How Do I Calculate This?
 

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?

Re: How Do I Calculate This?
 

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%

Re: How Do I Calculate This?
 

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

Re: How Do I Calculate This?
 

Yes.

Re: How Do I Calculate This?
 

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.

Re: How Do I Calculate This?
 

[ 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.

Re: How Do I Calculate This?
 

It would be 1 - .8^3 because you are playing against 3 players. And like drbst said, this is not really correct since they are not independent events. Also I want to point out that this answer is for at LEAST one player calling you.

Re: How Do I Calculate This?
 

[ QUOTE ]
It would be 1 - .8^3 because you are playing against 3 players. And like drbst said, this is not really correct since they are not independent events. Also I want to point out that this answer is for at LEAST one player calling you.

[/ QUOTE ]

Thanks. And I understand the last part, at least one player calling.

How much differentiation would you expect, in percent, between the two answers, since they are not independent events? Just so I have a rough idea.

Re: How Do I Calculate This?
 

[ QUOTE ]
How much differentiation would you expect, in percent, between the two answers, since they are not independent events? Just so I have a rough idea.

[/ QUOTE ]
In this case, you expect negative correlation. The maximum percentage of time you can be called is 60%, that's if no more than one player ever calls you. I would expect the actual answer to be between 49% and 60%, say 55%.

The other extreme, if the calls are completely positively correlated, is you could be called only 20% of the time, but always by all three other players.

Re: How Do I Calculate This?
 

[ QUOTE ]
[ QUOTE ]
How much differentiation would you expect, in percent, between the two answers, since they are not independent events? Just so I have a rough idea.

[/ QUOTE ]
In this case, you expect negative correlation. The maximum percentage of time you can be called is 60%, that's if no more than one player ever calls you. I would expect the actual answer to be between 49% and 60%, say 55%.

The other extreme, if the calls are completely positively correlated, is you could be called only 20% of the time, but always by all three other players.

[/ QUOTE ]

I'm a little confused. Sorry. Let me say what I understand. 1- .8^3=.488. This means that 48.8% of the time, nobody will call. Or that 51.2% of the time, at least one person will call.

As someone said, since these are not independent events, this answer is a little off. And you are saying it can be as much as 9% off? (51.2% up to 60%?) But more likely somewhere in the middle?

Is that how I should understand your answer?

Re: How Do I Calculate This?
 

If the differentiation is that large, am I able to reliably figure out the EV of pushing a certain hand into a set number of players and putting them on a range of hands, such as top 20%? Or does the fact that the probabilities are not independent mean that I cannot get a reliable enough answer to the question of what percentage of time everybody will fold given that calling range?

Re: How Do I Calculate This?
 

I see where the 80% comes from, but why do you cube it. I know on the surface you cube it because there are 3 players, but why?

Re: How Do I Calculate This?
 

[ QUOTE ]
I see where the 80% comes from, but why do you cube it. I know on the surface you cube it because there are 3 players, but why?

[/ QUOTE ]

You are interested in the probability that each player does not have a top 20% hand. If these are independent events, each person will fold 80% of the time. Probability of all three folding is .8^3. So the probability of getting at least one caller is 1-.8^3.

Now, if you want to think about these as dependent it depends critically on the number of the possible starting hands and what those hands include. This is a much harder problem than it seems because every time someone folds you assume they don't have a good hand, but that then throws out two cards that could be picked. Just assume that the first guy has real trash like 72o. That subtracts two bad cards. Then there are ~2*51 fewer bad hands to pick from. There are a couple of overlaps, but that's not big deal. Let's say those were all bad hands that could have been had. Again, not true (A7 and 77 are probably top 20%), but this will overestimate the calling percentage a little so its fine. The next guy is now picking from 51*50 cards, but he likes more than 20% of those. His percentage is 52/250 (that's .2*52/50). If his hand is crap, then the next guy is missing ~4*51 hands (again similar issues). His percentage is 52/240 (that's .2*52/48). So that gives us a percentage of 49.6% of the time all three fold. So that's pretty close to break even.