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#2
09-09-2005, 09:58 AM
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Re: Odds of hitting set with all 5 cards, not just flop.
[ QUOTE ]
LetYouDown, Thanks for the percentage. Can you please explain the formula? 1 - C(48,5)/C(50,5) = 19.18% By "explain", I just mean how do I enter that into a caluclator to get the same figure? [/ QUOTE ] If your calculator has a PRB or a nCr key then just use that and enter in the two variables. Otherwise... C(N,r) = N!/r!(N-r)! Im assuming you got at least a factorial button.. otherwise you are gonna be really doing this the loooooong way. [img]/images/graemlins/grin.gif[/img] 
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#3
09-09-2005, 08:26 PM
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Re: Odds of hitting set with all 5 cards, not just flop.
[ QUOTE ]
If your calculator has a PRB or a nCr key then just use that and enter in the two variables. Otherwise... C(N,r) = N!/r!(N-r)! Im assuming you got at least a factorial button.. otherwise you are gonna be really doing this the loooooong way. [/ QUOTE ] 1 - (45*44) / (49*50) is quite within reach of calculators without factorial buttons. In fact, 45x44 = 1980; 50x49 = 2450; 1 - 1980/2450 = 1 - 198/245 = 47/245 is quite doable with no calculator at all. As the other posters have mentioned, lots of things cancel. 48C5 = 48*47*46*45*44/(5*4*3*2*1) and 50C5 = 50*49*48*47*46/(5*4*3*2*1), and when you divide 48C5 by 50C5, the 5*4*3*2*1, the 48, the 47, and 46 all go away. Always pays to think before you start pushing buttons on a calculator.
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#4
09-09-2005, 11:29 PM
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Re: Odds of hitting set with all 5 cards, not just flop.
[ QUOTE ]
[ QUOTE ] If your calculator has a PRB or a nCr key then just use that and enter in the two variables. Otherwise... C(N,r) = N!/r!(N-r)! Im assuming you got at least a factorial button.. otherwise you are gonna be really doing this the loooooong way. [/ QUOTE ] 1 - (45*44) / (49*50) is quite within reach of calculators without factorial buttons. In fact, 45x44 = 1980; 50x49 = 2450; 1 - 1980/2450 = 1 - 198/245 = 47/245 is quite doable with no calculator at all. As the other posters have mentioned, lots of things cancel. 48C5 = 48*47*46*45*44/(5*4*3*2*1) and 50C5 = 50*49*48*47*46/(5*4*3*2*1), and when you divide 48C5 by 50C5, the 5*4*3*2*1, the 48, the 47, and 46 all go away. Always pays to think before you start pushing buttons on a calculator. [/ QUOTE ] What you just said is the "long method." Luckily lots of the bigger numbers cancel out in something like C(48,5) but you wouldnt be so lucky if it was like C(52,24) and you had to do it by hand. In the end, imo its a lot easier to do my method instead of doing a combination divided by a combination. Just dont be dumb and make simple mistakes like I did. [img]/images/graemlins/frown.gif[/img] (48/50)(47/49)(46/48)(45/47)(44/46) = 0.808 1 - 0.808 = 0.192 = 19.2%
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#5
09-10-2005, 11:28 AM
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Re: Odds of hitting set with all 5 cards, not just flop.
Hello, I am a newbie, and I was wondering what the variables in the equation below represent. I assume that r = the number of cards to come, and C(N,r) is the probablitiy of catching your set. What does N represent?
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#6
09-10-2005, 03:01 PM
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Re: Odds of hitting set with all 5 cards, not just flop.
[ QUOTE ]
Hello, I am a newbie, and I was wondering what the variables in the equation below represent. I assume that r = the number of cards to come, and C(N,r) is the probablitiy of catching your set. What does N represent? [/ QUOTE ] C(N,r) The C is just to tell you that this is a "combination" problem; this solves problems where order is NOT important and you want to find how many different ways "r" can come out from a sampling of "N." If you go back to LetYouDown's answer, he finds out what C(50,5) is. This means he is calculating how many different ways 5cards can uniquely come out from a sampling of 50cards. (Because OP already is holding two pocket cards) Btw, C(50,5) = 50!/45!5! = 2,118,760
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#7
09-09-2005, 10:01 AM
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Re: Odds of hitting set with all 5 cards, not just flop.
It would probably be easier with excel or google if you have those tools available. I've illustrated combinations below, sorry if it's vague. If you have trouble following any of it, let me know. Luzion's math is fine, he just had the wrong numbers the first time. You can use that approach if all you have is a calculator. Most scientific calculators have Combinations and Permutations.
C(N,R) = N!/(N-R)! * R! ...so: C(48,5) = 48!/(48-5)! * 5! X! = X * (X - 1) * (X - 2) * ... * 1 5! = 5 * 4 * 3 * 2 * 1 48!/43! has a lot of repeating numbers, so it's just (48 * 47 * 46 * 45 * 44)/(5 * 4 * 3 * 2 * 1). Hope this helps
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#8
09-09-2005, 10:07 AM
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Re: Odds of hitting set with all 5 cards, not just flop.
I just have the calculator on my computer, it doesn't have Combinations and Permutations.
I have been wondering the same thing about sets by the river and other variables, and wanted to know how to go about it on my own.
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#9
09-09-2005, 10:22 AM
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Re: Odds of hitting set with all 5 cards, not just flop.
Is google an option?
1 - 48 choose 5/50 choose 5 in google will give you the answer.
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