Quote:
Originally Posted by DrChesspain
Can anyone offer an opinion as to the value of this chip?
The value of the chip to the person holding it is going to depend on three factors - their position, their playing style, and how many hands are left to play before the end of the night. Let's model this with some slightly simplifying assumptions:
- A 6-handed table
- Every player has probability p of keeping the chip if they hold it UTG
- Every non-UTG player has equal probability of getting the chip when it is used
- Instead of "when the table breaks" as the end condition, let's use "in n hands"
What's the probability of straddle keeping the chip? I'm going to guess that a strong aggressive player would have a 40% chance six-handed, but by making it a variable, I can eliminate the guesswork and also let other people plug in the value they prefer.
With these assumptions, the only things that determine value to me are: Who has the chip; what position I am in for this hand; and how many hands are left to play. Rather than determining how much the chip is "worth" to each player, I'll compute the probability that any given player will have it at the end of the night.
When
n = 0 - after the last hand has been played - the chip held by the person holding it. This gives that person 100% and everybody else 0%.
When
n = 1 - at the start of the final hand - the chip is still 100% with the holder unless she is UTG. In that case, the chip is
p with the holder, and (1 -
p)/5 with everyone else. At
p = 0.4, it's 40% to the holder and 12% to the other players.
When
n = 2 and the holder is UTG, things get a bit more interesting. Using the 40% example, 40% of the time she keeps the chip; 48% of the time she loses it to a player who keeps it until the final hand; but 12% of the time she loses it to HJ, and it then goes into the pot again. So HJ only gets it 40% of 12% (4.8%), while everyone else gains an extra 12% of 12% (1.44%).
In general, this model is easy to compute by iterating 36 values: we need to keep track of each position's probability at the table on each hand, and there are 6 possible places for the chip to be relative to the button.
Interestingly, the chip does
not end up being distributed evenly to everybody in the long run, which is what I would have expected. For 40%, the values converge as follows (for the seats that people are sitting in on the final hand):
UTG 10/105
HJ 13/105
CO 16/105
BT 19/105
SB 22/105
BB 25/105
By trial and error, I have determined that for probability
p, the values converge as follows: UTG gets
p/(3(
p + 1)), and each move of one seat to the right from there gains (1 -
p)/(15(
p + 1)) over the previous seat. As far as just the straddle chip goes, you prefer to be sitting in the big blind at the end of the night.
Just to forestall the inevitable comments,
I am aware that this is only a model. I am aware that different positions are stronger and weaker. But as a first approximation, I think this is a decent place to start.