Calculation of when to call an all-in heads up (simple math)

[Proofrock]

I did this calculation a couple nights ago, and perhaps others will find it interesting. I imagine it has been done before, and probably more elegantly, but here's my first shot at it.

Situation: You are heads up in a tournament. Your opponent has just moved all-in. Do you call?

Assumptions:
(1) You know what your opponent holds (so you know your probability of winning the hand).
(2) Your probability of winning the tournament at any time is the fraction of total chips you have in your stack.
(3) You win more for first place than for second place.

Definitions: Let
fw = fraction of chips you have if you win the hand
fl = fraction of chips you have if you lose the hand
ff = fraction of chips you have if you fold the hand
p = probability of winning the hand
W = $ for 1st place
L = $ for 2nd place

calculation:
The idea is to calculate EV(call) - EV(fold). If this is greater than 0, then a call is warranted -- if it's less than 0, then fold.

EV = Sum of [(probability of hand outcome)*(sum of resulting probabilities of tournament outcome)]

EV(call) = p ( fw*W + (1-fw)*L) + (1-p) ( fl*W + (1-fl)*L)
= p*fw*(W-L) + (1-p)*fl*(W-L) + L
Similarly,
EV(fold) = ff(W-L) + L

So,
dEV = EV(call) - EV(fold)
dEV = (W-L)*[p*fw + (1 - p)*fl - ff]

Now, we are interested only in the sign of this expression -- if positive we call, if negative we fold. Since W-L is an overall positive factor, this becomes
dEV > 0 =>
p*fw + (1 - p)*fl > ff

Observation: The overall pay structure doesn't effect the decision at all, except by an overall scaling factor. Whether it is +EV to call is determined entirely by stack sizes and the probability of winning the hand.

Two cases:
(1) You're shortstacked (or more generally, you have fewer than 50% of the chips to start the hand). Then fl = 0, and you need only p > (ff / fw).

(2) You're chip leader. Then fw = 1, and you need
p(1 -fl) + fl > ff, or
p > (ff - fl)/(1-fl)

Example: There are 6000 total chips, you started the hand with 1700. You raised it to 800 preflop with KJ and opponent called. Flop is A 10 6 r, and opponent moves all-in with 10 5 o.
You have 10 outs, so p ~ 0.37, ff = 9/60, fw = 34/60, so ff/fw = 9/34. Therefore, p > ff/fw, and you should call.

COINFLIPS:
As shortstack, to take a coinflip you'd need 2*ff < fw. If there are no blinds, then ff = fw/2, so this is 0EV (as is intuitively expected, since it's a coinflip). However, the presence of the blinds means ff < fw/2 (now there's money in the pot), so it is profitable from a tournament perspective to take a coinflip when you're behind in chips.

To take a coinflip as chipleader you'd need
0.5 > (ff - fl)/(1-fl),
or 0.5(1 + fl) > ff.
BUT THIS IS ALWAYS THE CASE, since BEFORE you put in any blinds, ff = 0.5(1 + fl), and once blinds go in, ff < 0.5(1 + fl). Therefore, it is always +(tournament)EV to take a coinflip when you're ahead in chips (note: this discussion applies for an actual coinflip, not necessarily 22 v. AK, but in practice the blinds are typically large enough to make it correct).

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These cases can be further elaborated on to address preflop all-ins and the size of the blinds, but I will leave that for others to work out, or for another post if there is interest.

-cj

[AlphaWice]

".. probability of winning is the fraction of chips .."

any model with this assumption is unusable.

[J Chap]

[ QUOTE ]


Assumptions:
(1) You know what your opponent holds

[/ QUOTE ]

Damn - I'd be interested to see the model through which you've achieved Assumption (1).

[citanul]

[ QUOTE ]
[ QUOTE ]


Assumptions:
(1) You know what your opponent holds

[/ QUOTE ]

Damn - I'd be interested to see the model through which you've achieved Assumption (1).

[/ QUOTE ]

admissions:

i didn't read the OP past that point.

conclusions:

i have no idea why there would have to be a long mathematical post about what to do, once you know your cards and your opponents cards and you are heads up.

citanul

[citanul]

[ QUOTE ]
".. probability of winning is the fraction of chips .."

any model with this assumption is unusable.

[/ QUOTE ]

meh. you're going to have to defend yourself.

in the future, try to put, oh, any content in your posts.

the problem with that as an assumption is that if you're making that assumption, there's no reason to look at "strategy."

while probability of winning heads up is very closely tied to fraction of chips when you have large blinds relative to stack size, it is not perfect. when one stops to look at strategy, you're seeking to pick up any extra little percentage points above your fraction of the chips you can. if, given your opponent, this is a large change from fraction of chips in play, all the better.

citanul

[Proofrock]

Would you care to elaborate, please? This was the assumption that I had the hardest time with, but I think it's equivalent to saying "both players are equal poker players," (which of course isn't true but isn't a bad first approximation for a simple model). Am I overlooking something? Here's my reasoning, but it's incomplete:

Let's say I have 75% of chips and get involved in 2 coinflips. I win 1/2 of the time on the first coinflip. If I lose that first coinflip, I win the second coinflip 1/4 of the time => I win 75% of the time and lose 25% of the time. This same reasoning applies to all cass where 1 person has (2^(-n)) of the chips, so it's true for all of these situations, so at any rate we're looking at a function that is correct at all points (2^(-n)), (n, 0, infinity). Any reason it shouldn't generalize?

Thanks,
cj

[Proofrock]

Thanks for your reply - the reason for the calculation, and why I think it is relevant, though simplistic, is as follows:

I was heads up in a tournament, 6000 chips total, I had 1700 on SB before hand started, with blinds at 150/300. I raised to 800 with K J (perhaps it was a mistake to do anything but push here, but at any rate it's how I ended up in the next situation), and he called. The flop was A 10 6, and when he pushed I figured he'd hit the 10 or 6, leaving me 10 outs. But given that information, should I call or fold?

I recognize that there are many other important features, but in trying to determine whether or not I should call, I went with the simplest model that made sense to me. This isn't the kind of reasoning to use when in the middle of a heads up battle -- it's not a discussion of how to play the hand other than "opponent has moved in, I think he has xy, so should I call or fold?"

This being said, is there a better indication or model to use for making such decisions? This was meant to apply only to this special situation, but I am interested in a more robust model that could be used to evaluate the decisions that are more about the cards than the people.

-cj