Convex Quadrilateral from Random Chords
Place $10$ points on a circle and draw all $\binom{10}{2} = 45$ chords between them. You then pick $4$ of these chords uniformly at random.
What is the probability that the $4$ selected chords form a convex quadrilateral? (A convex quadrilateral is a four-sided polygon in which every interior angle is less than $180^\circ$.)
Hints
- Think about the denominator first: how many total chords are there among 10 points, and how many ways can you choose 4 of them?
- For the numerator, consider the geometric fact: given 4 points on a circle, how many distinct convex quadrilaterals can they form?
- Any 4 points on a circle determine exactly one convex quadrilateral (connect them in cyclic order). So the favorable count is simply $\binom{10}{4}$.
Worked Solution
How to Think About It: The question is really asking: out of all the ways to grab 4 chords from 45, how many of those selections happen to trace out a convex quadrilateral? The key geometric fact is that any 4 points on a circle determine exactly one convex quadrilateral -- you just connect consecutive points in cyclic order. Any other way of pairing those 4 points into edges would produce a self-intersecting figure, not a convex polygon. So the numerator is just the number of ways to choose 4 points, and the denominator is the number of ways to choose 4 chords.
Quick Estimate: There are $\binom{45}{4} \approx 150{,}000$ ways to pick 4 chords, but only $\binom{10}{4} = 210$ ways to pick 4 points. So even before computing we know the probability is roughly $210 / 150{,}000 \approx 0.0014$ -- a bit more than 1 in 700. Not surprising: grabbing 4 random chords and having them form a nice closed polygon is a very specific event.
Approach: Count favorable outcomes (4-chord sets forming a convex quadrilateral) and divide by total outcomes (all 4-chord subsets).
Formal Solution:
*Denominator.* There are $\binom{10}{2} = 45$ chords. The number of ways to choose 4 of them is
$$\binom{45}{4} = \frac{45 \times 44 \times 43 \times 42}{4!} = 148{,}995.$$
*Numerator.* A convex quadrilateral inscribed in the circle has 4 vertices, all lying on the circle, connected by 4 edges in cyclic order. Given any 4 of the 10 points, there is exactly one way to connect them as a convex quadrilateral: join consecutive points around the circle. (Any other pairing of the 4 points into 4 edges produces a self-intersecting figure.) So the number of favorable 4-chord selections is exactly
$$\binom{10}{4} = 210.$$
*Probability.* Dividing and simplifying (both are divisible by 105):
$$P = \frac{210}{148{,}995} = \frac{2}{1419}.$$
Answer: The probability is $\dfrac{2}{1419} \approx 0.00141$.
Intuition
The crux of this problem is a clean bijection: every convex quadrilateral inscribed in the circle corresponds to a unique choice of 4 vertices, and every choice of 4 vertices produces exactly one convex quadrilateral (by connecting consecutive points around the circle). This one-to-one correspondence turns a geometric question into pure counting. The reason it works so neatly is that convexity on a circle is automatic -- any polygon formed by connecting points in cyclic order on a circle is convex. Self-intersecting figures only arise when you connect the points out of order.
This type of reasoning -- reducing a geometric or probabilistic question to a combinatorial bijection -- shows up constantly in quant interviews. Whenever you see "probability of some geometric configuration," your first instinct should be to ask: how many ways can that configuration arise, and is there a clean way to count them? Here the answer is a ratio of two binomial coefficients, which is about as clean as it gets.