College Math Teaching

May 14, 2012

Probability in the Novel: The Universal Baseball Association, Inc. J. Henry Waugh, Prop. by Robert Coover

Filed under: books, editorial, elementary mathematics, pedagogy, popular mathematics, probability, statistics — collegemathteaching @ 2:31 am

The Robert Coover novel The Universal Baseball Association, Inc. J. Henry Waugh, Prop. is about the life of a low-level late-middle aged accountant who has devised a dice based baseball game that has taken over his life; the books main character has a baseball league which has played several seasons, has retired (and deceased!) veterans, a commissioner, records, etc. I talked a bit more about the book here. Of interest to mathematics teachers is the probability theory associated with the game that the Henry Waugh character devised. The games themselves are dictated by the the result of the throws of three dice. From pages 19 and 20 of the novel:

When he’d finally decided to settle on his baseball game, Henry had spent the better part of two months just working on the problem of odds and equilibrium points in an effort to approximate that complexity. Two dice had not done it. He’d tried three, each a different color, and the 216 different combinations had provided the complexity all right, but he’d nearly gone blind trying to sort the three colors on each throw. Finally, he compromised, keeping the three dice, but all white reducing the number of combinations to 56, though of course the odds were still based on 216.

The book goes on to say that the rarer throws (say, triples of one numbers) triggered a referral to a different chart and a repeat of the same triple (in this case, triple 1’s or triple 6’s (occurs about 3 times every 2 seasons) refers him to the chart of extraordinary occurrences which includes things like fights, injuries, and the like.

Note that the game was very complex; stars had a higher probability of success built into the game.

So, what about the probabilities; what can we infer?

First of all, the author got the number of combinations correct; the number of outcomes of the roll of three dice of different colors is indeed 6^3 = 216 . What about the number of outcomes of the three dice of the same color? There are three possibilities:

1. three of the same number: 6
2. two of the same number: 6*5 = 30 (6 numbers, each with 5 different possibilities for the remaining number)
3. all a different number: this might be the trickiest to see. Once one chooses the first number, there are 5 choices for the second number and 4 for the third. Hence there are 20 different possibilities. Or put a different way, since each choice has to be different: this is {{6}\choose{3}} = \frac{6!}{3! 3!} = \frac{120}{6} = 20

However, as the author points out (indirectly), each outcome in the three white dice set-up is NOT equally likely!
We can break down the potential outcomes into equal probability classes though:
1. Probability of a given triple (say, 1-1-1): \frac{1}{216} , with the probability of a given throw being a triple of any sort being \frac{1}{36} .
2. Probability of a given double (say, 1-1-2) is \frac{{{3}\choose{2}}}{216} = \frac{3}{216} = \frac{1}{72} So the probability of getting a given pair of numbers (with the third being any number other than the “doubled” number) would be \frac{5}{72} hence the probability of getting an arbitrary pair would be \frac{30}{72} = \frac{5}{12} .
3. Probability of getting a given trio of distinct numbers: there are three “colors” the first number could go, and two for the second number, hence the probability is: \frac{3*2}{216} = \frac{1}{36} . So there are {{{6}\choose{3}}} = 20 different ways that this can happen so the probability of obtaining all different numbers is \frac{20}{36} = \frac{5}{9} .

We can check: the probability of 3 of the same number plus getting two of the same number plus getting all distinct numbers is \frac{1}{36} + \frac{5}{12} + \frac{5}{9} = \frac{1 + 15 + 20}{36} = 1 .

Now, what can we infer about the number of throws in a season from the “three times every two seasons” statement about triple 1’s or triple 6’s?
If we use the expected value concept and figure that double triple 1’s has a probability of \frac{1}{216^2} = \frac{1}{46656} and getting either triple 1’s or triple 6’s would be \frac{1}{23328} and using E = np , we obtain \frac{n}{23328} = 3 which implies that n = 69984 throws per two seasons, or 34992 throws per season. There were 8 teams in the league and each played 84 games which means 336 games in a season. This means about 104 throws of the dice per game, or about 11.6 throws per inning or 5.8 throws per half of an inning; perhaps that is about 1 per batter.

Evidently, Robert Coover did his homework prior to writing this novel!


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