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

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