#1
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Another What are the Odds?
Given that two players each hold a pocket pair what are the odds that they split the pot in the way described below?
Example 1: Player 1 - AA Player 2 - QQ Board - KKKKA Example 2: Player 1 - JJ Player 2 - 44 Board - 7777Q Example 3: Player 1 - 88 Player 2 - 77 Board - TTTT9 As you can see there are MANY combinations for this to work. I'm not wondering the odds in the three specific examples above, they were mearly examples. I am wondering the odds that given two players each hold a different pocket pair they split the pot with 4-of-a-kind. I think I have described what I was thinking well enough to get my question across. |
#2
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Re: Another What are the Odds?
The odds are slim-to-none [img]/images/graemlins/laugh.gif[/img]
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#3
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Re: Another What are the Odds?
[ QUOTE ]
The odds are slim-to-none [img]/images/graemlins/laugh.gif[/img] [/ QUOTE ] Yeah, I was sort of figuring that much. |
#4
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Re: Another What are the Odds?
[ QUOTE ]
The odds are slim-to-none [img]/images/graemlins/laugh.gif[/img] [/ QUOTE ] What a douche bag.... [img]/images/graemlins/mad.gif[/img] |
#5
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Re: Another What are the Odds?
well, youd obviously need 4 of a kind to flop. so. my guess is it would be 4/50 + 2/47 + 1/46... which would be ~ 4%.
4% sounds pretty accurate to me for a situation like this to come up. i've never seen it. |
#6
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Re: Another What are the Odds?
[ QUOTE ]
well, youd obviously need 4 of a kind to flop. so. my guess is it would be 4/50 + 2/47 + 1/46... which would be ~ 4%. 4% sounds pretty accurate to me for a situation like this to come up. i've never seen it. [/ QUOTE ] Seriously, you're not funny. |
#7
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Re: Another What are the Odds?
[ QUOTE ]
I am wondering the odds that given two players each hold a different pocket pair they split the pot with 4-of-a-kind. [/ QUOTE ] It depends on the rank of the higher pocket pair. The higher it is, the less likely the pot will be split, since splitting requires that the kicker be greater than or equal to the higher pocket pair. If the higher pair is a pair of N's (J = 11, Q = 12, K = 13, A = 14) then the probability of a quad split is: { [(14-N)*4 + 2]*(N-3) + [(13-N)*4 + 2]*(14-N) } / C(48,5) That is, for each of the N-3 possible quads less than N, there are (14-N)*4 kickers greater than N, plus 2 of rank N. Then for each of the 14-N quads greater than N, there are (13-N)*4 kickers greater than N, plus 2 of rank N. <font class="small">Code:</font><hr /><pre> high pair P(split) odds-to-1 33 0.027% 3705 44 0.025% 4057 55 0.022% 4481 66 0.020% 5006 77 0.018% 5669 88 0.015% 6535 99 0.013% 7712 TT 0.011% 9407 JJ 0.0083% 12057 QQ 0.0060% 16786 KK 0.0036% 27617 AA 0.0013% 77831 </pre><hr /> |
#8
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Re: Another What are the Odds?
Finally an answer I can believe.
Thanks. Edit - Are the numbers you gave already assuming Player 2 has a pocket pair as well? |
#9
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Re: Another What are the Odds?
[ QUOTE ]
Are the numbers you gave already assuming Player 2 has a pocket pair as well? [/ QUOTE ] Yes. |
#10
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Re: Another What are the Odds?
Thanks, this re-assures my assumption that this will never happen to me again....
Holding JJ.... Flop J-5-5...detecting weakness I check and Both villans check Turn 5...Bet, Call, Call.... Flop 5 Check, bet, bet I fold they show weak 'A' and split the pot. Thanks, Bco |
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