[math-fun] Texas Hold 'em
Two cards are dealt to each of n>1 players in a game of Texas Hold 'em. For which values of n is it possible that all the two-card hands have an equal a priori probability of winning, assuming that everyone will stay in the hand until all the 5 common cards are dealt? -- Thane Plambeck tplambeck@gmail.com http://www.plambeck.org/ehome.htm
Two cards are dealt to each of n>1 players in a game of Texas Hold 'em.
For which values of n is it possible that all the two-card hands have an equal a priori probability of winning, assuming that everyone will stay in the hand until all the 5 common cards are dealt?
i'd be amazed if the answer were anything other than n=2, 3, and 4. deal every player the same denominations in non-matching suits. erich friedman
-----Original Message----- From: math-fun-bounces+andy.latto=pobox.com@mailman.xmission.com [mailto:math-fun-bounces+andy.latto=pobox.com@mailman.xmission .com] On Behalf Of Erich Friedman Sent: Tuesday, June 05, 2007 12:38 AM To: math-fun Subject: Re: [math-fun] Texas Hold 'em
Two cards are dealt to each of n>1 players in a game of Texas Hold 'em.
For which values of n is it possible that all the two-card hands have an equal a priori probability of winning, assuming that everyone will stay in the hand until all the 5 common cards are dealt?
i'd be amazed if the answer were anything other than n=2, 3, and 4. deal every player the same denominations in non-matching suits.
erich friedman
Well, then be amazed! n=23 works too. Choose virtually any set of 23 hands that together include all the cards except 4 aces and 2 kings. The board will be either AAAAK or AAAKK, and as long as no player holds KK, it will be a 23-way tie. n=22 also works, with all cards dealt except AAAAKKKK. Not common numbers to see at a hold-em table, but I've actually played (electronic) Texas Hold-em 23-handed. Andy Latto andy.latto@pobox.com
participants (3)
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Andy Latto -
Erich Friedman -
Thane Plambeck