# Probability of first player rolling double six first

A high school university entrance examination question:

Players A and B take turns in throwing two dice. The winner is the first player to throw a double six. Player A starts the Game.

1. Find the probability that Player A wins on the first throw. Attempt: Obviously 1/36.

2. What is the probability that Player A wins on the first or second throw? Attempt: Probability of win on the second throw is 35/36 times 1/6. Probability of win on first or second throw is therefore the sum of (i) and (ii), or 1/6(1+35/36).

3. Find the probability that Player A eventually wins the game. Attempt: I tried a geometrical progression using 35/36 as the common ratio, but am stuck. Can you give me a prompt?

• In part (ii), it looks like you've omitted B's first throw. – user21467 Dec 1 '13 at 9:10
• Where do you get the 1/6 from? This is all a big bunch of 1/36 and 35/36 probabilities. – aaaaaaaaaaaa Dec 1 '13 at 9:21
• Thank you! I had completely forgotten about the other player. – Guy Corrigall Dec 1 '13 at 9:27

You can solve this recursively. Noting that after A and B both didn't throw double six the game basically restarts, you get

$p={1\over 36} + {35 \over 36}{35 \over 36}p$.

Thus $p = \frac{{1\over 36}}{1-{35 \over 36}{35 \over 36}}= {36 \over 71}.$

• Very succint. Thank you. – Guy Corrigall Dec 1 '13 at 9:33

In the (iii) part we can do in the following way :-

Let the event A : A wins in his turn

and the event B : B loses in his turn

$P = P(A) + P(A'BA) + P(A'BA'BA) + \ldots$

$P=\frac{1}{36}+\frac{35^2}{36^2}*\frac{1}{36}+\frac{35^4}{36^4}*\frac{1}{36}+\ldots$

$P = \frac{1}{36}*(1+\frac{35^2}{36^2}+\frac{35^4}{36^4}+\ldots)$

$P = \frac{1}{36}*\frac{1}{1-\frac{35^2}{36^2}}$

$P = \frac{36}{71}$

• Yes, that was the GP I was looking for. – Guy Corrigall Dec 1 '13 at 9:32