Ttgjyuk

In a hockey-themed board game, players start the game in the penalty box. If rolling the same number on both dice is required to escape from the penalty box, and Piper, Quincy, and Riley take turns, in he order named, rolling a pair of standard six sided dice, what is the probability that Piper is the last player to escape from the penalty box?

My initial thought was that riley and quincy both escape with a 1/36 probability. So, we multiply the three together getting $frac136^3$, but this is clearly incorrect.

Then I took more thinking to this problem. To satisfy the problem we need Piper to lose her first throw, Quincy to win her throw, Riley to win her throw, then Piper will eventually win her throw. So we have $dfrac56 cdot dfrac16 cdot dfrac16 cdot = dfrac56^3$.

However, there other cases where Piper looses, Quincy looses, Riley looses, Piper looses, then Quincy and Riley wins, or where Riley wins before Quincy.

So this can account into what I'm thinking of an infinitely long geometric sequence that later converges to a rational number.

There are many, many more cases. To account for this, I have to keep analyzing all possibilities at each turn, and I believe that eventually I will get some geometric progressions. However, this seems tedious and error prone.

So then I thought that I could just denote the probabilities that I am interested in by letters and write some equations. For instance, if the probability of Piper escaping last is $p$, then after a turn in which nobody escapes the probability of Piper escaping last is still $p$.

This allows me to write the equation: $$p= 5over 6^3cdot p +textother cases .$$

But however, I cannot seem to find that "other cases" thing, and I even further doubted that the $p$ in the equation is stable.

But after some further examination, I find this is not so. I considered first the case with two players: Let $p$ be the probability that the player who starts first escapes first and $q$ the probability that the second player escapes first. I could write the equation $p=1/6+5/6cdot q$ because the probability of the first player escaping is 1/6 plus the probability of him not escaping (5/6) times the probability of him winning as second player, because now the other player rolls the dice first.

Another equation I could write is $p+q=1$ because one of them will escape eventually (with probability 1). Solving this I get $p=1/6+5/6(1-p)$ etc. which leads to $p=6/11$. This is incorrect.

And, after a long time of thoughts, I finally figured the way to write the infinite sum, $p=1/6 + (5/6)^2cdot 1/6+(5/6)^4cdot 1/6+dots$, because the probability of the first player escaping first is 1/6 (if he rolls a double) + (5/6)^2 (=probability of both not rolling a double) times 1/6 (rolling a double at the second turn) and so on...

The sum of that infinite progression is $$p=1/6cdot 1over 1-(5/6)^2$$ which still gives $6/11$, and is wrong again.

I cannot think of a way to continue. Help please?

Let $p$ be the probability that Piper escapes first. Her probability of escaping on her first roll by rolling doubles is $frac16$.

Let $q$ be the probability that Quincy escapes first. Then before Piper's first roll there is a probability of $frac 16$ that Quincy has no chance to escape first, and a probability of $frac 56$ 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, $q=frac5p6$.

Finally, let $r$ be the probability that Riley escapes first. Then before Piper's first roll there is a probability of $frac1136$ that Riley has no chance to escape first (because either Piper or Quincy or both roll doubles) and a probability of $frac2536$ that Riley's chance of escaping first is $p$ (because Riley is now in the same position Piper was in). Thus, $r=frac25p36$.

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

$$p+frac5p6+frac25p36=frac91p36=1 Rightarrow p=frac3691, q=frac3091, r=frac2591.$$

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 $frac3091$), you can use the same mode of analysis to determine the probability that Riley beats Piper out of the box ($frac611$), and if Riley leaves first (which will happen with probability $frac2591$), you can use the same mode of analysis to determine the probability that Quincy beats Piper out of the box ($frac511$). The answer should turn out to be:

$$frac3091cdotfrac611+ frac2591cdotfrac511=frac1801001+frac1251001=frac3051001.$$

At the end of Riley's $n$th turn (that is, after each player has had exactly $n$ turns), each player has a $left(frac56right)^n$ 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.

Given that the outcomes of a pair of dice are
1,1
2,2
3,3
4,4
5,5
6,6

1,2
1,3
1,4
1,5
1,6
2,3
2,4
2,5
2,6
3,4
3,5
3,6
4,5
4,6
5,6

The probability is 6/21 to throw a double pair.

There exists the probablity of each player throwing a double and therefore getting out of the penalty area in one round of dice throwing so getting back to Piper that is merely 4 throws of the dice if She is the last to get out so

( 6 / 21 ) X ( 6 / 21 ) X ( 6 / 21 ) X ( 6 / 21 ) =
0.00666389004581424406497292794669

or approx 0.0067% chance. ( WINDOWS CALCULATOR was used here. )

Regards,

John O.