How many clairvoyants would be needed before being certain that one of them would correctly predict the future of the World Series?
Suppose the World Series is about to begin. Some clairvoyants get together and decide to each predict different win-loss patterns for the seven possible games during the coming series. That is, one might predict that the first team will winn the first three games, then lose two, then win the next. How many clairvoyants would be needed before being certain that one of them would correctly predict the future of the World Series?
1 Answer | Add Yours
We assume each one predicts a different pattern.
There are a maximum of 7 games; the complexity occurs because any 4 wins is enough.
The possibilities for the eventual winner include:
(1) WWWW win the first 4 games.
(2) Win the first 3 games -- the possibilities include
(3) Win the first two games:
(4) Win the first game:
(5) Lose the first game:
(6) Lose the first 2 games:
(7) Lose the first 3 games
Thus for a particular team, there are 35 different win/loss patterns. If the psychics must also select the winning team, you would need 70 to guarantee a match.
A more mathematical approach:
For each game in the 7-game series, there are 2 possibilities; L or W. Thus there are `2^7=128` different patterns. But some are not allowed -- you cannot have a series with 7,6,or 5 wins or losses.
Now there is 1 way to get 7 wins and 1 way to get 7 losses. This is `_7C_7=1` where `_pC_r` is the number of combinations of p things taken r at a time.
There are 7 ways to get 6 wins, and 7 ways to get 6 losses. (Win all but game 1, game 2, etc...) This is `_7C_6=7`
There are 21 ways to get 5 wins and 21 ways to get 5 losses: this is `_7C_5=21` .
Thus there are 1+1+7+7+21+21=58 impossible series -- 128-58 = 70, the number of possible series from a particular team's perspective.
We’ve answered 319,203 questions. We can answer yours, too.Ask a question