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#21
07-10-2005, 05:24 PM
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Re: QQ utg check raised by 2
[ QUOTE ]
[ QUOTE ] Because I'm trying to improve my hand. If someone else flopped a set of 3's then they have 2 of my outs to improve. I know I have top set, but I also know I am behind a flush (95%) and need a boat to win. [/ QUOTE ] This is an incorrect way to view your outs. You don't know what your opponents have. In fact, it can be proven that this doesn't even matter; your odds don't change at all. It's 4am and I'm in a mathematical mood, so I don't know if this post is going to be useful for anyone, but I feel like doing it anyway. So here goes. I'll use the more simple example of a flush draw. The flop comes with 2 [img]/images/graemlins/diamond.gif[/img]'s. You hold 2 [img]/images/graemlins/diamond.gif[/img]'s, giving you a flush draw. There are 2 people in the pot against you (for simplicity I'll limit it to two; the calculations get too long for more than that). There are 47 unseen cards, and 9 unaccounted for [img]/images/graemlins/diamond.gif[/img]'s. We say you have 9 outs. 9/47 = 0.191489 which is your probability of hitting a diamond on the turn. With 2 other people, there's a decent chance that someone has one of your [img]/images/graemlins/diamond.gif[/img] outs. Lets say we take this route, and assume that one of our opponents does indeed have a [img]/images/graemlins/diamond.gif[/img], and we discount our outs to 8. Once you do this, you are no longer dealing with a deck that has 47, or 46 unseen cards. That is, we don't just calculate 8/46 and that's our chance of hitting our diamond on the turn. We are considering the odds of all our opponents cards when we say that someone must have a diamond. So the new number of unseen cards, that is, the cards we are free to draw at, is 43. (2 opponents, 2 cards each). Here's the correct way to do the new calculation. The probability that an opponent holds 0 diamonds is 38/47*37/46 = 0.6503. We'll call this p0. p2 (someone holds 2 diamonds) = 9/47*8/46 = .0333 and p1 = 1 - p0 - p2 = 0.3163. Call m the probability of there being a certain number of total diamonds between our two opponents. If there are 0 diamonds between our opponents, there are still 9 diamonds left in our new 43 card deck, and the probability that neither opponent has one is equal to p0*p0. So, m0 = 9/43*p0*p0. (Neither opponent has a diamond). Similarly, m1 = 8/43*p0*p1 + 8/43*p1*p0. (Either player has a diamond, but the other does not). m2 = 7/43*p0*p2 + 7/43*p1*p1 + 7/43*p2*p0 (Either one player can have two diamonds, or both players have one). m3 = 6/43*p1*p2 + 6/43*p2*p1 (one player has one and the other has 2) m4 = 5/43*p2*p2 (both players have 2 diamonds) Now we have a list of probabilities. If we calculate m0+m1+m2+m3+m4, which is the total probability of all of these events, we get a probability of 0.191489. Which, as it turns out, is exactly the same as the initial probability with 47 unseen cards - 9/47. It doesn't matter if you think your opponent has X card in his hand. Your odds of drawing do not change, because he will only have X card in his hand with some probability. And by considering that someone has that card in their hand, you have to take your opponents cards out of the deck when calculating odds. Edit: Had p1 and p2 probabilities around the wrong way. [/ QUOTE ] Dude. I've always suspected that something like this was the case, that the probability was at least approximately the same if you take everybodies cards into account. This is fantastic. Good work. Just to ease my mind, does the probability stay the same regardless of the amount of people in the hand? I mean, there's another person to hold a diamond or 2, but now we are doing all the calculations based on 41 unseen cards so it stays about the same, Right? 
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#23
07-10-2005, 10:36 PM
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Re: QQ utg check raised by 2
[ QUOTE ]
Just to ease my mind, does the probability stay the same regardless of the amount of people in the hand? I mean, there's another person to hold a diamond or 2, but now we are doing all the calculations based on 41 unseen cards so it stays about the same, Right? [/ QUOTE ] The probabilities stay exactly the same. If we consider three opponents, there are 6 cards to consider, and so there is a maximum of 6 diamonds that could be held by our opponents. The equations look like this (this time I'm factorising the equations with the numeric value out front): m0 = 9/41*p0*p0*p0 m1 = 8/41*(p0*p0*p1 + p0*p1*p0 + p1*p0*p0) m2 = 7/41*(p0*p1*p1 + p0*p0*p2 + p0*p2*p0 + p1*p0*p1 + p1*p1*p0 + p2*p0*p0) m3 = 6/41*(p0*p1*p2 + p0*p2*p1 + p1*p0*p2 + p1*p2*p0 + p2*p0*p1 + p2*p1* p0 + p1*p1*p1) m4 = 5/41*(p0*p2*p2 + p1*p1*p2 + p1*p2*p1 + p2*p0*p2 + p2*p1*p1 + p2*p2*p0) m5 = 4/41*(p1*p2*p2 + p2*p1*p2 + p2*p2*p1) m6 = 3/41*(p2*p2*p2) Summing these again gives the same probability of 0.191489. So it doesn't matter how many opponents you include when doing this. It's good that it works out like this, because it's much easier to just calculate 9/47. [img]/images/graemlins/tongue.gif[/img]
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#26
07-11-2005, 01:11 AM
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Re: QQ utg check raised by 2
I would Cap it. If the villains fire again on the river I'll call down, otherwise I bet.
_____________________________ Looking at some of your posts, am I the crazy one? I see people over play these very frequently, and you might boat up, and your qqqueens are going to hold up sometimes too.
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#30
07-11-2005, 01:26 AM
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Re: QQ utg check raised by 2
[ QUOTE ]
Here I'll say that giving credit for a flush isn't yet justified. We play like we have the best hand until we learn otherwise. [/ QUOTE ] So you don't think getting check raised by SB and BB is a clue I am behind? I'm not trying to argue or say you are crazy, but that was the first cap reply I've gotten.
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