Tom and Paul roll 2 dice alternatively starting with Tom. Consider they use two fair 6-faced dice. The player who rolls 6 first wins. They continue to roll until one of them wins. Find probability that Tom wins.

I have listed out the total possible outcomes below:

{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Here are all the possible outcomes of two dice rolls. And as one can see, only 5 of these comes out with a 6 first; so, I think the probability of Tom winning should be 5/36.

However, when I looked at the solution (This is a question from one midterm in one of previous semesters given at my school) says:

p is 5/36

$$P(Tom\, Wins)=\sum_{k=0}^{\infty }p(1-p)^{2k}=p\sum_{k=0}^{\infty }(1-p)^{2k}=$$ $$\frac{\frac{5}{36}}{1-\frac{31}{36}}=\frac{5}{5}$$

As one can see this answer is really rediculus! (Tom will win no matter what??)

I think that the key here is probably on the word "alternatively", but cannot figureout what has gone wrong here.

  • $\begingroup$ You’ve computed that sum incorrectly. Observe that the exponent is $2k$, not $k$, i.e., you’ve got a geometric series in $(1-p)^2$, not $1-p$. $\endgroup$
    – amd
    Commented Apr 3, 2020 at 19:50
  • $\begingroup$ I didn't come out with that answer. It is answer given by someone. I looked at it and was very confused. $\endgroup$
    – Amos Ku
    Commented Apr 4, 2020 at 1:37

1 Answer 1


The probability that Tom wins on the first roll is $\frac16$. He wins on the third roll if and only if he first rolls something other than a $6$, then Paul rolls something other than a $6$, and then Tom rolls a $6$; the probability of this is $\frac56\cdot\frac56\cdot\frac16$. In general, Tom wins on the $(2k+1)$-st roll if and only if the first none of the first $2k$ rolls is a $6$, and Tom rolls a $6$ on roll $2k+1$; this occurs with probability $\frac16\left(\frac56\right)^{2k}$. Thus, the probability that Tom wins is

$$\begin{align*} \sum_{k\ge 0}\frac16\left(\frac56\right)^{2k}&=\frac16\sum_{k\ge 0}\left(\frac56\right)^{2k}\\ &=\frac16\sum_{k\ge 0}\left(\frac{25}{36}\right)^k\\ &=\frac16\cdot\frac1{1-\frac{25}{36}}\\ &=\frac6{11}\;. \end{align*}$$

This is the formula in the solution that you read, with $p=\frac16$.

It should not be surprising that this is slightly more than $\frac12$: the fact that Tom goes first gives him an advantage, but it’s a small one, since the game is otherwise very symmetric.

One can also compute the desired probability without resort to infinite series. Let $p$ be the probability that Tom wins, so that Paul wins with probability $1-p$. On the other hand, the probability that Paul when Tom first rolls something other than a $6$ must be $p$, because at that point the game is effectively starting over with Paul as the first player. Thus, Paul wins with probability $\frac56p$, the probability that Tom rolls something other than a $6$ initially and Paul then rolls a $6$ before Tom does. In short, $1-p=\frac56p$, and solving for $p$ again yields $p=\frac6{11}$.

  • $\begingroup$ Thank you. I think your answer is good. I guess the solution given for this problem is probably wrong. $\endgroup$
    – Amos Ku
    Commented Apr 4, 2020 at 1:36
  • $\begingroup$ @AmosKu: You’re welcome. $\endgroup$ Commented Apr 4, 2020 at 1:42
  • $\begingroup$ I have given the question quite a bit of thought last night. I think the key is "alternatively". And whenever Tom rolls a 6, he wins. I think your answer is for the case where only one 6 face dice is rolled. I have edited my wording above, hoping to make the meaning of original question clearer. $\endgroup$
    – Amos Ku
    Commented Apr 4, 2020 at 17:44
  • $\begingroup$ @AmosKu: Alternatively makes no sense here; I suspect that the intended word is alternately. That simply means that Tom rolls his die, then Paul rolls his, and so on. This is exactly the same problem as if they were using a single die and trading it back and forth. $\endgroup$ Commented Apr 4, 2020 at 17:52
  • $\begingroup$ Possible. But I just double checked the original question. The word is " Alternatively". There are other errors in the writing of the test as well. I guess you are right. $\endgroup$
    – Amos Ku
    Commented Apr 4, 2020 at 20:07

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