How Do I Calculate This?

EasilyFound · #1 03-20-2006, 09:00 AM

This is probably simple for math wizzes, but I'm not one. In a four-handed game, I'm playing against three players who will only play the top 20% of hold'em hands. If I push, what is the chance that one player will call?

drbst · #2 03-20-2006, 09:11 AM

I think you mean the probability that not all will fold, right?

If we assume that the individual folding probabilities are independend -- they are not, but just because the calculation is much easier -- the probability is 1 - 0.8^4 = 59.04%

EasilyFound · #3 03-20-2006, 09:18 AM

So 41% of the time, I will win the blinds uncontested, correct?

drbst · #4 03-20-2006, 09:34 AM

Yes.

Dromar · #5 03-20-2006, 01:55 PM

I think it would actually be 1 - .8^3. He's one of the four people at the table.

1 - .8^3 = .488, so 48.8%, you'll be called. 51.2% you'll win uncontested.

EasilyFound · #6 03-20-2006, 06:32 PM

[ QUOTE ]
I think it would actually be 1 - .8^3. He's one of the four people at the table.

1 - .8^3 = .488, so 48.8%, you'll be called. 51.2% you'll win uncontested.

[/ QUOTE ]

Is this the correct way? Not being a mathematician, I didn't know whether it should be 1 - .8^3 or 1 - .8^4. I don't mean to doubt any of the posters above, but can someone confirm that is should be ^3 instead of ^4. That makes sense to me, but I don't know who is correct.