Enter your search terms Submit search form

Make the first roll.

1/36 of the possibilities gives A an instant win.

6/36 give B a chance on a second roll, and the other 29 - this is the key - are meaningless in every sense. With those 29 you start over.

So, the only outcomes of interest are those first 7 outcomes.

So we now look at the second roll, and here is the second key.

If neither A nor B wins by the second roll, then you are again back where you started. You start over.

So the question simplifies to this:

At what rate does A win when the game begins with one of those 7 rolls and does not last longer than 2 rolls?

There are 7*36 possibilities.

A wins 36 of those on the first roll, and another 6 on the second roll.

So A wins 1/6.

B wins 36 of them, all on the 2nd roll. So 1/7.

A/B = 7/6

And clearly A + B = 1

So the probability for A to win is 7/13.

Another way to formulate the same:

On the first roll we can either roll 12 (probability 1/36, A wins), something else than 7 or 12 (probability 29/36) which takes us back to the initial situation, or a 7 (probability 1/6) which leads to a new situation.

In this new situation we can roll 12 (probability 1/36, A wins), something else than 7 or 12 (probability 29/36) which also takes us back to same position as when the game began.

So we can set up an equation for the probability that A wins,

p = 1/36 + 29/36 × p + 1/6 × (1/36 + 29/36 × p)

Solving for p gives 7/13.