[gulebjorn]

X-posted from Poker Theory because I couldn't get any answers there...

I was just in a NLHE tourney where KQo pushed all-in after three limpers. I was sitting in the small blind with AQs and folded. Two limpers called, holding 44 and 33. Now let's say I called as well. What would you guess that the different probabilities of winning are for all four hands?

To get you started, here are some head-up matchups:

cards %win
4s 4c 50.67
Ah Qh 48.85

3s 3c 50.02
Ah Qh 49.44

Qs Kd 23.75
Ah Qh 75.15

4s 4c 78.48
3d 3h 17.83

So far, this is nothing special and most of you know these % by heart.

Now, what happens if we put these hands in a 4-way pot? I'm sure everyone sees that AQs and 44 are the big favorites, and that KQ and 33 don't stand much of a chance.

Before I let twodimes calculate it, I would have guessed that the probabilties of these hands in a 4-way pot would look something like this:

AQs 38%
KQo 13%
44 42%
33 8%

This distribution would mean that AQ and 44 remain a coinflip, with the pair as a slight favorite, AQ vs KQ is still 3:1, and 44 vs 33 remains 5:1.

But the twodimes result says:

cards %win
4d 4h 32.02
3s 3c 17.09
Kd Qh 17.90
Ac Qc 31.84

As you can see, suddenly AQs no longer dominates KQo as badly as it does headsup, and the probability of 33 winning in a 4-way pot remains almost exactly the same as heads-up. Also, I would expect AQ to do worse, because one Q is gone, and the K interferes with some of the straight possibilities.

So, anyone have an explanation for these differences?

And more importantly, does anyone know of an easy way to guesstimate these probabilities preflop? Lets say that in a tournament, there are two all-ins before you. You are holding AK and you assume that you are up against a small pair and some suited big cards. Is there an easy way of calculating the chance of you winning the race, if we know the numbers for the matchup of the individual hands?

Thanks a lot.