Ponder This Challenge - February 2024 - A Dice Game

This problem is based on a suggestion by Frank Mayer - thanks Frank!

Alice and Bob play the following weird dice game using a standard set of 6 D&D dice, i.e., dice with values

$1,\ldots,4$

$1,\ldots,6$

$1,\ldots,8$

$0,\ldots,9$

$1,\ldots,12$

$1,\ldots,20$

At each round, they throw the dice and sum their values. All six dice need to be rolled once each round, and it doesn’t matter who rolls the dice. What matters in deciding who wins is the sum.

Alice wins if the result is a prime number, and Bob wins if the result is a nonprime even number. Otherwise, the result of the round is a draw.

The winner of the game is the first player to reach $N$ consecutive wins.

Your goal: For $N=13$, find the probability that Alice wins the game.

A Bonus "*" will be given for finding the probability that Alice wins given that $N=300$ but the rules have changed and Bob now wins if the result is a nonprime odd number.

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Solution

• Click here to view the solution

  February 2024 Solution:

  This problem turned out to be more challanging than expected, due both for the (non intentional) tricky phrasing and the nontrivial mathematical structure involved.

  Some of you interpreted "$N$ consecutive wins" of Alice to be "a sequence of at least $N$ rounds in which $N$ are Alice wins and the rest are draws". This was not the intention - a draw certainly interrupts a sequence of wins.

  Others correctly calculated Alice's chance to win a round (0.24826171875, can be found by straightforward enumeration of all cases) and raised it to the power of N; this is wrong, since it answers the different question "If Alice and Bob play exactly $N$ rounds, what is the chance of Alice winning all of them?" whereas the real question is "given that Alice and Bob play indefintely, what is the chance that at one point of the game Alice will have $N$ consecutive wins?"

  The natural way to solve such a question is by using Markov chains. In this formulation, we consider all the states the game can reach; here the states are of the form "Alice has $k$ consecutive wins" and "Bob has $k$ consecutive wins". This can be modeled using a single integer $k$ (e.g. $k=-2$ means "Bob has 2 consecutive wins"). We have a simple transition matrix, with each row having at most three nonzero entries corresponding to Alice win, Bob win and draw.

  We are interested in the hitting probability for the state $h_N$ representing Alice's win; a standard result in the theory of Markov chains is that the hitting probability vector $h^X_i$ for a set $X$ of states is the minimal nonnegative solution to the system of linear equations

  $\begin{cases}h_{i}^{X}=1 & i\in X\\h_{i}^{X}=\sum_{j}p_{ij}h_{j}^{A} & i\notin X\end{cases}$

  Solving this system yields the winning probability for Alice (where $A$ is Alice's probability for winning a round and $B$ is Bob's)

  $\frac{\frac{A^N(1-B^N)}{1-B}}{\frac{A^N(1-B^N)}{1-B}+\frac{B^N(1-A^N)}{1-A}}$

  Plugging

  $A=0.24826171875, B=0.5, N=13$

  yields

  $0.0001675666801570852$

  and plugging

  $A=0.24826171875, B=0.25173828125, N=300$

  yields

  $0.015257532936794722$

Solvers

• *Lazar Ilic (1/2/2024 4:59 PM IDT)
• Lorenz Reichel (1/2/2024 5:07 PM IDT)
• *Andrew Gauld (1/2/2024 8:49 PM IDT)
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