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.
With probability 1/3, the same player wins the 1st two hands, and the game is over after 2 hands.
Otherwise, with probability 2/3, two different players win the first two hands, and the third is eliminated. The coins are distributed 3-3, and now we have a classical one-dimensional random "drunkard's" walk with absorbing endpoints (the game ends when a player has 0 or 6 coins) and probability 1/2 of gaining or losing 1 coin on each hand. I happen to know that the expected number of steps from this point is the product of the number coins held by each player, which is 3*3 = 9. So, the answer to the problem is 2 + (2/3)*9 = 8 expected hands. Final answer.
The number of expected steps in the two player game (9) could have been calculated. Let F(x) be the expected remaining steps if you have x coins out of 6.
F(0) = 0
F(1) = 1 + 1/2(F(0) + F(2)) = 1+ F(2)/2
F(2) = 1 + 1/2(F(1) + F(3))
F(3) = 1 + 1/2(F(2) + F(4))
but F(2) = f(4), so F(3) = 1 + F(2)
This gives F(2) = 1 + 1/2((1+ F(2)) + (1 + F(2)/2))
Simplifying gives F(2) = 8
Thus:
F(0) = 0
F(1) = 5 (note, this is 5*1)
F(2) = 8 (note, this is 4*2)
F(3) = 9 (note, this is 3*3)
Don't play poker while drunk (spoiler)
