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The game works as follows: Two $n$-sided dice are rolled. If the dice show unequal numbers, then the higher number is rerolled. If the dice show equal numbers, then that is the final result of the silly dice game. As $n \rightarrow \infty$, what is the probability of getting $m$ for some finite $m$? I know that the probability for $m=1$ is $1/2$ + $1/2n$ because when the dice are different and neither is $1$, the probability of getting $1$ as the result is $1/2$ because the chance of rolling the number on the other die and the chance of rolling $1$ are equal.

mathlander
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    What have you tried? – Henry Dec 02 '22 at 17:51
  • I figured out that it is $1/2$ for $m=1$. – mathlander Dec 02 '22 at 18:00
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    you can also pretty quickly see that $P(n)=\left(\frac{1}{n}\right)^2$ if $P(k)$ represents the probability of the final result being $k$, as the only way that $n$ can be the final result is if both players roll it on their first attempt – wjmccann Dec 02 '22 at 18:10
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    Simulations are not always correct. – mathlander Dec 02 '22 at 18:45
  • It should be at least $1/2$ for any given value of $n$ since rolling a $1$ and rolling the number on the other die have equal probabilities... – mathlander Dec 02 '22 at 18:53
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    I have an answer, but I think you have not provided enough context. Can you edit your question with your reasoning that led you to believe the probability was $1/2$ for $m=1$, plus any other efforts you made on this problem? – Mike Earnest Dec 02 '22 at 19:08
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    Suppose on the initial roll, die 1 is larger than die 2. So die 1 is rerolled. Suppose that now the new die 1 is smaller than die 2. Is die 2 now rerolled? That's what I imagine, but it should be clarified. In particular, this would mean that the argument leading to $\frac12+\frac1n$ is incorrect. – Greg Martin Dec 02 '22 at 19:29
  • I don't see how this would make my argument incorrect. – mathlander Dec 02 '22 at 19:34
  • It always works. If neither die is 1 and the dice show different numbers, then the probability is 1/2. – mathlander Dec 02 '22 at 19:40
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    @GregMartin mathlander's reasoning is sound. It's the same reason you get $1/2$ in the drunkard on an airplane problem. Suppose you currently have two different rolls, where the smaller roll is $s$. You reroll the larger die until it is in ${1,\dots,s}$, call the result $t$. Then there is a $1/s$ that $t=s$, so you end on $s$. There is a $1/s$ that $t=1$, in which case you surely output $1$. In all other cases, you move to a new state. Either way, the event of ending at $1$ is balanced against not ending at $1$. – Mike Earnest Dec 02 '22 at 19:43
  • My answer is $\frac{1}{m(m+1)}$ – acat3 Dec 02 '22 at 21:06
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    mathlander: Your $\frac12+\frac1n$ looks wrong for $n=1$ and $n=2$. I think your argument actually suggests $\mathbb P(M=1 \mid n) =\frac1n+ \frac12(1-\frac1n)=\frac{n+1}{2n}=\frac12+\frac1{2n}$, which would be consistent with $\mathbb P(M=m \mid n) = \frac{n+1}{n}\frac1{m(m+1)}$ for $1 \le m \le n$ and sums to $1$ while giving @RezhaAdrianTanuharja's limit of $\frac1{m(m+1)}$ as $n \to \infty$ – Henry Dec 02 '22 at 22:52
  • Thanks, that seems correct. – mathlander Dec 02 '22 at 22:53

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For simplicity, we call ending up with $m$ as a win. There are only three states of the game that result in win or at least a chance to win:

  1. Both dice show $m$.
  2. One die show $m$ and the other show a larger number.
  3. Both dice show different number larger than $m$.

Probability to Win from State #2

Say the probability is $p_{2}$. If we roll an $m$ we win, probability $\frac{1}{n}$. If we roll a number lower than $m$ we lose, probability $\frac{m-1}{n}$. If we roll a number larger than $m$ the game remains in state #2 , probability $\frac{n-m}{n}$.

$$ \frac{1}{n}+\frac{n-m}{n}\cdot p_{2}=p_{2}\implies p_{2}=\frac{1}{m} $$

Probability to Win from State #3

Say the probability is $p_{3}$. If the two dice becomes equal we lose, probability $\frac{1}{n}$. If we roll lower number than $m$ we lose, probability $\frac{m-1}{n}$. If we roll $m$ the games move to state #2, probability $\frac{1}{n}$. If we roll larger number than $m$ and not equal to the other die the game remains in state #3, probability $\frac{n-m-1}{n}$.

$$ \frac{1}{n}\cdot\frac{1}{m}+\frac{n-m-1}{n}\cdot p_{3}=p_{3}\implies p_{3}=\frac{1}{m(m+1)} $$

Remaining Things to Do

Simply calculate the probability of ending in any of the three states above. It's trivial to see that as $n$ becomes larger and larger, the probability to end in state #3 approaches unity.

acat3
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  • Indeed, let $p(h,e)$ denote the chance of ending with a score of $e$ if we have a die with value $h$ and a higher die that is being rerolled; coding the recursive calculation of these probabilities does confirm that $p(h,h) = \frac1h$ and $p(h,e) = \frac1{e(e+1)}$. Let $q(n,e)$ denote the chance of ending with a score of $e$ if we begin by rolling two $n$-sided dice; a similar program shows that $q(n,e) = \frac{n+1}{ne(e+1)}$ for all $1\le e\le n$. Simulations support these formulas. – Greg Martin Dec 02 '22 at 23:26