HMMT 二月 1998 · ADV 赛 · 第 10 题
HMMT February 1998 — ADV Round — Problem 10
题目详情
英文原题
- In the fourth annual Swirled Series, the Oakland Alphas are playing the San Francisco
Gammas. The first game is played in San Francisco and succeeding games alternate inlocation. San Francisco has a 50% chance of winning their home games, while
Oakland has a probability of 60% of winning at home. Normally, the series willstretch on forever until one team gets a three game lead, in which case they aredeclared the winners. However, after each game in San Francisco there is a 50%
chance of an earthquake, which will cause the series to end with the team that haswon more games declared the winner. What is the probability that the Gammas willwin?
解析
英文解析
- In the fourth annual Swirled Series, the Oakland Alphas are playing the San Francisco Gammas. Thefirst game is played in San Francisco and succeeding games alternate in location. San Francisco hasa 50% chance of winning their home games, while Oakland has a probability of 60% of winning athome. Normally, the serios will stretch on forever until one team gets a three-game lead, in which casethey are declared the winners. However, after each game in San Francisco there is a 50% chance of anearthquake, which will cause the series to end with the team that has won more games declared thewinner. What is the probability that the Gammas will win?
Answer: . Let F ( x ) be the probability that the Gammas will win the series if they are ahead by 34
x games and are about to play in San Francisco, and let A ( x ) be the probability that the Gammas will 73
win the series if they are ahead by x games and are about to play in Oakland. Then we have
3 A (1)
F (2) = +
4 4
6 F (0) 4 F (2)
A (1) = +
10 10
1 A (1) A ( − 1)
F (0) = + +
4 4 4
6 F ( − 2) 4 F (0)
A ( − 1) = +
10 10
A ( − 1)
F ( − 2) =
6 F (0) 4 F (2) A (1)4
Plugging A (1) = + into F (2) = + , we get 3
10 10 4 4
( )
3 1 6 F (0) 4 F (2)
F (2) = + +
4 4 10 10
9 F (2) 3 6 F (0) 5 F (0) = + ⇔ F (2) = +
10 4 40 6 6
6 F ( − 2) 4 F (0) A ( − 1)3
Plugging A ( − 1) = + into F ( − 2) = , we get
10 10 4
34 A ( − 1) 4 F (0) 2 F (0) = ⇔ F ( − 2) =
40 10 17
Now,
( ) ( )
1 1 6 F (0) 4 F (2) 1 6 F ( − 2) 4 F (0)
F (0) = + + + +
4 4 10 10 4 10 10
F (0) F (2) 6 F ( − 2)
This simplifies to F (0) = + + + . Then, plugging our formulas in, we get 1
4 4 10 40
( )
1 F (0) 1 5 F (0) 3 F (0)
F (0) = + + + + .
4 4 10 6 6 170
73 F (0) 1 34 = ⇔ F (0) = .
102 3 73
Since F (0) is the situation before the Series has started, the probability that the Gammas will win is .34
473