Question

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

• A. best of 3 games
• B. best of 5 games
• C. same probability in both
• D. cannot win

Answer 1

Answer: (A)

best of 3 games

The probability $p_1$ or winning the best of three games is = the probability of winning two games + the probability of inning three games
=${}^3{C}_2\left(0.6\right){\left(0.4\right)}^2{+}^3{C}_3{\left(0.4\right)}^3$
Similarly the probability of winning the best five games $p_2$ = the probability of winning three games + the probability of winning 5 games
${=}^5{C}_3{\left(0.6\right)}^2{\left(0.4\right)}^3{+}^5{C}_4{\left(0.6\right)\left(0.4\right)}^3{+}^5{C}_5{\left(0.4\right)}^5$
We have $p_1=0.288+0.064=0.352$
and $p_2=0.2304+0.0768+0.01024=0.31744$
As $p_1>p_2$
Therefore A must choose the best of three games