1971 AMC 12 Problem 23

Question

Teams

A

and

B

are playing a series of games. If either team has an equal chance to win any game, and Team

A

must win two games while Team

B

must win three games to win the series, then the odds favoring Team

A

to win the series are:

11

to

5

5

to

2

8

to

3

3

to

2

13

to

6

It is shorter to enumerate the ways Team

A

can lose

Before

A

’s second win,

B

can finish as BBB, ABBB, BABB, or BBAB

Team

A

loses in the sequences BBB, ABBB, BABB, and BBAB. Their total probability is

\frac18+3\left(\frac1{16}\right)=\frac5{16}.

Thus Team

A

wins with probability

\frac{11}{16},

so the odds in its favor are

11:5.

Therefore, the correct answer is A.

Problem 23 in other years