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Server Time: 12/1/2008 12:06:36 PM PACIFIC |
 The contradiction in Poker odds., Blade, 10. Oct 2003 19:10 | ||
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| Stay with me here but today I gave a lot of thought to how odds of making hands in poker are calculated. Specifically the Flush. The way we now do it is take the number of known desired suited cards subtracted from the total desired suited cards divided by the number of unknown cards. For example I have A8spades and the 5 10 are on the flop. So there are 9 remaining out of 47 cards or 9/47 = 1-5.2. Note though that this has a built in assumption that all 9 spades are available. However using the same mathematical principle involved in calculating these odds should require us to include the amount of spades that mathematically are not available. so in a ten handed game there are twenty cards dealt. there is a 1 in 4 chance of getting a spade. so we should assume that 5 spades will be dealt. we already know that we have 2 of them which leaves 3 that are taken. Notice that in order for us to be able to use the principle of odds & pott odds we have rely on the premise that what is most correct and therefore most profitable is that which is most mathematically likely. It is therefore incorrect to ignore that probability that 3 of our desired suit are in our opponents hands. Lets return to our orignal problem preflop with 2 spades there are 3 more assumed taken. 2 on the flop which leaves 6/44 or 1 in 7. or and increase of 40% which is significant. The effects of which are undetermined but playing hands below A10s which does not have the potential str8 draw or much EV from top pair alone could become an unprofitable play. This is strictly theory and I expect and would hope that it would be intensely challenged. | ||
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| Re: The contradiction in Poker odds., TrippH, 10. Oct 2003 20:05 | ||
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| dude, you're saying 6/44, taking away the spades you're giving to the other hands at the table from the numerator, but not taking away all the other cards they have from the denominator ... if you're going to bring their hands into it there aren't 44 cards left, there are only 29 after the flop | ||
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| Re: The contradiction in Poker odds., Blade, 10. Oct 2003 20:08 | ||
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| I am only taking away the 3 cards assumed to be the desired suit. Your point about subtarcting the remaining 7 cards may be valid. This I am uncertain.......Dude | ||
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| Re: The contradiction in Poker odds., Big_Loser1, 10. Oct 2003 20:20 | ||
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| TrippH is correct. If there are 20 cards out plus 3 on the board. You have two suited and one on the board. If you assume that out of the other 18 cards out 3 are of the same suit, then you naturally have to assume that the other 15 are not of that suit. That would leave 29 cards left and assuming that there are 7 of your suit left in the 29. I don't know if I like this theory, you are making to many assumptions. But if it were to be used you have to assume the other cards in play are not the suit you are looking for. | ||
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| Re: The contradiction in Poker odds., Blade, 10. Oct 2003 20:33 | ||
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| ahhhh. good point | ||
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| Re: The contradiction in Poker odds., Angel, 10. Oct 2003 20:39 | ||
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| Fermat and Pascal were kind enough in coming to a gamblers aid back in 1654 and create the mathematical theory of probability. I suspect that a short search on the net will lead to a really good basic intro of the work they did and how it will apply to the flush problem. While it's great Blade that you're willing to think outside the box and question established thoughts on a matter (and I mean that - no PC feel good BS here) you are wrong on this one. TrippH made a good observation but while it gets the answer closer - it can't quite be done like he suggested either. Your line of thinking on this is more like the idea that if I flip this coin 4 times and it comes up heads all four times - then if John Doe flipped a coin in the other room 4 times - we can assume it came up tails 4 times all four times because that would make it 50:50. Clearly this isn't so - and it uses a similar line of thinking. | ||
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| Re: The contradiction in Poker odds., TrippH, 10. Oct 2003 20:42 | ||
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| to be clear, I wasn't suggesting it be done that way either ... I have a pretty good grasp of probability theory ... I was merely trying to illustrate the error in his approach | ||
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| Re: The contradiction in Poker odds., TrippH, 10. Oct 2003 20:24 | ||
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| that's not what I'm talking about ... you used 6/44 to determine the odds for a flush to come on the turn after you subtracted the 3 theoretical spades from the opponents' hands ... if you do that though, you have to subtract all their cards from the unseen cards, dropping 44 down to 29, in which case you end up with roughly the same percentage | ||
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| Re: The contradiction in Poker odds., Snorbolus, 11. Oct 2003 08:01 | ||
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| Blade, I think that I can explain this to you in an intuitive way. It is because you are not asking: where are the remaining flush cards? You don't care about that. You are asking: will one of the remaining board cards complete my flush? This is a very different question. Think of it like this. If after the flop the dealer was to take 10 cards off the bottom of the deck and throw them away, would that change your chance of making your flush? No it would not. Those cards were not going to be dealt to the board anyway. The probability that you are interested in is the chance that either or both of cards 2 and 4 on the top of the deck (burn card =1, turn card =2, burn=3 river=4), are flush cards. That probability is set when the cards are shuffled. If the flush cards are not there it doesn't matter to you where else they are. Snorbolus | ||
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| Re: The contradiction in Poker odds., PairTheBoard, 11. Oct 2003 12:11 | ||
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| Best Answer | ||
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| Re: The contradiction in Poker odds., Blade, 11. Oct 2003 13:17 | ||
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| Great answer. Your right my line of thinking was off and intuitively this does make sense. Thank you and hopefully my post will incorrect did atleast provoke thought in an otherwise taken for granted area. | ||
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| Re: The contradiction in Poker odds., Bungus, 12. Oct 2003 16:42 | ||
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| I posed this question in a post several weeks ago. No one replied much, back then, but Snorbolus did a reat job debunking that theory. If you're going to assume there are 3 suits of your flush in other peoples hands, TrippH would be right in saying the rest are not of that suit, and have to be subtracted from 49 know cards. In either method, the probability of you completing your flush is within %2 of each other. Good thread | ||
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| Re: The contradiction in Poker odds., pt_Gatsby, 15. Oct 2003 12:55 | ||
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| This was more or less covered already, but I thought I would throw in my two bits... The way we now do it is take the number of known desired suited cards subtracted from the total desired suited cards divided by the number of unknown cards. That is exactly it... For example I have A8spades and the 5 10 are on the flop. So there are 9 remaining out of 47 cards or 9/47 = 1-5.2. Note though that this has a built in assumption that all 9 spades are available. Actually, that's not what it says. It says that there are 9 spades in 44 cards that are unknown. Beyond that becomes conjecture - or a model of something else... Lemme explain: so in a ten handed game there are twenty cards dealt. there is a 1 in 4 chance of getting a spade. so we should assume that 5 spades will be dealt. we already know that we have 2 of them which leaves 3 that are taken. Lets look at the probabilities at each step of the way: 13/52 cards - and to get double suited, 13/52*12/51. Lets assume that actually happens (using 'concrete information' to pair down the tree)... That leaves 11 cards in 50 for the suit. Assuming 'even' distribution between the players a certain number of spades is delt out - in exact proportion to how many there are in the deck (still 11/50). This would translate to: ~7/30 cards... and to be percise, it should be directly proportional to 11/50, or: 11/50 = 6.6/30. In other words - because the cards are evenly distributed in the probability theory, you can't assume any extra information, and therefore the probability of them having the spades is exactly the same as the deck holding them. Your odds cannot change on a projection that uses the same distribution information... Though, if you really want to really get a headache think how much of an advantage do you have if you get to look at the top ten cards of the deck? I still haven't figured out how to work that out... Came up in an arguement regarding collusion - I still maintain the difference in knowledge is minimal and the problem is betting advantage, but I can't find a way to express that... oh well. Probably better we have something to argue about rather than trying to find something new, heh. | ||
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