# What is the probability that a player is the first to roll five doubles?

The problem

Two players take turns rolling two dice. The first player to roll 5 (not necessarily consecutive) doubles wins.

What is the probability that the player going first will win?

What I've tried

I initially tried using infinite sums for all the possible outcomes of the game, but I couldn't properly keep track of all the different cases.

My other approach was similar to this answer for the first person to roll just one number, but I feel that it calculates the probability of five in a row, not five total:

$$p = P(\mathrm{player\,one\,wins\,through\,first\,5\,turns}) + P((\mathrm{player\,one,misses\,a\,double\,on\,one\,of\,the\,five\,turns}) \cap (\mathrm{other \,player\,also\,does\,not\,win}))$$ $$p = (⅙)^5 + (1 - (⅙)^5) * (1 - p)$$ $$p=7776/15551$$

• here is another, similar problem. The methods generalize, though of course it gets messier the more states you need to consider.
– lulu
Commented Jan 14, 2023 at 16:19
• @lulu: No need to consider states here :-) Commented Jan 14, 2023 at 17:11
• @joriki Interesting. For this sort of thing, states are almost always easier than the sums. Maybe not here...
– lulu
Commented Jan 14, 2023 at 17:14
• Why so many close votes ? It is a problem that possibly doesn't have a parallel here, and OP has searched, found no parallel, and tried but failed summing to infinity. Sometimes ( often ?) I find strange decisions on this forum . Commented Jan 15, 2023 at 19:02

The probability for player $$1$$ to win on the $$k$$-th roll is

$$\binom{k-1}{5-1}\left(\frac16\right)^5\left(\frac56\right)^{k-5}\left(\sum_{j=0}^4\binom{k-1}j\left(\frac16\right)^j\left(\frac56\right)^{k-1-j}\right)\;,$$

since there are $$\binom{k-1}{5-1}$$ ways to choose the first four double rolls (the fifth one is the $$k$$-th) and player $$2$$ may have rolled between $$0$$ and $$4$$ doubles in the meantime.

We need to sum this over $$k\ge5$$. I couldn’t get Wolfram|Alpha to handle the entire double sum in one go, but calculating it for each $$j$$ separately has the added advantage that we get the individual probabilities for player $$1$$ to win with player $$2$$ having rolled $$j$$ doubles. The result is

$$\frac{59560056}{2516421875}+\frac{9891714384}{138403203125}+\frac{914238016776}{7612176171875}+\frac{61841428690464}{418669689453125}+\frac{3450754831279896}{23026832919921875}\\[20pt]=\frac{11808349128075816}{23026832919921875}\;,$$

$$2.4\%+7.1\%+12.0\%+14.8\%+15.0\%=51.3\%\;,$$