# Frank Morgan's Math Chat - BRIDGE VOIDS

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 27 = 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?

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.

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