# What's the probability that Abe will win?

Abe and Bill are playing a fun game. Each of them roll a dice every turn. Abe succeeds if he rolls either a 1 or 2. Bill succeeds is he rolls either a 3, 4, or 5.

If they both succeed on a turn, then they tie the game. If exactly 1 player succeeds on a turn, then the succeeding player wins. If neither player succeeds on a turn, another turn occurs.

What's the probability that Abe will win? I originally thought that the answer was $\frac{2}{5}$, but I realized that the players could tie as well.

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Does Bill succeed if Abe rolls a $3$, $4$, or $5$? Or do they need to roll "their" number? –  André Nicolas Oct 4 '12 at 0:12
They both need to roll one of their numbers on their own dice. Good point. –  David Faux Oct 4 '12 at 0:13
A non-mathematical comment: dice is plural, the singular being die. –  Brian M. Scott Oct 4 '12 at 0:28

Out of 36 (6x6) combinations of die rolls:

Abe will win in 6 of them: (Abe rolls 1 or 2) AND (Bill rolls 1 or 2 or 6)

Bill will win in 12 of them: (Abe rolls 3 or 4 or 5 or 6) AND (Bill rolls 3 or 4 or 5)

The remaining lead to another roll, so they became irrelevant as only one of the 18 situation above finishes up the game.

Out of 18 possible, and equally likely, game ends, Abe wins 6 times, thus:

P{Abe wins} = 6/18 = 1/3

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Let $p$ be the probability that Abe eventually wins the game. This can happen in three ways:

(i) (Immediately) Abe succeeds and Bill fails.

(ii) Abe succeeds, and Bill does. Then they are tied. Given this, Abe's probability of ultimately winning is $p$.

(iii) Abe and Bill both fail. Then we are in a situation similar to (ii).

So $$p=\frac{2}{6}\cdot\frac{3}{6}+\left(\frac{2}{6}\cdot\frac{3}{6}\right)p+\left(\frac{4}{6}\cdot\frac{3}{6}\right)p.$$

Solve. We get $p=\dfrac{1}{3}$.

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