[CaptLego]

A hand I got into with SpaceMonkey in the QToC has me thinking... a little...

Basically, there was some money in the pot, I bet all-in, and he called.
What I'm thinking about is a different way to analyze the decision to call an all-in bet heads-up. (This could be generalized, but for now I'm trying to keep it simple.) I haven't carefully worked through the math yet to see if this makes any difference, but here's the direction of the thinking:

The normal way to make this decision (call the all-in bet vs. fold) is based on pot odds. You look at the ratio of the potsize to the betsize, and compare that to your probability of winning the hand.

I'm thinking that the probability of winning the hand is important, but instead of pot odds, the important ratio is the ratio of the potsize to your remaining chips. And rather than just the probability of winning the hand, you want to base your decision on the probability of winning the tournament.

So for example:
Let's call the first player "SM", with a chipstack of 7000.
The second player, "CL" has a chipstack of 5000.

After some rounds of betting, there's 6000 in the pot when CL goes all-in for his last 2000. Should SM call?

Option A:
fold. Then CL would have 8000 chips, and SM would have 4000. So CL would have a 2:1 chiplead. What's the prob ability of SM winning the tournament? In this case, let's say it's even money (YMMV), or 50%. (SM is better than CL)

Option B:
call.

Now if SM wins the hand, he wins the tourney. What's the probability of this? It depends on # of outs, etc. Let's say, just for the sake of argument, that he's getting exactly the correct pot odds for a call. The pot is 8000, and it takes 2000 to call, so the pot is laying 4:1. So SM needs a 20% chance to win to match those pot odds. Assume he has the 20% chance. So he has a 20% chance to win the tourney right here.

Then there's a (1 - .2) = 80% chance SM will lose that hand. In that case, he's got 2000 chips facing a stack of 10,000. CL would have a 5:1 chiplead. So if 1:2 is 50% to win the tourney, then 1:5 would be -- what, maybe 32% to win the tourney? (.32=.5^(log(5+1)/log(2+1))

So the probability of SM winning the tourney if he calls is:
.20 + .80*.32 = 46%

So in this case, an even-money bet based on pot odds actually reduces SM's probability of winning the tournament.

Suppose SM started with 10,000 chips. Should he more inclined, or less inclined to make that call? He's getting the same pot odds.
In this case,
Option A: fold, would leave him with 7000 chips vs. 8000. So he'd be a 62% favorite to win the tourney.
Option B: call If he loses, he'd be 5000 vs 10000 chips, or 50% to win the tourney, for a total call expectation of (.2 + .8*.5) = 60%

This is a closer decision, but it's still 62% for fold vs 60% for call, even though the pot odds are precisely correct.

On the other hand, suppose SM started with only 6000 chips.
Then:
Fold: he has 3000:8000 chips, or 44% chance to win the tourney.
Call: (lose means 1000:10,000 chips or 21% chance) so the probability of winning the tourney by calling is (.2 +.8*.21) = 37% , or 7% lower than folding.


Playing around with the numbers in this formula:

Against an evenly-match opponent, it makes no difference if you call this bet or not. (Makes sense, since it's an even-money bet.)

Playing against a weaker opponent:
This is a bad bet. But the bigger your chiplead is to start, the less the penalty for making this call.

Playing against a stronger opponent:
This is a good call. Further, the smaller your starting stack is, the greater gain you get in %chance to win the tournament. For example, with a starting chipstack of 6000 against an opponent over whom you'd need a 2:1 chipstack advantage for a 50% chance of a tourney win, you'd be doubling your chances to win the tourney by calling this bet.