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.
From the initial status of (2,2,2)  that is, two for each player, the next state of the system will be (4,1,1)  it doesn't matter who the 4 is.
From there, it can be an immediate end to the game at (6,0,0) or a continuation at (3,3,0). The latter case leads to an immediate next status (probability 1) of (4,2,0), which has a 1/2 probability of going back to (3,3,0) and the same probability of advancing to (5,1,0).
From the (5,1,0) state, there's a 1/2 probability of retreating to the (4,2,0) state and the same probability of closing out at (6,0,0).
(2,2,2)
 Each segment traversal is one hand.
1
 Each segment is marked with the
 probability of taking that path
V 2/3 from the preceding state of the match.
(4,1,1)>(3,3,0) x
  ^
1/3 1  The x, y and z are explained below
   as the expected remaining length of
V   1/2 the match from the given state that
++ V  each is shown next to.
End (6,0,0)Game (4,2,0) y
++ ^ 
^   1/2
  
 1/2  
 1/2  V
+(5,1,0) z
The complication is the two loops: (3,3,0)(4,2,0) and (4,2,0)(5,1,0) that are linked at the (4,2,0) state.
The transition from (2,2,2) to (4,1,1) is definite and requires 1 hand.
The 1/3 probability of immediate end to the match accounts for (1/3)*1 added to the expected number of hands.
That leaves the 2/3 probability branch that leads to (3,3,0).
Call the expected remaining hands when the state is (3,3,0) x.
Call the expected remaining hands when the state is (4,2,0) y.
Call the expected remaining hands when the state is (5,1,0) z.
Because of the probabilities involved in the transitions, and counting 1 for the cost of each transition:
z = (1/2)*1 + (1/2)*(y+1)
y = (1/2)*(z+1) + (1/2)*(x+1)
x = y + 1
Solving these gives
z = 5
y = 8
x = 9
The overall expected number of hands is the original 1 to get from (2,2,2) to (4,1,1) plus (1/3)*1, representing the immediate end of the match with the next hand with probability 1/3, plus 2/3 of (1+x):
1 + 1/3 + (2/3)*(1+9) = 8
The expected number of hands is 8.

Posted by Charlie
on 20120907 12:48:55 