# Probability of winning a game by rolling the die first

Two persons are playing a game where they take turns rolling a die (so A rolls first, then B, then A again and so on). The first person to roll a $$6$$ wins the game. What is the probability that the person who started the game (rolled the die first) wins?

This was the question that I was given, but I feel like the probability depends on the number of turns played? My approach was that the probability of rolling a 6 at any particular turn will be $$\frac{1}{6}$$. So, the probability of winning in $$n$$ turns will be the probability of not rolling a $$6$$ on the first $$n-1$$ turns (since if a $$6$$ had been rolled, that would have been the last, i.e. the $$nth$$ turn) and then rolling a $$6$$ on the $$nth$$ turn. The required probability would then be $$\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdot...(n-1\text{ times})\cdot\frac{1}{6} = \frac{5^{n-1}}{6^n}$$

Is this train of thought correct? Or have I misunderstood something?

• @lulu Since there are two players, we should have $p = \frac{1}{6} + (\frac{5}{6})^2p$. Feb 23, 2022 at 17:50
• @N.F.Taussig Ah, only two players. You are of course, correct. For some reason, I invited a third player in. Will edit.
– lulu
Feb 23, 2022 at 17:51
• Alternately, once A fails B is in the same position that A was in, so will win the same fraction of the time. For A to win he has to win on the first throw or have B lose. If A wins $p$ of the time $p=\frac 16+\frac 56(1-p)$ Feb 23, 2022 at 17:53

Clearly $$p_A + p_B = 1$$ (somebody wins). And if $$A$$ misses on the first throw, then $$B$$ is in the position that $$A$$ was originally in; so $$p_B = (5/6)p_A$$. Putting these together, you find that $$p_A=6/11$$ and $$p_B=5/11$$.

• This is the most elegant solution, exploiting the symmetry of the problem. Feb 25, 2022 at 9:01

To repair a badly botched comment:

A quick way to get the result is to note that the game resets if both players miss the mark. Thus, if $$p$$ is the answer we seek, we have $$p=\frac 16+\frac 56\times \frac 56\times p\implies p = \frac 6{11}$$

The probability of winning can be seen as a sum $$P = P_1+P_2+P_3+\dots$$. Here $$P_1$$ is the probability of winning at the first turn. Since you are the first to throw the dice, we simply have $$P_1=\frac{1}{6}\,.$$ Then $$P_2$$ is the probability of winning at the second term. For this you need to have rolled something which is not a 6 in the first term (which brings a factor $$5/6$$, than your opponent must have rolled something different from 6 (brings a factor $$5/6$$) and finally you must roll a 6. So you have $$P_2= \frac{5}{6}\times\frac{5}{6}\times\frac{1}{6} = \left(\frac{5}{6}\right)^2\frac{1}{6}\,.$$ Similarly $$P_3$$ means that: no 6 (you), no 6 (opponent), no 6 (you), no 6 (opponent), 6 (you). This brings $$P_3 = \left(\frac{5}{6}\right)^4\frac{1}{6}\,.$$ In general $$P_n= \left(\frac{5}{6}\right)^{2(n-1)}\frac{1}{6}\,.$$ Now you conclude that $$P = \sum_{n=1}^\infty P_n = \frac{1}{6}\sum_{n=0}^\infty\left(\frac{5}{6}\right)^{2n} = \frac{1}{6}\times\frac{1}{1-\frac{25}{36}}=\frac{6}{11}\,.$$

Note that in the sum one should exclude the term $$n=\infty$$, which is the event in which the game never ends. However $$P_\infty=0$$.

• yep I summed from one instead of from 0.. I'll fix it
– ECL
Feb 23, 2022 at 17:56
• While I agree with the overall steps, I found this answer confusing. It uses a slightly different definition of "turn" than the question does. Feb 24, 2022 at 10:29
• What is the sample space? How is it that $P_1$ and $P_2$, for example, have different denominators when they are part of the same probability model? Shouldn't the denominators be the same? That is, should the denominators be the size of the sample space which can't vary in a single probability model? Sep 18, 2022 at 5:23