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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.

enter image description here

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 $\frac15\times\frac13=\frac{1}{15}$.)

N. Owad
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Dan
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    Is it possible to infer whether a piece without a curved edge exists based on the number of vertices, edges and faces in your pizza? (Asking because there are lots of results about e.g. number of intersections of random chord diagrams.) – Jeroen van der Meer Jun 10 '23 at 15:23
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    @JeroenvanderMeer Likely no. The way the edges lie matter. EG With 6 vertices, if we connect 3 pairs of neighbors, there is no "center" piece. However, if we connect opposite points, then we can get a "center" piece. – Calvin Lin Jun 10 '23 at 18:21
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    @CalvinLin Yes, but those two situations do have different numbers of faces/edges/vertices though right? – Milten Jun 10 '23 at 19:04
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    @Milten Not completely. Each has $6$ vertices and $3$ edges. From those you can create five isoforms with $4, 4, 5, 6, 7$ faces each. Only in the last case do we get a center piece. So I suppose if we know the number of faces we can possibly infer the existence of a "center" piece, but beyond $6$ vertices it becomes much harder I suspect. However, I think OP's probabilities are on the right track; we just need to generalize to $n>3$. – Eric Snyder Jun 10 '23 at 21:53
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    Another possibility: could the existence of a center piece depend only on the combinatorial property of which of the $2n$ vertices (ordered clockwise) are paired with each other, independent of the exact locations of those vertices? – Greg Martin Jun 11 '23 at 04:47
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    @GregMartin I think that is definitely possible. I think that if any three chords all cross each other, you get a non-edge piece. And that tracks with OP's $1/15$ probability. But the question is, how to do it for $n>3$ ? – Eric Snyder Jun 11 '23 at 07:43
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    FWIW, running some sims suggests the probability of a center piece existing after $4$ (or fewer) cuts is almost certainly $\frac15$; for $n=5,6,7,8,9$ I get $37%, 55%, 69%, 81%, 89%$ respectively. Not sure what fractions those correspond to, however. – Eric Snyder Jun 11 '23 at 10:05
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    @EricSnyder, for $n=4$, it's $23/105$, if I count correctly. – zhoraster Jun 11 '23 at 13:05
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    @zhoraster You may well be correct; I realized after posting that I missed a configuration type: four chords that each cross two of the others in a cycle produces a center piece that isn't a triangle, and doesn't require three chords to all cross each other. Having that be $2/105$ chance seems completely reasonable. (Oh, it's two possibilities due to chirality. So that makes sense.) – Eric Snyder Jun 11 '23 at 20:51

1 Answers1

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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.

zhoraster
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  • Are you sure you've quoted the formula correctly? What you've written doesn't match Theorem 1 in the linked paper. I found this while trying to compute the probabilities of a central piece with a fixed number $n$ of cuts (which would be helpful information to add to this answer). – Greg Martin Jun 11 '23 at 17:10
  • @GregMartin, to make the formula work both for $m<n$ and $m=n$, I've transformed it a little bit, using that $a\cdot{a - 1 \choose b - 1} = b \cdot{a \choose b }$. – zhoraster Jun 11 '23 at 19:26
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    Can you edit your answer to include some numerical probabilities for individual $n$, of which $\frac1{15}$ is the first nontrivial one? – Greg Martin Jun 11 '23 at 23:03
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    @zhoraster If I understand correctly, you are saying that if $X$ is the number of cuts made to obtain a piece with no curved edge, then $P(X>n)=\frac{1}{(2n-1)!!}\sum_{m=1}^n \frac{1}{3n-m}{2n \choose m-1}{3n-m \choose n-m}$. But this formula gives $P(X>1)=\frac12, P(X>2)=\frac23, P(X>3)=\frac45$, but I believe the correct values should be $P(X>1)=1, P(X>2)=1, P(X>3)=\frac{14}{15}$. – Dan Jun 12 '23 at 01:52
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    I agree with what @Dan commented just above; the transformation doesn't seem to have worked. Running the formula from the paper's Thm $1$ gives $14$ for $n=3$, where yours gives $12$. Similarly, I get the $23$ you noted above from the paper's formula, but $19$ from yours. – Eric Snyder Jun 12 '23 at 03:20
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    Sorry, $2$ was missing before $m$ in $3n-2m$. It's ok now. – zhoraster Jun 12 '23 at 06:17
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    So the expectation is approximately $6.11446944632...$. Seems reasonable. – Dan Jun 12 '23 at 08:38
  • @Dan I get $5.1144694\dots$. I think the $1+$ may be a holdover from something else? There is no chord for $n=0$. – Eric Snyder Jun 12 '23 at 20:28
  • @EricSnyder There is no chord for $n=0$, but if we express the expectation in terms of sums of $P(X>n)$, we should still start with $P(X>0)$ as shown here. I'm still getting $6.1144694...$. – Dan Jun 12 '23 at 21:31
  • @Dan Oops, you're right. I have my output misaligned. Apologies! – Eric Snyder Jun 13 '23 at 04:24
  • I wonder if the expectation has a closed form. – Dan Jun 16 '23 at 01:44
  • @Dan, not likely. Even writing it as a single sum (or integral) is quite challenging... – zhoraster Jun 17 '23 at 06:32
  • This sum looks like a hypergeometric series and should be able to be analyzed with some of those common techniques, like Zeilberger's Algorithm. I am not expert but I added the tag so maybe someone else will be able to do this? – N. Owad Nov 10 '23 at 14:12