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 Poker Game (Posted on 2012-09-07)
A game of poker is played where there are 3 players and each player starts with 2 coins. At the beginning of each hand, every player bets 1 coin and the winner of the hand gets his own coin back in addition to all the bets placed by the other players (for example, in the case of 3 players each betting 1 coin, the winner will receive 2 coins in profit). Now assume the winner of a hand is determined at random (e.g. with 3 players the chance of winning is 1/3). Also assume that once a player reaches 0 coins he/she is out. Determine the expected number of hands that will be played in the match.

 No Solution Yet Submitted by Chris, PhD Rating: 4.0000 (3 votes)

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 The Drunkard's walk Comment 4 of 4 |
I first read about this in 1969.  In the problem, a drunkard is on a street which has a bar at one end (m steps away) and his home at the other end (n steps away).  Each of his steps is random, with probability 1/2 of stepping towards the bar, and probability 1/2 of heading towards home.  His walk ends when he reaches one end or the other.  His probability of ending up at the bar is easily proved to be m/(m+n).  His probability of ending up at home is n/(m+n).  The expected number of steps is mn (which is almost as easily proved).

The same problem, in a gambling setting, has two players, with m and n chips respectively.  If one chip is bet per hand, and each player is equally likely to win the hand, and they play until one player is out of chips, then what is the probability that the first player winds up with all the chips?  That the second player has all the chips?  What is the expected number of hands?  Same answers as above.

The problem is more interesting if the game is not necessarily fair, and the first player has probability p of winning.  If the second player is the casino, and the casino has many many more chips than the player, and the odds of winning favor the casino, then the chance that the player can wipe out the casino (instead of the other way around) is infinitesimal.  (I'm sorry, I don't remember the formulae.)

In the drunkard's walk, p< 1/2 corresponds to the case where the street is not level, and the drunkard is therefore more likely to take a step in one direction than the other.  Hopefully, it is his home (and not the bar) that is at the bottom of the hill.

Edited on September 9, 2012, 10:23 am
 Posted by Steve Herman on 2012-09-09 10:20:17

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