Question

Fischer and Spassky play a chess match in which the first player to win a game wins the match. After 10 successive draws, the match is declared drawn. Each game is won by Fischer with probability 0.4, is won by Spassky with probability 0.3, and is a draw with probability 0.3, independent of previous games.

Answer 1

the probability that Fischer wins the match in game "n" is:
\[(.4)(.3)^{n-1}\]
the probability that he wins match is the cumulative sum up to 10 games
\[\sum_{n=1}^{10}(.4)(.3)^{n-1} = \frac{4}{3} \sum_{n=1}^{10} (.3)^n = \frac{4}{3}(\frac{.3(1-.3^{10})}{1-.3}) = \frac{4}{7}(1-.3^{10})\]
Similarly, the probability Spassky wins match is:
\[\sum_{n=1}^{10}(.3)(.3)^{n-1} = \sum_{n=1}^{10} (.3)^n = \frac{.3 (1-.3^{10})}{1-.3} = \frac{3}{7}(1-.3^{10})\]
Finally the probability that match ends in a draw is (.3)^10