frequency of J+ pocket pairs |
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Posted Thu May 26, 2005 4:47 pm GMT by ixroby
Pre-flop, what is the probability that any one or more of my nine opponents holds a pocket pair, jacks or better, when I don't hold such a pair?
This should be easy but I'm getting contradictory answers with different methods.
Given 24 such pairs and 1225 possible hands, and working backwards from all nine opponents holding such a pair, to eight holding such a pair, etc., I get
C(24,9)+
C(24,8)*1200+
C(24,7)*C(1200,2)+
C(24,6)*C(1200,3)+
C(24,5)*C(1200,4)+
C(24,4)*C(1200,5)+
C(24,3)*C(1200,6)+
C(24,2)*C(1200,7)+
24*C(1200,8)
which yields 16.25%
However, when I figure it arithmetically, I get 23.25%. (Jacks or better coming up one out of every four hands seems counter-intuitive.)
I.e.,
24 / C(50,2) = 1.96% +
24 / C(482) * 98.04% = 2.09% +
24 / C(462) * 95.91% = 2.22% +
24 / C(442) * 93.59% = 2.37% +
24 / C(422) * 91.06% = 2.54% +
24 / C(402) * 88.27% = 2.72% +
24 / C(382) * 85.19% = 2.91% +
24 / C(362) * 81.78% = 3.12% +
24 / C(342) * 77.97% = 3.34%
To make things worse, I ran a simulation in Excel with VBA, and it gave me 23%.
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Posted Thu May 26, 2005 4:50 pm GMT by ixroby
The character combination eight right-parenthesis is displaying as a smiley for some reason.
Posted Mon May 30, 2005 9:04 pm GMT by suitedaces84
An approximation of this is...
-probability of oppoent getting one jack or better 16/50
then completing the pocket pair 3/49
-the probability of one oppoent getting a pocket pair of jacks or better is .02 or so.
-the probability of none of your oppoents getting JJ or better is .98^9 = .833 or 83.3%
-the probability of one or more of your opponets getting JJ or better is about 16.7%
-of course this # is only an approximation, because one player's cards have some dependance on cards that have already been dealt.
Given your cards there are (50c2*48c2....34c2) different combinations of cards your nine oppoents may have.
Given you have two cards of T or worse there are (24*9)(48c2*46c2...34c2) ways for one or more of your oppoents to be holding pocket jacks or better.
Divide the bottom # by the top # and you should get .176 or 17.6%.
For anyone trying to figure out what's going on here.
X! = X*(X-1)*(X-2)...4*3*2*1
XcY = X!/(Y!)(X-Y)!
The significance of this is it allows one to determine how many combinations are possible. If you pick 8 cards out of a full deck there 52c8 possible combinations of cards you could pick out (assuming order does not matter).
Posted Tue May 31, 2005 10:03 am GMT by AAceman
.....why?
Posted Tue May 31, 2005 11:24 am GMT by suitedaces84
Given your cards your first opppoent has 50c2 different hands he could be dealt. Given your cards and your oppoent's cards there are 48c2 different hands your second oppoent could be dealt...and 34c2 different hands your ninth oppoent could be dealt. Multipling these numbers gets you the total number of hands your oppoents can be dealt.
There are 6 ways to get any pocket pair (sc, sh, ch, dc, sd, dh). And there are 4 pocket pairs of J or better. So there are 24 ways to get a pocket pair of J or better. There are 24 ways for your first oppoent to get pocket jacks or better. If your first oppoent gets pocket jacks or better there are 48c2 different hands your second oppoent could have...and 34c2 different combinations your ninth oppoent could have. And each one of the nine could get pocket jacks or better so this # must be multiplied by nine.
So the final answer is...
(9*24)(48c2*46c2...34c2)/50c2*(48c2*46c2...34c2)
the (48c2*46c2...34c2) cancels out (because it's in both the top and the bottom)
This leaves...
9*24/50c2
Which is...
9*24/(50!)/(48!*2!)
50!/48! = (50*49*48...*3*2*1)/(48...3*2*1)
the (48...3*2*1) cancel out and 50!/48! = 50*49
and 2! = 2*1 = 2
50c2 = 50*49/2
9*24/50*49/2 = .176 or 17.6%
Posted Mon Jun 06, 2005 1:37 am GMT by gamble44pro
You don't hold such pairs, but the odds are depending on the number of jacks, queens, kings or aces you hold.
Assuming you hold none of them, the probability for one opponent (a specific one) to hold a pair of jacks (or any other pair of mentioned values) is C(4,2)/C(50,2) = 6/1225. The odds for at least one opponent to hold JJ will be:
P= 6*n/1225 - 6*C(n,2)/(C(50,2)*C(48,2)). Just do the calculation for n=9.
The same odds stand for any other pair.
You can add them for the event "one opponent to hold JJ, QQ, KK or AA" (these are independent events), but you cannot add them for the event "at least one opponent to hod JJ, QQ, KK or AA".
Posted Tue Jun 07, 2005 7:38 am GMT by Jackal
how about raising to see if anyone plays back. If you have been observing your opponents betting patterns you would have an idea if they have a par by their preflop betting. Don't be gutless raise!
Think about it, if a player goes 4 bets preflop he has a monster or it's play money. Forget odds.
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