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### Predicting Divisional Winners

```
Date: 07/29/99 at 15:53:28
From: Jackie
Subject: Probability

There are seven teams in each of the two divisions of American League
baseball, and six teams in each of the two divisions of the National
League. A sportswriter "predicts" the winner of the four divisions by
choosing a team completely at random in each division.

A. What is the probability that the sportswriter will predict at least
one winner correctly?

B. Suppose that the sportswriter eliminates from each division one
team that clearly has no chance of winning, and predicts the winner
from each division from the remaining teams. Assuming that the
eliminated teams do not end up surprising anyone, what is the
writer's chance of predicting at least two winners?

C. Suppose the writer puts all 26 teams in a hat and draws four teams
at random. Does this increase or decrease the writer's chance of
picking at least one winner?
```

```
Date: 07/29/99 at 16:26:46
From: Doctor Anthony
Subject: Re: Probability

(A)

P(at least one winner) = 1 - P(no winner)

= 1 - (6/7)^2*(5/6)^2

=  1 - 900/1764

=  24/49

=  0.489796

(B)

P(at least 2 winners) = 1 - [P(0) + P(1)]

P(0) = (5/6)^2*(4/5)^2  = 400/900  =  4/9

P(1) = yes x  no x no  x no  = 1/6 x 5/6 x 4/5 x 4/5 = 4/45
=  no x yes x no  x no  = 5/6 x 1/6 x 4/5 x 4/5 = 4/45
=  no x  no x yes x no  = 5/6 x 5/6 x 1/5 x 4/5 =  1/9
=  no x  no x no  x yes = 5/6 x 5/6 x 4/5 x 1/5 =  1/9
------------
Total = 2/5

So probability at least 2 winners = 1 - [4/9 + 2/5]

= 7/45

(C)

There are 4 winning teams and 22 not winning.

P(at least 1 correct) = 1 - P(0 correct)

C(22,4)
= 1 -  ---------
C(26,4)

7315
=  1 -  -----
14950

=  1527/2990

=  0.5107

Compare this with 0.489796 for at least one winner by the first
method, and we see that gives a higher probability.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

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