# Playing chess until one party wins or ties 5 times

Mathematics Asked by Daisy Otaku on January 2, 2021

Players $$A$$ and $$B$$ decide to play chess until one of them wins or ties $$5$$ times in a row.

Assume games are independent, where $$A$$ wins with probability $$P(A) = 0.3$$, $$B$$ wins with probability $$P(B) = 0.25$$, and they draw if $$P(D) = 0.45$$ on each game.

If the game ends in a draw another game will be played until they tie 5 times. Find the probability $$A$$ wins before $$B$$.

The chance they tie the first five games in a row is $$0.45^5$$. If they do, they quit and neither A nor B wins a game. If they do not tie the first five, A will win the first decided game with probability $$frac {0.3}{0.3+0.25}=frac 6{11}$$. The probability A wins without B having won (so it is clear that we accept cases where A wins and B never does) is $$frac 6{11}left(1-0.45^5right)$$. Similarly, the probability B wins without A having won is $$frac 5{11}left(1-0.45^5right)$$

"Find the probability A wins before B" could also require that they both win at least one game but I don't read it that way.

Answered by Ross Millikan on January 2, 2021