What is the probability that the World Series will be won in
4 games?
5 games?
6 games?
7 games?
For the sake of this exercise assume that each team has an equal probability
of winning any one game. The Series is won when any team wins four games.
Note that four numbers are sought and they necessarily must add to 1.0.
For those who like more of a challenge, suppose Team A is favored over Team
B. That is, on the average, Team A will win 2/3rds of the time and Team B
will win 1/3rd of the time.
Off the top of your head, how do you think it would affect the four numbers
for the even odds case? Which ones would go up and which ones would go down?
The answer will be published next month. If your curiosity is simply unbearable
and you would like to see it sooner, it can be viewed at
http://pages.prodigy.net/_wsn/puzzle.htm
The Solution
The answers for the even odds case are 4 games: 1/8;
5 games: 1/4; 6 games: 5/16 and 7 games: 5/16.
The probability of going four games is simply 2/16 or 1/8. If you play four
games, there are 16 ways they can come out. Of those 16, two involve one or
the other team winning four in a row.
We recall that the combination of n things taken r at a time is C(n:r) = n!
/ ( r! * (n-r)! ), but you don’t need to really know this to solve the problem.
To compute the probability
of going 5 games, we note there are 32 ways five games can turn out. There are
ten situations where one or the other team has four wins. Simply list all 32
and note this or compute the combination of 5 things taken four at a time is
5! / 4! or 5. Double that for the two teams is 10. Two of these 10 we eliminate
because they are the case where a team wins the first four games in a row. (The
Series was over in four games.) So eight other possibilities remain. The probability
of going 5 games is 8/32 = 1/4. In the case of six games, we have
64 possible outcomes. Of these, 30 involve four wins by one or the other team.
Of these 30, we eliminate 10 because they represent one team or the other winning
the first four in a row, or losing one and winning four. (The series going 4
or 5 games.) This leaves 20 favorable events for a probability of 5/16. Finally
there are then 20 situations where we have a tie at the end of six games. -
The combination of six things taken three at a time 6! / ( 3! * (6-3)!) So the
probability of going to seven games is again 20/64 or 5/16. For
those into computer programming, the problem can be quickly solved by using
the Monte Carlo Method.
The program in pseudo-code goes as follows: (You can write it in the language
of your choice.)
PROGRAM WSERIES
DIMENSION floating array KEEP_TRACK[7] and INITIALIZE to 0. //clear KEEP_TRACK
SET integer NO_CASES to 10000 // or a suitable large number of your choice
LOOP over NO_CASES
SET integers TEAMA and TEAMB to 0 LOOP over I = 1 to 7
IF (RANDOM 0 or 1) equals 0, increment TEAMA , OTHERWISE increment TEAMB
IF TEAMA equals 4 or TEAMB equals 4
THEN increment KEEP_TRACK[I] and
BREAK from this LOOP
END LOOP
END LOOP
Print over I = 1,7, I, KEEP_TRACK[I] / NO_CASES
END PROGRAM
Output from program will be approximately 1, 0 2, 0 3, 0 4, 0.125
5, 0.25 6, 0.3125 7, 0.3125
For the case where Team A is favored 2/3 - 1/3 over Team B, we alter one statement
in the above program as follows:
IF (RANDOM 0 or 1 (P(0) = .666667)) equals 0, increment TEAMA , OTHERWISE increment
TEAMB The answers will
now be approximately
1 0 2 0 3 0 4 210168 5 295786 6 274988
7 219058
So the likelihood of going 4 or 5 games goes up, six games goes down slightly
and seven games goes down significantly.