UPF - Odds

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Odds, The Fish, 23. Jan 2003 14:12
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Hello there! I was just wondering if someone with knowledge of odds could go over a few things I have been working out. Sorry to bring up the dry topic of probability.....but I am taking a little break from playing at the table to study opening hands....anyway .....here are some hands and probabilities.......if anyone knows these to be incorrect please correct me! Or if they are correct plz let me know. 1). Flopping 3 specific ranks of cards. (e.g./ holding A 2) Chances of flopping 3 4 5: 444 / 52C3 = 0.2896% Side: Say you have 4 5 in the hole, thus you can flop a straight 4 ways: (A 2 3), (2 3 6), (3 6 7), (6 7 8) is it correct to say that your probability of flopping a straight is 0.28964 OR about 1.1584% ? In other words I added the probability of each separate event. Is this correct? 2. Holding two suited cards (say A K). Probability of flopping a four-flush: (11C2 39) / 52C2 = 9.7% I am more interested that the process is right, if I made some small calculation problem, it would be nice to know, but far more important is the process. So anyone who knows if my process is correct or incorrect please let me know. Thanks a lot! Ben
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Re: Odds, Louis, 23. Jan 2003 18:25
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im afraid your method is wrong the real way is much more complex but ill try to explain it. there are 11 to give you a flush when you hold two suited cards and 50 cards you havent seen and 5 cards to come. this problem is known as hypergeometric distribution and to work it out you need to know that 5! means 12345 and so also 4! means 1234 and also that M(nCr)R means M!/((M-R)!R! for example 4(nCr)2 = 4!/(2!2!) which means (1234)/((12)(12)) which equals 6 and also N(nCr)0 allways equals 1. Now to work out the probability you must use this formula ((D(nCr)x)((N-D)(nCr)(n-x)))/(N(nCr)n) where in this case D is the number of cards that can help you N is the total number of cards n is the number of cards to come and x is the nuber of cards you need so the formula would now look like this D=11 N=50 n=5 and x=3 ((11(nCr)3)(50-11(nCr)5-3))/(50(nCr)5) which becomes 122265/2118760 BUT we must also remember that if 4 or 5 cards of our suit fall then we also make our flush so to work the proper probability out you must substitute x=4 into the formula and add that to the value for when x=3 (122265/2118760) and add the value for when you substitute x=5 so the final equation looks like this (122265/2118760)+((11(nCr)4)(39(nCr)1))/(50(nCr)5) + ((11(nCr)5)*1)/(50(nCr)5) which ends up as (122265+12870+462)/(2118760) which equals .064 or 6.4 percent
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Re: Odds, The Fish, 23. Jan 2003 21:07
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Thank you for your reply. Well it seems like you know a thing or two about probability. But I believe you are patronizing me a bit. Notice that 52C3 is a combination ........you have it written as M(nCr)R (this is what I meant, unfortunately my notation is not as proper as yours. So when I write 52C3, what I mean is 52 = M and that R = 3. You might write it as: 52(nCr)3. If you would prefer this in factorial form it becomes: 52! / [(52-3)!3!] which reduces to 525150/321 = 525150 / 6. Also, there are actually 3 cards to come because notice I was asking what are the chances of "flopping" (getting these specific cards on the flop, the first three community cards). To put it more simply, the probability for ANY event is: # of desired outcomes / total # of outcomes possible I am not trying to be a smart ass, but at the same time I don't want you to think I am an idiot. I should have been more clear though. Thank you for taking the time to explain though. Your post was helpful to me. But my question remains: someone please check my work! Louis I believe you would have been right if I had asked what is the probability of me ending up with these cards after all 5 community cards are out, but right now I am interested in the flop alone. And I was wrong about the denominator, thank u! It should be 50C3 or 50(nCr)3 (in ur case) as you have it. So I believe I would need 11(nCr)2. Because I also asked for the probability that I flop a four-flush, not a flush. Again, thank you for your post! It was very helpful. And I am not trying to be a smart ass, I have said before, I am just trying to clarify some things, which I should have done b4! Here are my questions re-stated: 1). Flopping 3 specific ranks of cards. (e.g./ holding A 2) Chances of flopping 3 4 5: 444 / 50C3 OR (444) / 50(nCr)3 Side: Say you have 4 5 in the hole, thus you can flop a straight 4 ways: (A 2 3), (2 3 6), (3 6 7), (6 7 8) is it correct to say that your probability of flopping a straight is 0.28964 OR about 1.1584% ? In other words I added the probability of each separate event. Is this correct? 2. Holding two suited cards (say A K). Probability of flopping a four-flush: (11C2 * 39) / 50C3 OR [11(nCr) 2 * 39] / [50(nCr)3] thanks again, Ben
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Re: Odds, Mano, 24. Jan 2003 14:41
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1). Flopping 3 specific ranks of cards. (e.g./ holding A 2) Chances of flopping 3 4 5: An easy way to do this would be to look at it one card at a time. For the first card there are 12 (4 of each rank) remaining cards out of the 50 left that help you. For the second card there would be 8 (4 of each rank excluding the one that flopped on first card) of remaining 49, and for the last card 4 of remaining 48, so the total probability would be: (12/50)(8/49)(4/48) = .003265 or 3.265% Side: Say you have 4 5 in the hole, thus you can flop a straight 4 ways: Each of the different flops are mutually exculsive (flop these exact 3 ranks) so you could just multiply above probability by 4. > 2. Holding two suited cards (say A K). Probability of flopping a four-flush: You can either use an approach as I did above, or a combinatorial method as you attempted. You just need to be careful to cover all possible outcomes exactly once. I will calculate this for you using two mehtods (I am assuming you want the probability of a four flush only, and do not want flopped flushes included). Mehod 1 (combinatoric method - does not pay attention to order): Using your notation there are 50C3 possible 3 card combinations on flop. There are 11C2 different ways to pick 2 of your suit, with 39 possibilites for the remaining card, so the total probability should be: 39(11C2)/(50C3) = .1094 or 10.94% Method 2: There are 3 ways of flopping 2 of a suit - first 2 cards that suit then another card, first card of the suit followed by unsuited card followed by suited card, and unsuited card followed by two suited cards. If we add up the probability of all 3 of these occurences we have the total probability of flopping 2 of our suit. For flopping first 2 cards suited and last card unsuited the probability would be: (11/50)(10/49)(39/48) = .03648 For flopping suited, unsuited, suited: (11/50)(39/49)(10/48) = .03648 And similarly for unsuited followed by 2 suited cards: (39/50)(11/49)*10/48) = .03648 Adding these 3 results (which, by the way are all identical - just shuffling order on numerators) we get total probability of .1094, identical to the results calculated using combinatorics. Sorry to be so long winded, but I hope this helps with how to set these type of problems up.
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Re: Odds, Mano, 24. Jan 2003 14:48
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The reason your numbers are coming up wrong is that you are using the wrong numbers for combinations on the flop. In both instances there are 50 cards left of which 3 will be flopped, so you need to use 50C3 instead of the 52C3 you used in your first example and the 52C2 you used in your second example. Other than that, your equations would have worked.
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