Question

Suppose the probability for A to win a game against B is 0.4. If A has an option of playing either a "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 a draw)

• A. Best of 3 games
• B. Best of 5 games
• C. Both have equal chances of A winning.
• D. Can't be determined.

Answer 1

The correct option is A

Best of 3 games

$p_1$ = probability of winning the 'best of 3 games'
= last two terms in the expansion of
$(0.6+0.4{)}^3$
${=}^3{C}_2\left(0.6\right).(0.4{)}^2+(0.4{)}^3=0.288+0.064=0.35200$

$p_2$ = probability of winning the 'best of 5 games'
= last three terms in the expansion of
$(0.6+0.4{)}^5$
${=}^5{C}_3(0.6{)}^2.(0.4{)}^3{+}^5{C}_4\left(0.6\right).(0.4{)}^4+(0.4{)}^5=.2304+.0768+.01024=0.31744$