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...

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

Top Answer

embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

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
WWWLW
WWWLLW
WWWLLLW

(3) Win the first two games:

WWLWW
WWLWLW
WWLWLLW
WWLLWW
WWLLWLW
WWLLLWW

(4) Win the first game:

WLWWW
WLWWLW
WLWWLLW
WLWLWW
WLWLLWW
WLWLWLW
WLLWWW
WLLWLWW
WLLWWLW
WLLLWWW

(5) Lose the first game:

LWWWW
LWWWLW
LWWWLLW
LWWLWW
LWWLLWW
LWWLWLW
LWLLWWW
LWLWWLW
LWLWLWW
LWLWWW

(6) Lose the first 2 games:

LLWWWW
LLWWWLW
LLWWLWW
LLWLWWW


(7) Lose the first 3 games

LLLWWWW

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.

 

Sources:

We’ve answered 318,983 questions. We can answer yours, too.

Ask a question