Easy call, or foolish call?

[Jason Strasser]

Ok, I may be wrong, but this is how I look at it.

If you fold, and the other SS wins, you are going to get third most of the time. In the long run, this will happen 50% of the time. Assuming they both have mediocre hands, they will probably be in a coinflip spot most of the time.

So 50% third, 50% second by folding. Roughly. Thats (using 100 buy in) an avg of $250.

If you call, lets say you have a 20% chance of winning the hand, a 40% chance of losing to the big stack, and a 40% chance of losing to the small stack. (lets assume if u win the hand u have a 50% chance at first, 50% chance at second)

(.2)(400*) + (.4)300 + (.4)200 = 80 + 120 + 80 = $280

* = amount u make if u win the hand and win 50% of the time

Now this decision is close, but you could often be greater than a 20% favorite to win this hand on average. I think often you will be closer to 25%, maybe even approaching 30% in the long run.

I hope that explains my reasoning.

[ddubois]

Chip EV wise, there is no doubt this is correct. You are getting better than 3:1 on your call. So if your opponents showed QQ and JJ this would be absolutely correct

Two middle pairs is something of an optimistic scenario however. It's basically making an assumption that hero's king will be good, sans a set. A more common scenario is something like middle pair and overcard, where hero needs to both hit his king and dodge an ace, which drops his chances down to like 20%.

If they showed you AK and 33. Ok, fine, maybe u can let it go, but this is a call everyday for me.

This is a dramatized scenario of dominated K and middle pair, but it needn't be quite so dramatic to have a similarly crushing effect on hero's win rate. With merely KT and 77 he would be down under 10%. Ax and Kx would put hero around 15% win rate.

As it turns out, hero was in much-better-than-average shape with both opponents being on Ax, 32%ish, but I'm not sure 730:2314 is +chipEV over the range of two all-ins.

[imcastleman]

Folding:
Two situations:
You have 730, button has 1968, and BB has 5302 left. Call this A.
You have 730, button has 0, and BB has 7270 left. Call this B.

Calling:
Four situations:
You have 0, button has 0, and BB has 8000 left. Call this C.
You have 0, button has 2352, and BB has 5648 left. Call this D.
You have 3044, button has 0, and BB has 4956 left. Call this E.
You have 692, button has 2352, and BB has 4956 left. Call this F.

Assumptions here:
P(1st|A) = .05
P(2nd|A) = .15
P(3rd|A) = .8
P(1st|B) = .15
P(2nd|B) = .85
P(3rd|B) = 0

P(A) = .5
P(B) = .5
P(1st) = P(A)*P(1|A) + P(B)*P(1|B) = .5*.05 + .5*.15 = .025+.075 = .1
P(2nd) = P(A)*P(2|A) + P(B)*P(2|B) = .5*.15 + .5*.85 = .075 + .425 = .5
P(3rd) = P(A)*P(3|A) + P(B)*P(3|B) = .5*.8 + .5*0 = .4


P(1st|C) = 0
P(2nd|C) = 1
P(3rd|C) = 0
P(1st|D) = 0
P(2nd|D) = 0
P(3rd|D) = 1
P(1st|E) = .45
P(2nd|E) = .55
P(3rd|E) = 0
P(1st|F) = .05
P(2nd|F) = .15
P(3rd|F) = .8

If each event is equally likely, then
P(C) = .25
P(D) =.25
P(E) =.25
P(F) = .25

P(1st) = P(C)*P(1|C) + P(D)*P(1|D) + P(E)*P(1|E) + P(F)*P(1|F) = .25*(0+0+.45+.05) = .25*.5 = .125
P(2nd) = P(C)*P(2|C) + P(D)*P(2|D) + P(E)*P(2|E) + P(F)*P(2|F) = .25*(1+0+.55+.15) = .25*1.7 = .425
P(3rd) = P(C)*P(3|C) + P(D)*P(3|D) + P(E)*P(3|E) + P(F)*P(3|F) = .25*(0+1+0+.8) = .25*1.8 = .45

1st pays: 5
2nd pays: 3
3rd pays: 2

With the given assumptions, then the expected values for folding and calling are as follows:

EV(Folding) = 2.8
EV(Calling) = 2.8.

Since I do not believe you are ahead of either the small blind nor the button, then I think you should fold this every time. The expected value on calling will only go up if you are in the lead of one or the other and the expected value of folding goes up if the small blind is ahead of the button which I also think is likely. So, I believe you must fold here.

[slogger]

It does, but when you consider that you no better than 40% likely to win even after winning that hand, then I think you see one of the reasons why I think the fold is correct (that, and I think SB will be better than a coinflip more often than the other way around).

Thanks for the follow-up!

[Jason Strasser]

I guess I will back off this a little. I make this call every time, but it is close. The "bad" advice I was referring to before was some of the justifications for folding which seemed absurd.

Goot post.

[fnord_too]

I like this analysis. I'm not totally convinced yet but...

There is a good possibility that both players have an ace only (at this stage a ace is huge), or one has a small pair and the other an ace. Against the two actual hands, I win 31.5% of the time, with the SS winning 22.9% of the time. With me not there the SS only wins like 29%, but if the cards are the other way around I cut 18% off his chance of survival.

Say Im facing A8o and 77, then 77 is a 57 to 43 favorite without me, with me it's A8 35.4%, K3 23.5%, 77 41.1%.

All this suggests that getting out of the way may be a lot better than I originally thought. This is a much richer problem than I originally thought, thanks for all the great posts!