## Frank Morgan's Math Chat |

September 20, 2001

**OLD CHALLENGE**. High school seniors will soon be applying to colleges. Can
you think of a better way to match up students and colleges?

**ANSWER**. Perhaps not. The current system of applications and offers, however
ungainly, well respects the individuality of the colleges and applicants.

**OLD BRIDGE CHALLENGE**. In the September 2001 *Bridge Bulletin* of the
American Contract Bridge League (www.acbl.org), Betty Moore asks for the
probability in a bridge deal that each of the four hands has a void in a
different suit.

**ANSWER**. Joseph DeVincentis computes the probability as .0009101%, or about
1 in 100,000. Here's a heuristic way to guess the answer. First estimate
the chance that the first player is void clubs, the second diamonds, the
third hearts, and the fourth spades. For the first player, each card has
about a 3/4 chance of avoiding clubs, so the probability of a void in clubs
is very roughly (3/4)^{13}. The probability that all four players have their
respective voids is roughly (3/4)^{52}. Since there are 24 ways to decide
which player is void which suit, the probability that each player has a
void in a different suit is very roughly 24 times (3/4)^{52}, which is about
1 in 130,000, very close to the actual 1 in 100,000. Such accuracy is a
result of errors canceling each other. The probability that a single card
avoids a particular suit is sometimes less than 3/4: after several of the
first player's cards have avoided clubs, there are more clubs left, and it
becomes harder to continue to avoid clubs. On the other hand, the
probability is sometimes greater than 3/4: the abundance of clubs left
makes it easier for the second player to avoid diamonds. These errors
apparently cancel each other out rather well.

DeVincentis's rigorous computation is based on the considering the distributions of the cards in each suit, which must be one of:

11-1-1,
10-1-2, 9-1-3, 9-2-2, 8-1-4, 8-2-3,
7-1-5, 7-2-4, 7-3-3, 6-1-6, 6-2-5, 6-3-4, 5-3-5, 5-4-4,

plus permutations of
these, and a few other possibilities if additional voids beyond the required
four are allowed. He calculates the number of ways for the cards in each
suit to end up distributed each of these ways (for example, there are 13x12
= 156 ways to distribute the cards in an 11-1-1 distribution). He finds
all the combinations of distributions for the four suits that add up to 13
cards for each player and calculates the number of ways of dealing each
one. Finally, he adds up all the results and divides by the total number of
bridge deals, 52!/(13!^{4}). His computer finds that 482886422450433777597120
of 53644737765488792839237440000 bridge deals have the required four voids,
if he assumes no other voids existed, for a probability of .000900156%.
When he adds in hands with additional voids, he gets
488220815199432625023384 deals with the four required voids, and the
probability increases to .0009101%.

**QUESTIONABLE MATHEMATICS**. Derek Smith writes: It seems that John Blubaugh
has been banned by the American Contract Bridge League (ACBL) for shuffling
in a way that keeps the ace of spades on the bottom of the deck until it is
dealt to his partner. A Dallas magician, Norman Beck, was hired by the ACBL
to interpret videotapes of some of John's deals. "Guilty" was his verdict,
although I hope he wasn't
basing it on the probability he gave the reporter for the Wall Street
Journal (July 17, 2001, "Bridge Was His Life, until John Blubaugh was
Called a Cheat"):

"The odds of a card starting at the bottom of the deck, and being there again seven shuffles later are 1.026 trillion to one, Mr. Beck says in an interview."

Walter Wright comments: "The quote seems absurd on the face of it. For
example, with the simplifying
assumption that the particular card is equally likely to be in any place in
the deck before the shuffle, and equally likely to be any place in the deck
after the shuffle, the odds of a particular card starting at the bottom and
ending up at the bottom are 2,704 to one (52x52 to one). [The odds of
*staying* on the bottom for seven shuffles are just 2^{7} = 128 to one.] What
possible assumptions could this Mr. Beck have made in order to boost the
odds to 1.026 trillion to one?"

Readers are invited to submit more examples of questionable mathematics.

**NEW QUESTION**. I remember noticing as a kid, contemplating cutting
diagonally across a square lawn, that the diagonal was about one and a half
times as long as the side. (The actual ratio is the square root of two,
about 1.4.) Do you have an early mathematical memory?

Copyright 2001, Frank Morgan.

Send answers, comments, and new questions by email to
Frank.Morgan@williams.edu, to be eligible for* Flatland *and other book
awards. Winning answers will appear in the next Math Chat. Math Chat
appears on the first and third Thursdays of each month. Prof. Morgan's
homepage is at www.williams.edu/Mathematics/fmorgan.

THE MATH CHAT BOOK, including a $1000 Math Chat Book QUEST, questions and answers, and a list of past challenge winners, is now available from the MAA (800-331-1622).