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.

Luzion · #7 03-20-2006, 07:03 PM

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.

EasilyFound · #8 03-20-2006, 07:10 PM

[ 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.

AaronBrown · #9 03-20-2006, 11:04 PM

[ 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.

EasilyFound · #10 03-20-2006, 11:59 PM

[ 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?