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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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