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- From: columbus@strident.think.com (Michael Weiss)
- Newsgroups: sci.math,rec.puzzles
- Subject: Aces and Queens (P.S. to: The paradox of the second ace)
- Date: 16 Dec 92 10:03:20
- Organization: Thinking Machines Corporation, Cambridge MA, USA
- Lines: 28
- Message-ID: <COLUMBUS.92Dec16100320@strident.think.com>
- References: <1gj5grINNk05@crcnis1.unl.edu>
- <1992Dec15.012404.24027@galois.mit.edu>
- <1992Dec15.055832.26324@galois.mit.edu>
- <COLUMBUS.92Dec15111255@strident.think.com>
- NNTP-Posting-Host: strident.think.com
- In-reply-to: columbus@strident.think.com's message of 15 Dec 92 11:12:55
-
- Vaughn Pratt and I both posted simple combinatoric arguments that
- probability (a), below, is less than probability (b):
-
- (a) P(2nd ace | any ace), that is, the probability that a randomly chosen
- bridge hand has a second ace, given that it has at least one ace
-
- (b) P(2nd ace | ace of spades), that is, the probability that a randomly chosen
- bridge hand has a second ace, given that it has the ace of spades
-
- Vaughn was less lazy than I and worked out the actual probabilities. The
- difference is not small; (a) is approximately 37%, while (b) is
- approximately 56%. (I wonder, is this something any decent bridge player
- would know by instinct?) This paradox is discussed in one of Martin
- Gardner's collections.
-
- Here is a variant I haven't seen elsewhere. Consider:
-
- (c) P(ace | queen), that is, the probability that a randomly chosen
- bridge hand has an ace, given that it has a queen
-
- (d) P(ace | queen of spades), that is, the probability that a randomly chosen
- bridge hand has an ace, given that it has the queen of spades
-
- Despite what you might guess, (c) and (d) are *not the same*. They are
- close, it is true: one is about 67%, the other about 68%. (The exact
- figures are 55883/83300 and 10673305/15672338, if I haven't made any
- mistakes in arithmetic.) You may want to amuse yourselves deciding which
- of these answers is (c) and which is (d).
-