Dice rolling probability game

Let $p$ be the probability that Piper escapes first. Her probability of escaping on her first roll by rolling doubles is $\frac{1}{6}$.

Let $q$ be the probability that Quincy escapes first. Then before Piper's first roll there is a probability of $\frac {1}{6}$ that Quincy has no chance to escape first, and a probability of $\frac {5}{6}$ that Quincy's chance of escaping first is $p$ (because if Piper misses then Quincy is now in the same position Piper was in). Thus, $q=\frac{5p}{6}$.

Finally, let $r$ be the probability that Riley escapes first. Then before Piper's first roll there is a probability of $\frac{11}{36}$ that Riley has no chance to escape first (because either Piper or Quincy or both roll doubles) and a probability of $\frac{25}{36}$ that Riley's chance of escaping first is $p$ (because Riley is now in the same position Piper was in). Thus, $r=\frac{25p}{36}$.

We also know $p+q+r=1$. Thus,

$$p+\frac{5p}{6}+\frac{25p}{36}=\frac{91p}{36}=1 \Rightarrow p=\frac{36}{91}, q=\frac{30}{91}, r=\frac{25}{91}.$$

Edited to add: This doesn't completely answer the question because you asked for the probability that Piper is last to leave, not first to leave. But if Quincy leaves first (which will happen with probability $\frac{30}{91}$), you can use the same mode of analysis to determine the probability that Riley beats Piper out of the box ($\frac{6}{11}$), and if Riley leaves first (which will happen with probability $\frac{25}{91}$), you can use the same mode of analysis to determine the probability that Quincy beats Piper out of the box ($\frac{5}{11}$). The answer should turn out to be:

$$\frac{30}{91}\cdot\frac{6}{11}+ \frac{25}{91}\cdot\frac{5}{11}=\frac{180}{1001}+\frac{125}{1001}=\frac{305}{1001}.$$

At the end of Riley's $n$th turn (that is, after each player has had exactly $n$ turns), each player has a $\left(\frac56\right)^n$ probability to still be in the box.

If Piper is the last to get out, and this happens on her $n+1$st turn, then:

• at the end of Riley's $n$th turn, Piper was still in the box,
• at the end of Riley's $n$th turn, Quincy was not still in the box,
• at the end of Riley's $n$th turn, Riley was not still in the box, and
• on her $n+1$st turn, Piper rolled the same number on both dice.

You can work out the probability that all four of those events occurred.

If Piper was the last to get out, then it happened on her second turn ($n=1$), or her third turn ($n=2$), or her fourth ($n=3$), ... a list of possibilities for $n = 1$ to infinity, and all are mutually exclusive events so their probabilities can simply be added.

You do not get a geometric series out of this, but with a little manipulation you can get three geometric series, evaluate each one and add them together.