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January 21, 1999

**OLD CHALLENGE.** In his Contract Bridge syndicated column on December 31, Phillip Alder says that the odds against getting a "Yarborough--a hand with no card higher than a nine" are 1827 to 1, and that the odds against two players getting a Yarborough on the same deal are about 182 million to 1. Is he right? (A bridge deck consists of 52 cards, AKQJT98765432 in each of four suits. Each of four players is randomly dealt 13 cards.)

**ANSWER.** The odds against a particular player getting a Yarborough are indeed about 1827.04 to 1, but the odds against some two players geting Yarboroughs are only about 91 million to 1, half of Alder's figure. (The odds against two particular players getting Yarboroughs are six times as great: about 547 million to 1. Alder's figure of 182 million gives the odds of one particular player and some other player having Yarboroughs.) Joe Shipman observes that, "Enough Bridge is played that double Yarboroughs must have occurred in tournaments several times by now; one-third of the time the players who both get Yarboroughs will be partners." As a matter of fact, Alder's column reports just such a hand during the Spring Nationals in Reno, Nevada, last March; the two players were not partners.

To compute the odds that a particular player gets a Yarborough, note that 32 of the 52 cards are nine or less. The chance that the first card is nine or less is 32/52; given that, the chance that the second card is nine or less is 31/51; and so on. The chance that all 13 cards are nine or less is

32 | 31 | 30 | 29 | 28 | 27 | 26 | 25 | 24 | 23 | 22 | 21 | 20 |

__ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ |

52 | 51 | 50 | 49 | 48 | 47 | 46 | 45 | 44 | 43 | 42 | 41 | 40 |

or about 1/1828, for odds of about 1827 to 1. The chance that two particular players both have Yarboroughs, that their 26 cards are nine or less, is

32 | 31 | 30 | 29 | 28 | 27 | 26 | 25 | 24 | 23 | 22 | 21 | 20 | 19 | 18 | 17 | 16 | 15 | 14 | 13 | 12 | 11 | 10 | 9 | 8 | 7 |

__ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ | __ |

52 | 51 | 50 | 49 | 48 | 47 | 46 | 45 | 44 | 43 | 42 | 41 | 40 | 39 | 38 | 37 | 36 | 35 | 34 | 33 | 32 | 31 | 30 | 29 | 28 | 27 |

or about 1/547 million. John Robertson marvels that this answer is 1/547,255,474.5000 *exactly*. (There are enough big numbers in the denominator to cancel almost all of the numbers in the numerator.) Using the symbol n! ("n factorial") to stand for the product of all natural numbers less than or equal to n, this can be abbreviated as (32!/6!)/(52!/26!). Since there are six possible pairs of players (1&2, 1&3, 1&4, 2&3, 2&4, and 3&4), the chance that some two players both have Yarboroughs is about 6/547 million or about 1/91 million. (There are not enough small cards for three players all to have Yarboroughs.)

Other winning answers by Eric Brahinsky, Jean-Pierre Carmichael, John Snygg, and Al Zimmerman.

WHEN ORDER MATTERS. An important question studied in mathematics is when the order in which you do things matters. First multiplying by 9 and then dividing by 5 is the same as first dividing by 5 and then multiplying by 9. For example,

- (10x9)/5 = 90/5 = 18,

and

- (10/5)x9 = 2x9 = 18.

(This can be useful, since the second way is easier.) On the other hand, first adding 32 and then multiplying by 9/5 is different from multiplying by 9/5 and then adding 32. For example,

- (28+32)(9/5) = 60(9/5) = 108,

while

- (28x9/5)+32 = 50.4+32 = 82.4.

This led to a mistake in *The Christian Science Monitor* (January 12, 1999, p. 14) in instructions for converting temperature from Celius to Fahrenheit. The article said to add 32 and multiply by 9/5, whereas the correct proceudre is first to multiply by 9/5 and then to add 32. We hope they do not try first to put on their shoes and then to put on their socks!

**NEW CHALLENGE** (Derek Smith). You have two one-hour fuses: lighting one end of a fuse will cause it to burn down to the other end in exactly one hour's time. You know nothing else about the fuses; in particular you don't know how long any segment of a fuse will burn, only that an entire fuse takes one hour. How can you tell when exactly 45 minutes have passed?

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.

Copyright 1999, Frank Morgan.