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# Suppose the probability for A to win game B is 0.4. If A has an option of playing either a “best of 3 games” of a “best of 5 games” match against B, which option should be choose so that the probability of his winning the match is higher? (No game ends in a draw).

Jitender Pal
6 years ago
Hello Student,
The probability p1 (say) of winning the best of three games is = the prob. of winning two games + the prob. of winning three games.
= 3 C2 (0.6) (0.4)2 + 3 C3 (0.4)3 [Using Binomial distribution]
Similarly the probability of winning the best five games is p2 (say) = the prob. of winning three games + the prob. of winning 5 games.
= 5 C3 (0.6)2 (0.4)3 + 5 C3 (0.6) (0.4)3 + 5 C5 (0.4)5
We have p1 = 0.288 + 0.064 = 0.352
And p2 = 0.2304 + 0.0768 + 0.01024 = 0.31744
As p1 > p2
∴ A must choose the first offer i.e. best of three games.

Thanks
Jitender Pal