## Frank Morgan's Math Chat |

December 6, 2001

**OLD CHALLENGE** (Al Zimmermann). In the baseball World Series (best of 7),
which should be more difficult: to come back from being behind 0-2, or to
come back from being behind 2-3?

**ANSWER** (Jonathan Falk, John Shonder). It depends on whether our probability
p of winning a game is greater than or less than about 64%. At p = 64.04%,
our probability of a comeback is 41% in either
scenario. If p > 64%, we prefer the longer series following 0-2. If p <
64%, we take our chances with the shorter series following 2-3.
Essentially, time favors the better team, as long as it is better enough.

Here's the computation. Let p denote the probability of winning, so
1-p is the probability of losing. The probability of coming back from 2-3
(by winning two games in a row) is exactly p^{2}. The probability of coming
back from 0-2 is p^{4} + 4 (1-p) p^{4}, since we must win four games, either
consecutively (probability p^{4}) or also losing one game (probability (1-p)
p^{4}), which can happen four ways (losing the third, fourth, fifth or sixth
game). These two probability formulas are equal when p is (1 + √17)/8,
about 64%.

Incidentally, in one of the problems in his new book "Duelling Idiots and Other Probability Puzzlers," Paul Nahin calculates the probability that the World Series will require 4, 5, 6 or 7 games. He shows that over the years, the proportions are almost exactly what we would expect for two evenly matched teams. Thus p should probably be equal to 0.5 in the above equations, and it is harder to come back from 0-2.

Joseph DeVincentis agrees, pointing out that even the best major league baseball teams win only about 70% of their games when competing against a broad mixture of teams, good and bad, from across the league. (Seattle, which set a record this year with 116 wins, won only 71.6% of its games.) The teams that make it to the World Series are presumably somewhat better than the average, and so should be expected to be more evenly matched than a 64% chance of one team winning any given game.

A number of respondents mentioned other important factors, such as the home field advantage.

**QUESTIONABLE MATHEMATICS**. Eric Brahinsky reports that in the comic strip
Foxtrot, a football quarterback about to hike the ball goes:

Hutt one! Hutt two! Hutt three! Five! Seven! 11! 13! 17! 19! 23! 29! 31! 37! 41! 43! 47! 53! 57! 61! 67! .... And Deion Sanders thought HE was Prime Time!

Including 1 as a prime is an error of convention, but including 57 = 3 x 19 is a serious error. The cartoonist probably meant instead to write 59, which is prime.

Michael Benedict (Quincy High School Mathematics Department) spotted this report on the Brazilian soccer star Pele in the Wenatchee World (Wenatchee, WA) newspaper in October:

"During a 1957-1977 career, Pele played in 375,000 games and scored 281,000 goals."

There are about 7300 days in 20 years. This means that Pele would have to average approximately 51 games per day to achieve this kind of feat!

And then there's this quote submitted by Howard Waldman:

"We're going to turn this team around 360 degrees."

--Basketball star Jason Kidd, after being drafted to the Dallas Mavericks.

Of course, 360 degrees just brings you right back where you started.

Readers are invited to submit more examples of questionable mathematics.

**PUTNAM EXAM**. The notorious six-hour Putnam Exam contest for American
college students was held last Saturday. The median score is often 0. For
more information, including questions and answers, see
http://www.math.niu.edu/~rusin/problems-math/

**ATTACKING HYPERQUEENS**. Frequent contributor Al Zimmermann has announced an
Internet-based computer programming contest on the Attacking Hyperqueens
problem; see http://members.aol.com/DrMWEcker/Contest.html.

**NEW CHALLENGE** (Joe Shipman). This past Halloween was a "blue moon," the
second full moon in a month. Assuming that the interval between full moons is exactly 19/235 of a year
(about 29.5 days), blue moons occur 7 times in 19 years, but I read
that there had not been one on Halloween for forty-odd years. What is the
maximum number of years that can go by before a full moon occurs on
Halloween?

Copyright 2001, Frank Morgan.

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