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#1
10-23-2007, 06:01 PM
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Re: odds question
You are correct if we make the assumption that nobody would fold with a King in their hand. We could then say with certainty that there are only 39 unseen cards. My assumption that any of the five opponents will play any King actually has nothing to do with my calculation, now that I look at it.
1 - c(37,10)/c(39,10) = 1 - 348330136/635745396 =~ 0.452 I think that making the assumption that nobody would fold with a King is incorrect since there are many unplayable hands that include a King, even at 3/6. I understand your point though. It is certainly more likely that the hands that folded did not contain a King than hands that did contain a King. I have seen these points debated in the Probability forum and I'll see if i can dig some up. 
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#3
10-23-2007, 07:19 PM
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Re: odds question
[ QUOTE ]
You are correct if we make the assumption that nobody would fold with a King in their hand. We could then say with certainty that there are only 39 unseen cards. My assumption that any of the five opponents will play any King actually has nothing to do with my calculation, now that I look at it. 1 - c(37,10)/c(39,10) = 1 - 348330136/635745396 =~ 0.452 I think that making the assumption that nobody would fold with a King is incorrect since there are many unplayable hands that include a King, even at 3/6. I understand your point though. It is certainly more likely that the hands that folded did not contain a King than hands that did contain a King. I have seen these points debated in the Probability forum and I'll see if i can dig some up. [/ QUOTE ] I don't think we even need to make that assumption. The fact is that 16 cards were dealt out, each with an even chance of being a King with a maximum limit of two kings being dealt out in the 16 slots. From there, the events are NOT RANDOM. Folding is NOT a random event. I agree that some people would fold some hands with Kings in it. Further, I believe that people are making an analytical mistake when they say that the fewer limpers they face, the less likely it is that they will face a King. Inherent in that statement is the assumption that non-King hands will be folded at an equal rate to King hands, which clearly cannot be true. I submit the best we can say is that there is a 45% chance that at least one King got dealt out against you. From there, I don't believe stat geeks can truly help because events from that point forward in the problem presented are not random. I am interested in seeing if I am right or wrong in how I have structured the math in this issue.
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#4
10-23-2007, 07:50 PM
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Re: odds question
I seemed to have missed your main question.
[ QUOTE ] Ricks, assumining a 9 seat table, how does the fact that all 8 opponents were given two cards that could have been Kings impact the appropriate formula? [/ QUOTE ] If we are unaware of their cards and no one has acted, thus we make no assumptions, it has nothing to do with the formula except that we have to consider the number of cards dealt to them. After someone has acted you can make assumptions, which certainly supports your assertion that events are no longer random. I don't disagree with this. Edit: added "except that we have to consider the number of cards dealt to them."
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#5
10-23-2007, 08:28 PM
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Re: odds question
[ QUOTE ]
From there, I don't believe stat geeks can truly help because events from that point forward in the problem presented are not random. [/ QUOTE ] Perhaps binomial distribution . [img]/images/graemlins/smile.gif[/img]
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#6
10-23-2007, 09:18 PM
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Re: odds question
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#7
10-24-2007, 12:07 AM
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Re: odds question
[ QUOTE ]
Look at this page The absence of aces might be the same problem as the absence of Kings? [/ QUOTE ] It is because the we have the same number of each, 4. And I think the math can be altered to search for 2 cards in a deck of 47 post flop.
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