixed. However, the distribution of points is symmetric with respect to rotations by multiples of π/2? The quadrants are symmetric under rotations by 90°, because the diameters are perpendicular and centered. But note: The quadrants are defined by the diameters; the diameters are at 0° and 90°. The distribution of points is uniform over each quadrant. Since the quadrants are arcs of equal length, the entire configuration is symmetric under rotations that map quadrants to quadrants (i.e., rotations by multiples of π/2). However, the diameters themselves are fixed lines; but the chords are drawn without any directional bias, so the system is symmetric under the dihedral group of order 8 (rotations by 90° and reflections). So we can use symmetry to reduce cases.

Thus we can fix chord1 to be, say, from Q1 and Q2 (adjacent). Because all adjacent pairs are equivalent by rotation (since Q1-Q2 is adjacent, Q2-Q3 is adjacent, etc.). Similarly, opposite pairs (Q1-Q3) are equivalent. So we can compute the probability of intersection for chord1 = (1,2) and chord2 = any pair, and then multiply by the probability of chord2's pair (which is uniform over 6). And similarly if chord1 is opposite, we can compute separately, and then combine because chord1's pair is also uniformly chosen. But we can compute overall expectation by symmetry: The probability of intersection given two chords is symmetric, so we can compute by averaging over all possibilities.

Thus we need to compute:
P_intersect = E[P(intersection | quadrants of chord1, chord2)] = sum_{p,q} (1/36) * P_intersect_given_quadrants(p,q), where p,q range over 6 pairs.

We can compute P_intersect_given_quadrants(p,q) by integration. Since the arcs are intervals of length π/2, we can compute the probability that two chords cross given that each endpoint lies in specified arcs.

We need to compute for each combination of quadrant pairs. There are 36 combos, but many are symmetric. We can categorize based on the relative positions of the arcs. Let's define quadrants as intervals: Q1 = [0, π/2), Q2 = [π/2, π), Q3 = [π, 3π/2), Q4 = [3π/2, 2π). For chord1, its two endpoints are in two quadrants; similarly chord2.

We can compute P_intersect for given quadrants by considering the order of the four points. Since the points are uniform in intervals, we can compute the probability that the four points, when sorted by angle, alternate between chord1 and chord2. This is a classic problem: given two pairs of intervals (possibly overlapping? They are disjoint but adjacent), we need the probability that the order is alternating.

Because the intervals are disjoint (except boundaries, measure zero), the ordering of points from different intervals is determined by the intervals' positions. For instance, Q1 < Q2 < Q3 < Q4 in terms of angle range (since Q1: 0-90, Q2:90-180, Q3:180-270, Q4:270-360). So the natural order of quadrants is 1,2,3,4. So any point from Q1 has angle less than any point from Q2? Not strictly: Q1 angles in [0,π/2), Q2 in [π/2,π). So yes, any angle from Q1 is less than any angle from Q2 because the maximum in Q1 is π/2 (exclusive) and minimum in Q2 is π/2 (exclusive? Actually boundaries: Q1 includes 0 but excludes π/2; Q2 includes π/2? Typically, if we define quadrants as open intervals, the boundaries have measure zero, so we can treat them as open intervals. So any point in Q1 is < any point in Q2, because the right endpoint of Q1 is π/2, left endpoint of Q2 is π/2, but since points are random, probability of hitting boundary is zero. So we can assume strict inequalities: points from Q1 are less than points from Q2, which are less than points from Q3, which are less than points from Q4. So the order of points across quadrants is deterministic: all points from Q1 come first, then Q2, then Q3, then Q4, in that order. However, points within a quadrant can be in any order relative to each other, but the quadrant order is fixed. So the relative order of the four endpoints is determined by the quadrants they belong to: points from lower-numbered quadrants always precede points from higher-numbered quadrants. This simplifies the crossing condition.

Thus for two chords, each with endpoints in two quadrants, we have four endpoints each in some quadrant. The quadrants have a total order: 1<2<3<4. The ordering of the four points (by angle) will be: first, all points from Q1 (if any) in some order among themselves; then points from Q2; then Q3; then Q4. Within each quadrant, the points are uniform, so their order among themselves is random. But the crossing condition depends on the order of the four labeled points. Since quadrants have a fixed order, we can analyze the alternating condition based on which quadrants the endpoints belong to.

For two chords to intersect, the endpoints must alternate around the circle. With the fixed quadrant ordering, the alternation condition translates to a condition on the quadrant assignments and possibly the relative order within quadrants if multiple points share the same quadrant. Because if two points fall in the same quadrant, their order is not fixed by quadrant ordering; they can be in any order. However, the quadrant ordering ensures that points from different quadrants are ordered by quadrant number. So we need to compute, given the quadrant assignment of the four endpoints (i.e., which quadrants each endpoint belongs to), what is the probability that the chords intersect, considering the uniform distribution within each quadrant.

The quadrant assignment is determined by the chords: chord1 has two quadrants (i1,j1), chord2 has (i2,j2). The four endpoints: chord1 has endpoints in Q_{a} and Q_{b} (where a,b are i1,j1), chord2 has endpoints in Q_{c} and Q_{d} (c,d = i2,j2). Note that the quadrants may repeat: e.g., if chord1 uses Q1 and Q2, and chord2 uses Q2 and Q3, then we have endpoints: one in Q1, two in Q2, one in Q3. So the set of quadrants with multiplicities: some quadrants may have 2 endpoints, some 1, some 0. The order of points is: all points in Q1 (maybe 0 or 1 or 2) come first, then all in Q2, then Q3, then Q4. Within each quadrant, if there are two points, their order is random (uniformly either order). So the full order is a permutation of the four points consistent with the quadrant ordering: points in lower quadrants precede points in higher quadrants; within a quadrant, points can be in any order.

We need to compute, given this structure, the probability that the chords intersect. That is, the probability that the four points, when ordered, have the pattern alternating between chord1 and chord2. Since chords are labeled, we need to compute probability over the random positions within quadrants (and also over the selection of quadrants? But we condition on a specific quadrant assignment, which is part of the random process; we'll then average over assignments). So for each assignment of quadrants (a,b) for chord1 and (c,d) for chord2, we can compute P_intersect_given_assignment.

Then overall P_intersect = average over (a,b) and (c,d) each uniformly over 6 pairs.

We need to compute these probabilities. Let's enumerate possible quadrant pair types. Since quadrants are 1,2,3,4. For chord1, possible pairs: adjacent (1-2,2-3,3-4,4-1) and opposite (1-3,2-4). Similarly for chord2.

We can use symmetry to reduce the number of distinct assignments. Because of rotational symmetry, we can fix chord1 to be a specific pair, say (1,2) (adjacent). Then chord2 can be any of the 6 pairs. However, due to reflection symmetry, some of those might be equivalent. But we can compute for each possible chord2 pair relative to chord1=(1,2). Since the quadrants are ordered, we need to consider chord2 pairs: (1,2), (2,3), (3,4), (4,1), (1,3), (2,4). But note that (4,1) is adjacent (since 4 and 1 are adjacent across 0/2π). Since we fixed chord1=(1,2), chord2=(4,1) is adjacent but involves quadrant 1 which chord1 also uses. So it's distinct from (1,2). Similarly (2,3) is adjacent and shares quadrant 2; (3,4) adjacent shares none? (3,4) shares none with (1,2). (1,3) opposite shares quadrant 1; (2,4) opposite shares quadrant 2? Actually (2,4) shares quadrant 2? Yes, chord2=(2,4) uses Q2 and Q4, chord1 uses Q1 and Q2, so share Q2. So we need to compute for each.

Because of symmetry, we can compute P_intersect for each pair (p,q) by analyzing the order. However, we can also compute overall probability by a more direct method: Consider the random angles of the four endpoints, each drawn uniformly from the circle, conditioned on each chord's endpoints being in different quadrants. Instead of enumerating all quadrant assignments, we might compute the probability that two chords intersect by integrating over the angles with the condition. Since the condition "different quadrants" is a condition on each pair, we can use inclusion-exclusion? But perhaps enumeration is manageable.

Given the symmetry, we can compute P_intersect by using the following: For four points uniformly and independently on the circle, the probability that they are in "general position" and that two chords formed by pairing them in a specific way intersect is 1/3. But here, the pairing is fixed, but the points are not uniformly distributed over the whole circle; they are conditioned on each chord's endpoints being in different quadrants. So we need to compute the conditional probability given that condition. This can be expressed as:
P(intersection | each chord endpoints in diff quadrants) = P(intersection and each chord endpoints in diff quadrants) / P(each chord endpoints in diff quadrants). Since chords are independent, the denominator is (3/4)^2 = 9/16. And numerator = P(intersection and condition). So we could compute the probability that two random chords (with endpoints uniform i.i.d. on circle) intersect AND each chord's endpoints lie in different quadrants. Then divide by 9/16. That might be easier because we can compute the joint probability by integration over all four points uniformly on circle, and then check conditions. Since the condition is about quadrants, we can compute using symmetry.

Let's attempt this approach: Consider four independent random points on the circle, each uniformly distributed. Label them A,B for chord1, and C,D for chord2. But careful: The chords are defined by pairing: chord1 connects A and B; chord2 connects C and D. So we have four i.i.d. uniform points on [0,2π). Then we want to compute the probability that:
- A and B are in different quadrants.
- C and D are in different quadrants.
- The chords intersect.

Then P_intersect_conditional = (numerator) / (9/16). And we can compute numerator by integrating over all angles, using symmetries. Since the points are i.i.d. uniform, we can compute the probability that all three events occur. This might be easier than enumerating quadrant assignments because we can use continuous symmetry.

Alternatively, we can compute directly the probability that two chords intersect given the condition. Because the condition "different quadrants" is a restriction that might increase the intersection probability. We'll compute using integration.

Let's define the circle parameterization: angles θ in [0,2π). Quadrants: intervals: I1=[0,π/2), I2=[π/2,π), I3=[π,3π/2), I4=[3π/2,2π). Uniform distribution: each point's angle is uniform on [0,2π). For any four i.i.d. points, the order is uniform over all permutations (since the points are i.i.d. continuous, the probability that any particular ordering occurs is 1/4! = 1/24, because all orderings are equally likely). Actually, for i.i.d. continuous points on a circle, the probability that a specific ordering (like A, B, C, D in order) is 1/24? But careful: Because the circle is symmetric, we can condition on the ordering of the angles. For independent uniform points, the joint distribution of angles is symmetric. The probability that they are in a particular cyclic order (ignoring starting point) is 1/ (4-1)! = 1/6? Actually for points on a circle, the order is defined up to rotation. Usually, given n independent uniform points on a circle, the probability that they appear in a specific cyclic order (e.g., A, B, C, D clockwise) is 1/(n-1)! = 1/6 for n=4. Because we can fix the position of the smallest angle as reference, then the relative order of the remaining 3 points is uniform among 3! = 6. So the probability that the cyclic order is (A, B, C, D) (starting from A) is 1/6. But careful: The ordering is linear if we pick a starting point. Usually, if we take the angles and sort them, the order (as a permutation) is uniformly random among all 4! = 24 possibilities, but because the circle is rotationally symmetric, the probability of a specific order (like A, B, C, D) when we list in increasing order from 0 is 1/24? Actually, if we take the angles and sort them in increasing order, the resulting permutation of labels is uniformly random among the 4! permutations. Since the points are i.i.d. continuous, the order statistics are uniform over permutations. Indeed, the joint distribution of the sorted angles is order statistics, and the labels are equally likely to be any permutation. So the probability that the sorted order is (label1, label2, label3, label4) in that specific sequence is 1/24. But we often condition on the order because it's easier. However, for crossing condition, we need to know whether the order is alternating between the two chords. The condition for intersection is that in the cyclic order, the labels A and B are separated by C and D (i.e., the order is A, C, B, D or A, D, B, C up to rotation). That event has probability 1/3 for i.i.d. points, as we said. This is because given four random points, there are 3 ways to pair them into two chords (if we consider all possible pairings), and exactly one pairing yields crossing. But here the pairing is fixed, so the probability that the fixed pairing (A-B, C-D) yields crossing is 1/3. This is a known result: For four random points on a circle, the probability that the chords (A,B) and (C,D) intersect is 1/3. Indeed, it's equal to the probability that the order is alternating. And this can be derived: Given four random points, the probability that they are in alternating order (A, C, B, D) is 1/3? Let's compute: Among the 24 possible linear orders (when sorted), how many result in alternating? There are 4! = 24 orders. The alternating orders for fixed labeling: For chords (A,B) and (C,D) to intersect, the order (in a linear sense, not cyclic) must be either A, C, B, D or A, D, B, C (and also cyclic rotations? But we consider sorted order from 0 to 2π, which is linear; however, because the circle is linear with a cut at 0, the alternating condition must hold in the circular sense. But if we consider the sorted order, we must account for the fact that the interval between the largest and smallest angle wraps around. The standard result: For four random points on a circle, the probability that two specific chords intersect is 1/3. This can be derived by noting that there are three possible pairings of the four points into two chords; exactly one pairing yields crossing. Since the points are i.i.d., the pairing (A-B, C-D) is as likely to be the crossing one as any other, so probability 1/3. This argument does not rely on order statistics; it's combinatorial: Given four distinct points, there are exactly three ways to partition them into two pairs. The chords intersect if and only if the pairing is the "crossing" one. If the points are in general position (no three collinear), exactly one of the three pairings yields intersecting chords (the other two yield non-intersecting). Since the points are random and unlabeled, the pairing (A-B, C-D) is one of the three equally likely pairings? But the points are labeled; the labeling defines the pairing. However, given the four points, the pairing (A,B) and (C,D) is fixed; but if we consider the random positions of A,B,C,D, the probability that this fixed pairing equals the crossing pairing is 1/3 because the crossing pairing is equally likely to be any of the three possible pairings. This reasoning holds if the points are i.i.d. and the labeling is independent of positions. So indeed, P(intersection) = 1/3. So we can use that.

Now we want the probability that two chords intersect given that each chord's endpoints are in different quadrants. Let's denote event E1: A and B in different quadrants; E2: C and D in different quadrants; I: chords intersect. Then we want P(I | E1 & E2) = P(I & E1 & E2) / P(E1 & E2). Since chords independent, P(E1 & E2) = P(E1)*P(E2) = (3/4)^2 = 9/16. So we need numerator = P(I & E1 & E2). Compute numerator = probability that four i.i.d. uniform points on circle satisfy: A and B in different quadrants, C and D in different quadrants, and chords intersect. Then conditional probability = numerator / (9/16). So we can compute numerator.

Because of symmetry, we can compute numerator by integrating over the angles. Since the points are i.i.d., we can compute the probability by conditioning on the order of the four points. However, the quadrants are arcs, so we need to consider the quadrant assignments. Might be easier to compute via enumeration over quadrant assignments and compute the probability that given the quadrant assignment, the chords intersect (which we already need) and then weight by probability of that quadrant assignment under the i.i.d. uniform distribution, but also accounting for the condition that each chord endpoints are in different quadrants? Wait, we want numerator = P(I and E1 and E2). Under i.i.d. uniform, we can compute this by summing over all assignments of quadrants to the four points (each point independently assigned to one of four quadrants with probability 1/4 each). Then for each assignment, we compute the probability that the chords intersect given those quadrant assignments (and uniform within quadrants). Then numerator = sum over assignments of P(quadrant assignment) * P(I | quadrant assignment). But careful: For i.i.d. uniform points, the quadrant of each point is independent uniform over {1,2,3,4} with probability 1/4 each. So the probability of a specific quadruple of quadrants (qA,qB,qC,qD) is (1/4)^4 = 1/256. However, the events E1 and E2 require qA≠qB and qC≠qD. So numerator = sum over assignments with qA≠qB and qC≠qD of (1/256) * P(I | quadrants). Then conditional probability = numerator / (9/16) = (numerator * 16/9). But we can also compute P(I | E1&E2) directly by averaging over assignments with the conditional distribution: given qA≠qB and qC≠qD, the conditional probability of each assignment is (1/256)/(9/16) = (1/256)*(16/9)=1/(144)? Actually there are 4^4=256 total assignments. Among them, number with qA≠qB? For each point, quadrant independent. Number of assignments with qA≠qB: 4*3 *4*4? Actually total assignments: 4^4 = 256. For fixed qA,qB with qA≠qB: there are 4*3 = 12 choices for (qA,qB). For qC,qD unrestricted: 4*4=16. So total assignments with qA≠qB: 12*16 = 192. Similarly qC≠qD: also 192. But the conjunction qA≠qB and qC≠qD: assignments with qA≠qB and qC≠qD: count = (4*3)*(4*3)=12*12=144. So there are 144 assignments satisfying both conditions. The conditional probability of a specific assignment given E1&E2 is 1/144 (since uniform over those 144). So indeed, if we compute average over these 144 assignments of P(I | assignment), then that average is exactly P(I | E1&E2). So we can compute by enumerating all assignments (qA,qB,qC,qD) with qA≠qB and qC≠qD, compute P(I | assignment) (which depends only on the quadrants and the distribution within), then average over the 144 assignments. That's equivalent to our earlier plan of averaging over chord1's pair and chord2's pair, but with the nuance that chord1's quadrants are (qA,qB) and chord2's are (qC,qD). However, note that the assignments treat A,B as chord1's endpoints and C,D as chord2's endpoints. So the pair for chord1 is (qA,qB) (unordered) and for chord2 is (qC,qD). Since qA≠qB, qC≠qD. So the assignments correspond to ordered pairs of quadrants for each chord? Actually (qA,qB) is ordered but the chord is unordered; however, the probability of a particular unordered pair given qA≠qB is uniform over unordered pairs? Let's check: Under the conditional distribution (qA≠qB), the probability that {qA,qB} = {i,j} (i≠j) is: for ordered (i,j) with i≠j, probability 1/12 each; but unordered pair {i,j} appears for two ordered pairs (i,j) and (j,i), so probability 2/12 = 1/6. So indeed uniform over unordered pairs. So our earlier approach of considering unordered pairs for each chord with uniform probability 1/6 is consistent. And for chord2 similarly. So we can proceed with enumerating unordered pairs for each chord: 6 each, total 36 combos. But we need to compute P(I | chord1 pair = p, chord2 pair = q). However, note that within a given unordered pair, there are two ordered assignments (which point is in which quadrant) but that might affect the probability? Because the crossing condition may depend on the actual quadrants, but since the endpoints are unordered, the probability of intersection given the unordered pair should be the same regardless of labeling of which endpoint is in which quadrant? However, the crossing condition depends on the positions of the four points; the distribution of angles given unordered pair (i,j) is that we pick two points uniformly from Q_i and Q_j, but the assignment of which point is A vs B doesn't matter because chord is unordered. So indeed the probability of intersection given unordered pairs should be the same for both ordered assignments (since the distribution is symmetric). So we can compute for each unordered pair combination.

Thus we need to compute for each unordered pair p and q (each from the 6), the probability that two chords intersect, given that chord1 endpoints are uniformly in Q_i and Q_j (i≠j), chord2 endpoints uniformly in Q_k and Q_l (k≠l). Then average over p,q each with weight 1/36.

So we need to compute these probabilities. There are 36 combos, but many are equivalent due to symmetries. We can categorize based on the relationship between the two pairs. Let's define the quadrants as positions 1,2,3,4 around the circle. For chord1, we can fix its pair to be (1,2) (adjacent) without loss of generality, because we can rotate the whole configuration. However, note that the quadrants are fixed relative to the diameters; but due to rotational symmetry of the problem (the diameters are fixed, but the quadrants are symmetric under rotations by 90°, which map quadrants to quadrants, and also reflections). The distribution of points is uniform over each quadrant, and the diameters are fixed lines. However, the crossing condition for chords depends on the positions of endpoints relative to the diameters, but the diameters are the boundaries of quadrants. So the quadrants are arcs. Rotating the entire configuration by 90° would map Q1->Q2, Q2->Q3, etc., and also map the diameters to themselves? Actually the diameters are at 0° and 90°, so rotating by 90° swaps the two diameters? But the diameters are still perpendicular; the configuration is symmetric under rotations by 90° because the quadrants are rotated but the diameters also rotate? Actually if we rotate the entire disk by 90°, the diameters become the vertical and horizontal? Initially diameters are at 0° and 90°; after rotating by 90°, one diameter becomes at 90° and the other at 180°, which is still a pair of perpendicular diameters but the quadrants are shifted. However, the problem statement says "Alex divides a disk into four quadrants with two perpendicular diameters". So the quadrants are defined by those diameters. If we rotate the disk, the diameters rotate, but the definition of quadrants rotates accordingly. The random selection of points is on the perimeter, but the quadrants are fixed by the diameters. So the system is not rotationally symmetric in the sense that the diameters are fixed. However, because the points are uniformly distributed over each quadrant, the system is symmetric under the group of rotations that map the quadrants to themselves (i.e., rotations by multiples of 90°). Rotating by 90° sends Q1->Q2, Q2->Q3, Q3->Q4, Q4->Q1; this maps the diameters? The diameters are at 0° and 90°; after rotation by 90°, the diameters become at 90° and 180°. But the new diameters are still perpendicular, but the quadrants are now defined by these new diameters? However, the problem statement defines the quadrants initially with two perpendicular diameters. If we rotate the whole disk, we would have a different set of diameters, but the problem is not symmetric under arbitrary rotation because the diameters are fixed as the ones Alex drew. But the random points are chosen uniformly on the perimeter; the quadrants are fixed arcs. So the configuration has only the symmetry of the square: rotations by 90° about the center map the quadrants onto each other, and also map the diameters onto each other (the horizontal diameter becomes vertical, and vertical becomes horizontal, but they remain the same set of two lines). So indeed, the system is symmetric under the dihedral group D4 (rotations by 90° and reflections). Because the diameters are at 0° and 90°, rotating by 90° sends the horizontal diameter to the vertical diameter, and the vertical to the horizontal (but the set of two lines is unchanged). The quadrants are permuted. So yes, the entire configuration is symmetric under rotations by 90°. So we can fix chord1's pair to, say, (1,2) (adjacent) by applying a rotation if needed. But we must be careful: when we fix chord1 to (1,2), we are effectively using the symmetry that the probability for any adjacent pair is the same. Similarly, for chord2's pair, we can consider its relation to chord1's pair up to rotation. So we can reduce the number of distinct cases for chord2 relative to chord1=(1,2). Since rotations by multiples of 90° map quadrants cyclically, but they also map chord1's pair to other adjacent pairs. So we can fix chord1=(1,2) and then consider chord2's pair from the set { (1,2), (2,3), (3,4), (4,1), (1,3), (2,4) }. However, due to reflection symmetry (e.g., swapping left and right), some of these may be equivalent. For instance, (2,3) and (4,1) might be symmetric by reflection? Let's examine.

We have chord1=(1,2). Now chord2=(2,3): shares quadrant 2. chord2=(4,1): shares quadrant 1. Are these symmetric? Possibly by a reflection that swaps Q1 and Q2? Not exactly. But we can compute probabilities for each distinct case. Because the quadrants are ordered, we can compute the probability for each case by analyzing the order of points given the quadrant intervals. Since there are only a few cases, we can compute directly.

We need to compute P_intersect for given quadrant sets. Let's denote the quadrants for chord1: A in Q_i, B in Q_j. For chord2: C in Q_k, D in Q_l. Without loss, we can consider the order of quadrants. Because quadrants are ordered 1<2<3<4, the order of points by angle is determined by quadrant numbers: all points from the smallest quadrant come first, then next, etc. Within each quadrant, if there are two points, their order is random (uniformly either). So we can compute the probability that the chords intersect given this structure.

Let's formalize: Let the four endpoints be labeled: chord1: points X and Y; chord2: points U and V. Their quadrants: q(X), q(Y), q(U), q(V). The actual angles are uniform within their respective quadrants. Because quadrants are disjoint intervals, the ordering of points from different quadrants is fixed: any point from a quadrant with smaller index comes before any point from a quadrant with larger index. Within the same quadrant, if two points fall in the same quadrant, their order is random (since they are independent uniform on that interval). So we can compute the probability that the chords intersect by considering the permutation of the four labels consistent with the quadrant ordering.

Define the sorted order (by angle) as a sequence of labels, with the rule that points from quadrant 1 appear first (in some order if multiple), then quadrant 2, etc. The chords intersect if and only if the order is alternating: the labels of chord1 and chord2 appear interleaved. More precisely, if we list the four labels in order, the chords intersect if the sequence is of the form X, U, Y, V or X, V, Y, U (or any cyclic rotation) but since we have a fixed starting point (the smallest angle), we need to consider the cyclic condition. However, because the quadrants are ordered and the circle is not cut at 0? Actually we have a natural starting point: the smallest angle is from the quadrant with smallest index. Since quadrants are ordered, the smallest quadrant among the four points determines the first point. But there might be a wrap-around if the quadrants include Q4 and Q1? Since Q4 is the largest angles (270-360) and Q1 is the smallest (0-90), but note that 0 is the same as 360, so there is a wrap at 0. In our ordering, we treat angles modulo 2π. Typically we can map angles to [0,2π). Q1 = [0,π/2), Q2 = [π/2,π), Q3 = [π,3π/2), Q4 = [3π/2,2π). So Q4 angles are near 2π (just below 2π) and Q1 angles are near 0. There is a discontinuity: points from Q4 have large angles (close to 2π) and points from Q1 have small angles (close to 0). In a circular order, after Q4 comes Q1 (since after 2π we wrap to 0). So the natural cyclic order is: Q1, Q2, Q3, Q4, and then back to Q1. However, when we sort angles in increasing order from 0 to 2π, we get Q1 points first, then Q2, then Q3, then Q4. This linear order does not reflect the wrap-around between Q4 and Q1. But the cyclic order is important for crossing condition because the alternating condition is cyclic. For example, consider points: one from Q4, one from Q1, and two from Q2. In linear order from 0, the Q1 point comes first (since it's near 0), then Q2 points, then Q4 point last. But in the circle, the Q4 point is actually just before Q1 if we go clockwise from Q4 through 2π to 0. So the cyclic order might be different: Q4, then Q1, then Q2, etc. So we must be careful: The ordering we get by sorting angles in [0,2π) gives a linear order that does not account for the wrap. However, the crossing condition depends on the cyclic order. We can handle this by considering the circular order of points. Since the points are on a circle, we can pick a reference point (e.g., the point with smallest angle) and then consider the order of the other points as we go clockwise from that point. But the smallest angle is from some quadrant; because of wrap, points from Q4 have angles close to 2π, so they are larger than points from Q1; so the smallest angle is always from Q1 if there is any Q1 point; otherwise from Q2, etc. So the linear order from 0 to 2π gives a cyclic order if we consider that after the largest angle (close to 2π), we wrap to the smallest angle. So the cyclic order is: the sorted list in increasing order (from smallest to largest) followed by the smallest again. So the cyclic order is essentially the sorted order with the wrap from largest to smallest. So we need to consider the sorted list and then treat the pair (largest, smallest) as adjacent in the cyclic order. The alternating condition must hold on the circle, meaning that if we go around the circle, the labels should alternate. This can be analyzed by considering the sorted order and checking if the two chords intersect. There's a known criterion: For four points on a circle, chords (a,b) and (c,d) intersect if and only if exactly one of a and b lies in the open arc from c to d (the shorter arc) when going clockwise. Equivalent to: when the points are sorted in circular order, the chords intersect if and only if the two points of one chord are not both consecutive in the cyclic order. More precisely, if the cyclic order is a, c, b, d (or a, d, b, c) then they intersect. So we need to compute the probability that the cyclic order is alternating.

Given the quadrant intervals, we can determine the possible cyclic orders based on which quadrants the points lie in and the random ordering within quadrants. Because quadrants are arcs that are contiguous in the circle: Q1 (0-90), Q2 (90-180), Q3 (180-270), Q4 (270-360). The cyclic order of the quadrants is Q1, Q2, Q3, Q4, then back to Q1. So points from Q4 come before Q1 in the clockwise direction if we go past 360? Actually clockwise direction: Typically, increasing angle is counterclockwise. Let's adopt standard: angles increase counterclockwise from 0 at right. So going counterclockwise from 0, we go through Q1 (0-90), Q2 (90-180), Q3 (180-270), Q4 (270-360). Then back to 0. So the cyclic order in terms of increasing angle is Q1, Q2, Q3, Q4, then back to Q1. So if we list points in increasing angle order, we get Q1 points first, then Q2, then Q3, then Q4. The cyclic order is that list, with the Q1 points also coming after Q4 if we continue. So the adjacency between Q4 and Q1 is important: the largest angle point (from Q4) is followed by the smallest angle point (from Q1) in the cyclic order. So the cyclic order is essentially the sorted order, but with the wrap connection between the last and first.

Thus to compute the cyclic order, we need to consider the sorted list of angles in [0,2π). The cyclic order is that list, then the first again. So the labels in cyclic order are: first the points from the smallest quadrant (maybe Q1, if any), then points from next quadrant, etc., ending with points from Q4 (if any), then back to the first. So we need to compute the probability that this cyclic order is alternating between chord1 and chord2.

We can handle this by considering the sorted list and then checking the pattern, taking into account the wrap. Because the sorted list is linear but the wrap connects the end to the beginning. So the alternating condition on the circle means that if we write the sorted list (from smallest to largest) and then append the first element at the end, the resulting sequence of length 5 (with first repeated) should have alternating labels? Actually simpler: For four points on a circle, the chords intersect if and only if, when you go around the circle, the two endpoints of each chord are separated by the endpoints of the other chord. Equivalent: in the cyclic order, the two chords' endpoints appear interleaved. This can be checked by looking at the sorted order: If we take the sorted list (in increasing angle) and consider the order, we can determine if the chords intersect by seeing whether the two points of one chord are in the same "half" relative to the other chord? There's a known condition: For four points in order (p1, p2, p3, p4) around the circle, chords (p1,p3) and (p2,p4) intersect. So if we label the points in cyclic order, the chords intersect if the endpoints of one chord are in positions 1 and 3, and the other chord's endpoints are in positions 2 and 4. So the alternating pattern is exactly that: the chords correspond to the two possible pairings of the four positions in cyclic order: the "crossing" pairing is (first, third) and (second, fourth). So for a given cyclic order, the chords intersect if and only if the labels of the two chords are assigned such that one chord's labels are at positions 1 and 3, the other at positions 2 and 4 (or vice versa). In other words, the chords intersect if and only if in the cyclic order, the labels of chord1 and chord2 alternate.

Thus, given the cyclic order of the four labeled points, the chords intersect if and only if the labels are in alternating order. Since the cyclic order is random, we need the probability that the cyclic order is alternating given the quadrant constraints.

Now, because quadrants impose that points from lower-numbered quadrants come before points from higher-numbered quadrants in the sorted order, and the wrap only connects Q4 to Q1, the cyclic order is determined by the sorted order and the wrap. So we can compute, for a given set of quadrants (with multiplicities), the possible cyclic orders and their probabilities given the uniform distribution within quadrants.

We can break down by the multiset of quadrants for the four endpoints. For each combination of chord1 pair (i,j) and chord2 pair (k,l), we have a multiset of quadrants: the quadrants that appear, with multiplicities. Because the quadrants are ordered, we can compute the probability of intersection by analyzing the ordering of points within each quadrant and the wrap.

Let's define the sorted order (by angle) as the order of points when we sort angles in [0,2π). Because angles are continuous, ties probability zero. So the sorted order is a permutation of the four labels, but with the constraint that points from quadrant 1 come before points from quadrant 2, etc., because any point from Q1 has angle less than any from Q2, etc. So the sorted order is: first all points from Q1 (in some random order among themselves), then all from Q2, then all from Q3, then all from Q4. If there are no points from a quadrant, we skip. So the sorted order is determined by the multiplicities: e.g., if we have one point in Q1, two in Q2, one in Q3, then the sorted order is: Q1 point (label), then the two Q2 points in some order, then Q3 point. The Q4 point? none. So the cyclic order is then this sorted list, and then wrap to the first point after the last point (which is from Q3, then goes to Q1). But note: In the cyclic order, the adjacency between Q3 and Q1 is not natural because Q3 is at 180-270, Q1 at 0-90; the wrap goes from Q3 (largest angle) to Q1 (smallest) via 360. So indeed, after the last point (from the highest quadrant present), the next in cyclic order is the first point (from the lowest quadrant). So the cyclic order is: (points from Q1) then (points from Q2) then ... then (points from highest quadrant), then back to Q1 points. So the cyclic order is a cycle that goes through the quadrants in increasing order, but wraps from highest to lowest.

Thus, the cyclic order is essentially the sorted order with the wrap connecting the last to the first. So we can compute the alternating condition by considering the sorted list and the wrap.

We need to compute, for each configuration of quadrant multiplicities, the probability (over the random ordering within quadrants) that the cyclic order is alternating. Because the points within a quadrant are uniformly distributed, their order among themselves is uniformly random (all permutations equally likely). So given the multiset of quadrants (which points belong to which quadrants), the probability that the cyclic order is alternating can be computed combinatorially.

Let's denote the labels: for chord1: points X and Y; for chord2: points U and V. The quadrants: q(X), q(Y), q(U), q(V). We'll denote the quadrants as numbers 1-4. We'll compute the probability that the cyclic order is alternating. Because the distribution of angles within each quadrant is uniform and independent, the order of points within a quadrant is uniformly random. Also, points from different quadrants are ordered by quadrant index. So we can treat the cyclic order as: first, the points from the smallest quadrant present (say Q_min) in some order; then points from the next quadrant, etc., until the largest quadrant present; then wrap to the points from the smallest quadrant again. So the cyclic order is a cycle that goes through the quadrants in order. The relative order of points within a quadrant is random; between quadrants it's fixed.

We need to compute the probability that this cycle alternates between chord1 and chord2 labels.

We can compute by considering the sorted list (linear order) and then the wrap. For alternating, the cyclic order must be such that the labels appear in alternating fashion. Because there are only four points, we can enumerate all possible cyclic orders given the quadrant structure.

Let's first enumerate the possible quadrant assignments (multiset) for the four points. There are many, but we can group them by the number of points in each quadrant. Since each chord contributes two points from distinct quadrants, the possible multiplicities are limited. Let's list all possibilities for (i,j) and (k,l). We'll consider chord1 pair (i,j) and chord2 pair (k,l). The quadrants may be 1,2,3,4.

We can categorize by the pattern of overlaps: the two chords may share 0, 1, or 2 quadrants. They could share both quadrants (i.e., both chords use the same pair), share exactly one quadrant (e.g., chord1 uses {1,2}, chord2 uses {2,3} shares Q2), or share no quadrants (e.g., chord1 {1,2}, chord2 {3,4} shares none). Also possible: chord1 {1,3} and chord2 {2,4} share none? Actually {1,3} and {2,4} have no common quadrant. So categories: same pair, adjacent pairs that share one quadrant, opposite pairs that share one quadrant? Let's list systematically.

Given chord1 fixed as (1,2), the possibilities for chord2:
- (1,2): same pair.
- (2,3): shares Q2.
- (3,4): shares none? (1,2) vs (3,4): no overlap.
- (4,1): shares Q1.
- (1,3): shares Q1? Actually (1,3) shares Q1? chord1 has Q1 and Q2, chord2 has Q1 and Q3, so share Q1.
- (2,4): shares Q2? chord1 has Q1,Q2; chord2 has Q2,Q4, share Q2.

So for chord1=(1,2), chord2 can share 0,1,2 quadrants. Specifically:
- shares 2: only when chord2=(1,2) (same pair).
- shares 1: (2,3), (4,1), (1,3), (2,4). That's four possibilities.
- shares 0: (3,4). Also maybe (something else)? (1,3) shares Q1, so not 0; (2,4) shares Q2; (2,3) shares Q2; (4,1) shares Q1. So only (3,4) has no overlap. Also (something like (1,4)? But (1,4) is adjacent? Actually (1,4) is adjacent? Quadrants 1 and 4 are adjacent across the 0° boundary, so they are considered adjacent (sharing the horizontal diameter). That is included as (4,1) we already have. So (1,4) is same as (4,1). So only (3,4) is disjoint.

Thus we have three categories: same pair, sharing one quadrant, disjoint. For each, we need probability of intersection.

But also chord1 could be opposite pair, e.g., (1,3). But by symmetry, we can compute for chord1=(1,2) and then also consider chord1 being opposite, and then average appropriately. However, since chord1 itself is uniformly chosen over 6 pairs, we need to compute overall expectation. We can compute by fixing chord1 and averaging over chord2, but we need to account for the fact that chord1's pair may be adjacent or opposite with probabilities 4/6 and 2/6 respectively. However, due to symmetry, we might compute the overall probability by averaging over chord1's pair as well. It might be easier to compute directly by enumerating all 36 combos with the actual probabilities using the conditional distribution of quadrants given that each chord's endpoints are in different quadrants. Since the conditional distribution is uniform over unordered pairs for each chord, we can compute the overall expected intersection probability as (1/36) * sum_{p,q} P_intersect(p,q). So we need to compute P_intersect for each combination. We can compute these values by analysis using combinatorics and integration.

Given the symmetry, we can compute P_intersect for each type of pair relationship, and then count how many combos of that type. Let's categorize all unordered pair types (p,q). There are 6 pairs total. Let's label pairs: A1 = {1,2}, A2 = {2,3}, A3 = {3,4}, A4 = {4,1} (adjacent), and O1 = {1,3}, O2 = {2,4} (opposite). For two chords, the pair types can be classified based on the relationship between the two unordered pairs. The possible relationships:
- Both pairs are the same: e.g., both {1,2}. There are 6 such combos (p=q). For each of the 6 pairs, there is 1 combo where they are equal. So count = 6.
- Both pairs are adjacent but share exactly one quadrant? Or both adjacent and share no quadrant? Need to categorize all 36 combos. Instead, we can list all combos of two unordered pairs from the 6. There are 36 total ordered pairs (p,q). Since unordered pairs are distinct? We treat ordered combos because we average over 36 (p,q). Since each chord's pair is chosen independently, the probability of a specific ordered pair (p,q) is 1/36. So we need to compute for each ordered pair. However, due to symmetry, many ordered pairs will have the same probability because of rotational symmetry. We can group them by the relative configuration. Let's define the configuration by the set of quadrants involved and the overlaps.

We can categorize based on the multiset of quadrants for the four endpoints. This multiset is determined by the two pairs. For two pairs (p,q), the multiset of quadrants (with multiplicities) can be:
- If p = q: then we have two points in each of the two quadrants that constitute p. So multiset: two points in quadrant i, two points in quadrant j, where {i,j} is the pair. For adjacent pair, i and j are adjacent; for opposite pair, i and j are opposite.
- If p and q share exactly one quadrant: then one quadrant appears twice (the shared quadrant), and the other two quadrants each appear once. For example, p={1,2}, q={2,3} yields quadrants: 1:1, 2:2, 3:1, 4:0. So multiset: {1:1, 2:2, 3:1}. Similarly, p={1,2}, q={4,1} yields quadrants: 1:2,2:1,4:1. So pattern: shared quadrant appears twice, the other two quadrants (one from each pair) each appear once. Note that the two non-shared quadrants could be adjacent or opposite relative to each other? They are the "other" quadrants from each pair. For p={1,2} and q={2,3}, the non-shared quadrants are 1 and 3. Are 1 and 3 opposite? Yes, 1 and 3 are opposite (since they are separated by 2 and 4?). Actually quadrants 1 and 3 are opposite (across the center). For p={1,2} and q={4,1}, non-shared are 2 and 4. 2 and 4 are also opposite? 2 and 4 are opposite? Yes, quadrants 2 and 4 are opposite. For p={1,2} and q={1,3}, non-shared are 2 and 3. 2 and 3 are adjacent (since 2 and 3 share vertical? Actually 2 and 3 are adjacent, sharing the horizontal diameter? Q2 and Q3 are adjacent across the boundary at π? Q2: 90-180, Q3:180-270, they are adjacent, sharing the diameter at 180°). For p={1,2} and q={2,4}, non-shared are 1 and 4. 1 and 4 are adjacent? Q1 and Q4 are adjacent across 0°? Yes, they share the horizontal diameter. So when sharing one quadrant, the non-shared quadrants could be either adjacent or opposite, depending on which pair we have. So we need to distinguish further.

- If p and q share no quadrants: then the two pairs are disjoint. There are two subcases: the two pairs could be such that the four quadrants are all distinct. For adjacent pairs, possible disjoint pairs: {1,2} and {3,4} are disjoint. Also {2,3} and {4,1}? That would be {2,3} and {4,1}: quadrants 2,3 and 4,1: all four distinct? 1,2,3,4 all appear. So that's also disjoint. Also {1,2} and {4,1} shares Q1, so not disjoint. So disjoint pairs are those where the two pairs have no common quadrant. Which pairs are disjoint? For adjacent pairs, there are 4 adjacent pairs, but some pairs share a quadrant? Adjacent pairs share a quadrant if they are neighboring pairs: e.g., {1,2} and {2,3} share 2; {2,3} and {3,4} share 3; {3,4} and {4,1} share 4; {4,1} and {1,2} share 1. So disjoint adjacent pairs are those that are not neighboring: e.g., {1,2} and {3,4} are disjoint; {2,3} and {4,1} are disjoint? Check: {2,3} and {4,1} have quadrants 2,3,4,1: all distinct, so yes disjoint. Also {1,2} and {2,4}? {2,4} shares 2, so not disjoint. {1,2} and {1,3} shares 1, not disjoint. So disjoint pairs can also involve opposite pairs: e.g., opposite pair {1,3} and adjacent {2,4}? {1,3} and {2,4} are disjoint because they cover all four quadrants. Also {1,3} and {2,3} shares 3, not disjoint. {1,3} and {4,1} shares 1. So disjoint combos include: (adjacent, adjacent) with pairs that are opposite in the cycle (like {1,2} and {3,4}) and ({2,3} and {4,1}); also (adjacent, opposite) where the opposite pair consists of the two quadrants not in the adjacent pair? For example, {1,2} and {3,?} opposite pair {3,?} would be {3,1}? That shares 1, not disjoint. {1,2} and {?} opposite pair that is disjoint would be {3,?} but only opposite pairs are {1,3} and {2,4}. {1,2} and {2,4} shares 2, so not disjoint. {1,2} and {1,3} shares 1. So {1,2} has no disjoint opposite pair. However, {2,3} and {1,3} shares 3; {2,3} and {2,4} shares 2; {3,4} and {1,3} shares 3; {3,4} and {2,4} shares 4; {4,1} and {1,3} shares 1; {4,1} and {2,4} shares 4. So adjacent pairs are not disjoint from any opposite pair? Wait, consider {1,2} and {3,4} are disjoint; {2,3} and {4,1} are disjoint; those are both adjacent-adjacent disjoint. Also consider opposite-opposite: {1,3} and {2,4} are disjoint (they cover all four quadrants). Also opposite-adjacent can be disjoint if the adjacent pair's quadrants are both not in the opposite pair? For opposite pair {1,3}, which adjacent pairs are disjoint? Adjacent pairs that do not contain 1 or 3: that would be {2,4}? But {2,4} is opposite, not adjacent. Adjacent pairs: {2,3} contains 3, so shares; {3,4} contains 3; {4,1} contains 1; {1,2} contains 1. So none. So opposite-adjacent always share one quadrant because opposite pair consists of two quadrants that are opposite, and each adjacent pair contains exactly one of those two? Let's check: For opposite pair {1,3}, adjacent pairs: {1,2} shares 1; {2,3} shares 3; {3,4} shares 3; {4,1} shares 1. So yes, always share exactly one quadrant. For opposite pair {2,4}, adjacent pairs: {1,2} shares 2; {2,3} shares 2; {3,4} shares 4; {4,1} shares 4. So always share one. So opposite-adjacent combos always share exactly one quadrant. Similarly, adjacent-adjacent combos can share 0,1, or 2 quadrants: share 2 when same pair; share 1 when neighboring pairs (e.g., {1,2} & {2,3}); share 0 when opposite pairs in the cycle (e.g., {1,2} & {3,4}). And opposite-opposite combos: share 2 when same pair; share 0 when different opposite pairs ({1,3} & {2,4}) are disjoint; share 1? Could two different opposite pairs share a quadrant? Opposite pairs are {1,3} and {2,4} are disjoint; {1,3} and {1,3} same; there are only two opposite pairs, they are disjoint. So opposite-opposite combos: either same (share 2) or the two different (share 0). Also note: adjacent-opposite combos always share exactly 1. So we have categories:
- Category A: p = q (same pair). Subcategories: adjacent pair (4 such), opposite pair (2 such). So total 6 combos (ordered pairs where p=q). But careful: For ordered pairs, p=q gives 6 combos (since each of the 6 pairs, p=q). For each such combo, we need P_intersect.
- Category B: p and q share exactly one quadrant. This includes:
   - adjacent-adjacent neighboring pairs: e.g., ( {1,2}, {2,3} ) and ( {2,3}, {1,2} ) etc. Also ( {2,3}, {3,4} ), etc. Also ( {3,4}, {4,1} ), ( {4,1}, {1,2} ). How many ordered combos? There are 4 adjacent pairs; each has 2 neighboring adjacent pairs? Actually for each adjacent pair, there are two adjacent pairs that share exactly one quadrant with it? For {1,2}, the neighboring pairs sharing one quadrant: {2,3} shares 2, and {4,1} shares 1. That's 2. So for each of the 4 adjacent pairs, there are 2 neighboring adjacent pairs. So total ordered combos where both are adjacent and share exactly one quadrant: 4 * 2 = 8. But also consider opposite-adjacent combos: for each opposite pair (2 of them), each adjacent pair (4) shares exactly one quadrant with it? Actually for opposite pair {1,3}, the adjacent pairs that share exactly one quadrant: {1,2} (shares 1), {2,3} (shares 3), {3,4} (shares 3), {4,1} (shares 1) — that's 4. So each opposite pair has 4 adjacent pairs that share one quadrant. So ordered combos (opposite, adjacent) and (adjacent, opposite) both count. Since ordered pairs, we count both directions. So number of ordered combos where one is opposite and the other adjacent: For opposite pairs: 2, each with 4 adjacent, so 2*4 = 8 combos where chord1 opposite and chord2 adjacent. Similarly, chord1 adjacent and chord2 opposite: 4*2 = 8 combos. So total opposite-adjacent combos = 16. But careful: Are these all distinct? Yes. So total share-one-quadrant combos = adjacent-adjacent share-one (8) + opposite-adjacent (16) = 24. However, we must also consider adjacent-opposite already counted. So total 24 combos. Are there any other share-one-quadrant combos? Could adjacent-opposite be considered separately? Already included. Could opposite-opposite share one quadrant? No, the two opposite pairs are disjoint, so they share none. So only these.

- Category C: p and q share no quadrants (disjoint). This includes:
   - adjacent-adjacent disjoint: pairs like {1,2} & {3,4}, and {2,3} & {4,1}. How many ordered combos? For each unordered pair of disjoint