2024 AMC 10A Problem 17

Two teams are in a best-two-out-of-three playoff: the teams will play at most

3

games, and the winner of the playoff is the first team to win

2

games. The first game is played on Team A’s home field, and the remaining games are played on Team B’s home field. Team A has a

\tfrac23

chance of winning at home, and its probability of winning when playing away from home is

p.

Outcomes of the games are independent. The probability that Team A wins the playoff is

\tfrac12.

Then

p

can be written in the form

\tfrac12\!\left(m - \sqrt{n}\right),

where

m

and

n

are positive integers. What is

m + n?

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Solution

Team A takes game

1

at home with probability

\tfrac23,

and each away game with probability

p.

It can win the playoff three disjoint ways: win games

1, 2;

win

1,

lose

2,

win

3;

lose

1,

win

2, 3.

Adding those,

\tfrac23 p + \tfrac23(1 - p)p + \tfrac13 p^2 = \tfrac12.

This cleans up to

2p^2 - 8p + 3 = 0,

p = \tfrac{4 - \sqrt{10}}{2} = \tfrac12\!\left(4 - \sqrt{10}\right).

Then

m = 4,

n = 10,

and

m + n = 14.

Thus, E is the correct answer.