What's the probability that Abe will win the dice game?

Solution 1:

Let $P$ be the chance that Abe wins, then we have

$$P=\frac{1}{3}+\frac{1}{6}P$$

Solving P from this equation we get

$$P=\frac{2}{5}$$

Solution 2:

You are right. You can just ignore rolls of $6$ as they leave you back in the same situation. To formalize this, the chance Abe wins on turn $n$ is $\frac 13 \left(\frac 16 \right)^{n-1}$ and the chance that Bill wins on turn $n$ is $\frac 12 \left(\frac 16 \right)^{n-1}$. You can sum these if you want.

Solution 3:

I agree with you that $\dfrac{2}{5}$ is obvious. End of story. But if you really want to sum a series, abbreviate by $A$ the event "$1$ or $2$" and by $S$ the event "$6$." Then Abe can win in various ways. These are $A$ (wins immediately), $SA$ (get a $6$, then win), $SSA$, $SSSA$, and so on.

These have probabilities $\dfrac{2}{6}$, $\:\dfrac{1}{6}\cdot\dfrac{2}{6}$, $\:\dfrac{1}{6}\cdot\dfrac{1}{6}\cdot\dfrac{2}{6}$, and so on. So we want to sum the series $$a+ar+ar^2+ar^3+\cdots,$$ where $a=\dfrac{2}{6}$ and $r=\dfrac{1}{6}$.

By the usual formula for the sum of an infinite geometric series, this is $\dfrac{a}{1-r}$, which simplifies to $\dfrac{2}{5}$.

Solution 4:

The key point to realise is that if a 6 is rolled, then the probability of each player winning after that roll is the same as it was before. So if $p_A$ and $p_B$ denote the probability of each player winning, then we have $$ p_A = \frac 26 + \frac 16 p_A\\ p_B = \frac 36 + \frac 16 p_B $$ which is easy to solve.