#1
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face pairs?
if i have pocket JJ-KK, what are the odds that someone has a higher pocket pair?
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#2
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Re: face pairs?
Let's assume a full 10 player table for the following calculations:
For KK atleast one person will have a higher pair = 9 * 6/1225 - (9c2 / 50c4) = .043925 For QQ things get more difficult. Let's use a different method: When you hold 2 cards, how many ways are there to distribute the other 9 hands? D = 50c18 * 17!! = 622114224555779000000 (excel rounds off) The problem becomes how many of these will contain a higher pair? Sometimes there will be 4 higher pocket pairs... p(iiii) = 3 * 3 * 42c10 * 9!! = 12514622485365 Sometimes there 3 pocket pairs... p(iii)= 2 * 3 * 6 * 44c12 * 11! = 7892555247436860 Now 2 pocket pairs... more complications... the pairs could be of the same rank or different ranks... different rank... p2(ii) = 2 * 3 * 6 * 46c14 * 13!! = 1166970668728160000 same rank... p1(ii) = 2 * 3 * 46c14 * 13!! = 194495111454694000 Now 1 pocket pair... p(i) = 2 * 6 * 48c16 * 15!! = 54847621430223700000 Now what to do with these numbers? There will be exactly two rank against us R2 = p2(ii) - p(iii) + p(iiii) = 1159140686593150000 and exactly one rank R1 = p(i) - p1(ii) - 2*R2 = 52334844945582700000 so with QQ the odds are (R1 + R2) / D = .085987402 For JJ there are even more possibilities for combinations of hands that will contain a higher pocket pair, so because I don't like typing... Just the answer = .127125134 edited for typos |
#3
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Re: face pairs?
[ QUOTE ]
When you hold 2 cards, how many ways are there to distribute the other 9 hands? D = 50c18 * 17!! = 622114224555779000000 (excel rounds off) [/ QUOTE ] Not sure this is right. Don't you have to use multi-nomial here? 50!/((2!)^9 * 32!) I might be misunderstanding your approach, but I think that's right. |
#4
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Re: face pairs?
ok... 50c18 is how to choose 18 cards to go into the remaining 9 hands
17!! is the number of ways to arrange those 18 cards into 9 hands of 2 17!! = 17*15*13*11*9*7*5*3*1 arranging 6 cards into 3 hands... 5!! ways = 5*3*1 = 15 arranging 8 cards into 4 hands... 7!! = 7*5*3*1 = 105 etc |
#5
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Re: face pairs?
hmmmm....
this gives a different answer from multi-nomial, tho. why would that be? i think there is something not right about this method, but i havn't really thought about it. i'm fairly sure multi-nomial is correct, tho. how sure are you about your method? why would multi-nomial not work here? |
#6
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Re: face pairs?
Are you thinking of the general formula?
n! / ((n/2)! * 2^(n/2)) First we choose the 18 cards from 50... then apply the formula 18! / (9! * 2^9) = 34459425 this is the same as 17!! = 17*15*13*11*9*7*5*3*1 = 34459425 This mumber must be multiplied by 50c18 |
#7
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Re: face pairs?
[ QUOTE ]
Are you thinking of the general formula? n! / ((n/2)! * 2^(n/2)) First we choose the 18 cards from 50... then apply the formula 18! / (9! * 2^9) = 34459425 this is the same as 17!! = 17*15*13*11*9*7*5*3*1 = 34459425 This mumber must be multiplied by 50c18 [/ QUOTE ] No. I am thinking of the multi-nomial formula, which I recall as being the way to choose groups if size n1,n2,...nr from a total of N things, where n1+n2+...nr=N. The formula is: N!/n1!*n2!*...nr! You can also think of it like this. First choose 2 from 52. Now choose 2 more from the remaining 50. Now choose 2 more from the remaining 48, etc... (52*51/2) * (50*49/2) * .... This will simplify in this case to the multi-nomial forumula above. So I'm still pretty sure it is correct. Do you still see some error?? |
#8
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Re: face pairs?
[ QUOTE ]
This will simplify in this case to the multi-nomial forumula above. So I'm still pretty sure it is correct. Do you still see some error?? [/ QUOTE ] Mickey, pzhon pointed out that I am calculating ordered sets of hands, while you are calculating unordered sets. So your method is correct. Using multi-nomial could work too, but then we'd have to do all the other calcs on ordered sets as well. gm |
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