Probability – Average Cuts Needed for a Pizza Piece with No Curved Edge

circlescombinatoricsexpected valuehypergeometric functionprobability

On a circular pizza, we make a random straight cut by choosing two uniformly random points on the perimeter and cutting through them.

On average, how many times must the pizza be randomly cut, to get a piece with no curved edge?

(In other words, on average, how many random chords must be drawn on a circle, to get a polygon, if each random chord is drawn by connecting two uniformly random points on the perimeter of the circle?)

In the following example, a piece with no curved edge is obtained by the fifth random cut.

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My attempt

I made a pizza cutting simulator on desmos. It seems that the average should be around $7$ or so.

I tried to find the probability that a piece with no curved edge is obtained by the $n$th random cut. But I only worked out that the probability for $n=3$, is $\frac{1}{15}$. (Consider the six points that define three cuts. Start with one point: it must be paired with its opposite point, which has probability $\frac15$. Then one of the remaining points must be paired with its opposite point, which has probability $\frac13$. The remaining two points must be paired together. So the probability is $\frac15\times\frac13=\frac{1}{15}$.)

Best Answer

Write the expectation as $$ \mathbb{E}[X] = \sum_{n=0}^{\infty} \mathrm{P}(X> n). $$ The latter probability is that for $n$ cuts, there is no "central" piece. Note that, thanks to symmetry, all possible $(2n-1)!!$ pairings of the $2n$ endpoints are equally likely. So we need to count the number of pairings such that there is no central piece. In other words, we need to count the number of chord diagrams of size $n$ such that their intersection graph has no cycles, i.e. is a forest. The latter number, according to Hüseyin Acan Forests of chord diagrams with a given number of trees, is equal to$^*$ $$ \sum_{m=1}^n \frac{1}{3n-2m}{2n \choose m-1}{3n-2m \choose n-m}. $$ Therefore, the expectation in question is $$ 1+ \sum_{n=1}^\infty \frac{1}{(2n-1)!!}\sum_{m=1}^n \frac{1}{3n-2m}{2n \choose m-1}{3n-2m \choose n-m}; $$ I haven't found a way to simplify it even to a single series.


$^*$ Quite surprisingly, the sequence is not in the OEIS. I'm not registered there, so I would be grateful if someone submits it.

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