#1
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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?
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#2
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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% |
#3
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Re: How Do I Calculate This?
So 41% of the time, I will win the blinds uncontested, correct?
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#4
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Re: How Do I Calculate This?
Yes.
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#5
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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. |
#6
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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. |
#7
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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.
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#8
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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. |
#9
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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. |
#10
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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? |
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