Probability of winning a game by rolling the die first

Question

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?

Answer 1

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 $$p=\frac 16+\frac 56\times \frac 56\times p\implies p = \frac 6{11}$$

Answer 2

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 $P_\infty=0$.

Answer 3

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 $p_B=5/11$.