A prisoner and two dice

Question

A prisoner is offered the following deal. They will be given a pair of standard 6-sided dice to roll. If they can roll a total of 42 or more without rolling a double then they will be set free. Otherwise if they roll a double before reaching 42 then they will be executed. If they don't accept the deal then they will stay in prison forever. Should they accept the deal?

Answer 1

The prisoner should

accept the deal if they value a 31.03744% chance of freedom over a 100% chance of confinement.

I calculated this number with some basic dynamic programming—using the chances that the prisoner would be alive at any given score $n$ and combining those numbers (weighted by the probability of rolling each non-double result) to calculate further chances. Run or view it online to see the rather simple algorithm.

Answer 2

Using the same idea as @AxiomaticSystem's answer, but calculating it exactly:

The odds of hitting a given running total $n$ without being executed can, when $1 \le n \le 44$, be expressed as the recursive equation:

\begin{align} p_n = &\frac{2}{36} \cdot p_{n-11} + \frac{2}{36} \cdot p_{n-10} + \frac{4}{36} \cdot p_{n-9} \\\\ & + \frac{4}{36} \cdot p_{n-8} + \frac{6}{36} \cdot p_{n-7} + \frac{4}{36} \cdot p_{n-6} \\\\ & + \frac{4}{36} \cdot p_{n-5} + \frac{2}{36} \cdot p_{n-4} + \frac{2}{36} \cdot p_{n-3} \end{align}

Each of these terms is the probability of a single roll moving the total from from $n-k$ to $n$, without rolling a double, times the probability of reaching $n-k$ in the first place.
We can define $p_{n \lt 0} = 0$ (as negative totals can never be reached) and $p_0 = 1$ (as that is the starting value, which will always be reached).

When $n \gt 44$ the probability function changes, as no more rolls are made once the sum has reached 42. This eliminates any terms involving $p_{n\ge42}$. This changes the probability function to:

\begin{align} p_{45} &= \frac{2}{36} p_{34} + \frac{2}{36} p_{35} + \frac{4}{36} p_{36} + \frac{4}{36} p_{37} + \frac{6}{36} p_{38} + \frac{4}{36} p_{39} + \frac{4}{36} p_{40} + \frac{2}{36} p_{41} \\\\ p_{46} &= \frac{2}{36} p_{35} + \frac{2}{36} p_{36} + \frac{4}{36} p_{37} + \frac{4}{36} p_{38} + \frac{6}{36} p_{39} + \frac{4}{36} p_{40} + \frac{4}{36} p_{41} \\\\ p_{47} &= \frac{2}{36} p_{36} + \frac{2}{36} p_{37} + \frac{4}{36} p_{38} + \frac{4}{36} p_{39} + \frac{6}{36} p_{40} + \frac{4}{36} p_{41} \\\\ p_{48} &= \frac{2}{36} p_{37} + \frac{2}{36} p_{38} + \frac{4}{36} p_{39} + \frac{4}{36} p_{40} + \frac{6}{36} p_{41} \\\\ p_{49} &= \frac{2}{36} p_{38} + \frac{2}{36} p_{39} + \frac{4}{36} p_{40} + \frac{4}{36} p_{41} \\\\ p_{50} &= \frac{2}{36} p_{39} + \frac{2}{36} p_{40} + \frac{4}{36} p_{41} \\\\ p_{51} &= \frac{2}{36} p_{40} + \frac{2}{36} p_{41} \\\\ p_{52} &= \frac{2}{36} p_{41} \\\\ p_{53} &= 0 \\\\ p_{54} &= 0 \\\\ &\dots \end{align}

From this we can finally calculate the probability we are interested in. As we are interested in the probability of hitting any sum $n \ge 42$, the probability we are interested in is

$$P = \sum_{n=42}^{\infty} p_n = \sum_{n=42}^{52} p_n$$

Which we can calculate using numerical programming to be

$$P=\frac{2560089501515713}{8248389385254061} \approx 0.3103744721475$$

This is the exact probability that the prisoners win, which is in agreement with @AxiomaticSystem's approximate answer.