ACES vs KINGS twice in a button
June 13th, 2015 at 5:31:41 PM
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Hello, i want to find the propability of the follow situation
7 players table, had kings on button , player X has aces and we gamble the pot pre flop.
4 or 5 hands after, i am utg , i hold kings again , Same player HAS aces again!
i want to know whats the propability to happen this thing?:p
Cant calculate myself , and i want to see the way to build this function
7 players table, had kings on button , player X has aces and we gamble the pot pre flop.
4 or 5 hands after, i am utg , i hold kings again , Same player HAS aces again!
i want to know whats the propability to happen this thing?:p
Cant calculate myself , and i want to see the way to build this function
June 16th, 2015 at 7:25:52 AM
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arsonist
Not sure how you build the function, but I know 100% of the time when I get aces
near the end of a tournament, I lose.
dicesetter
Not sure how you build the function, but I know 100% of the time when I get aces
near the end of a tournament, I lose.
dicesetter
June 16th, 2015 at 9:21:04 AM
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The probability of you having KK on a particular hand is 4/52 (the probability that your first hole card is a King) x 3/51 (the probability that, given your first one is a King, the second one is a King as well) = 1/221. The probability that a particular player has AA given that you have KK is 4/50 x 3/49 = 6/1225. The probability of both = 1/221 x 6/1225 = about 1 in 45,120.
Remember, this is just the probability of it happening the second time (when you were the button after it happened the previous time you were on the button, if I read it correctly) with the same person getting the AA.
Assuming the probability of you having KK and any other player having AA is six times that, then even if you were the button on 50 hands, the probability of it happening at all is about 1/150, so the probability of it happening while you were on the button and then it happening with the same player the next time is about 1 in 6.768 million. Note that if you are the button 50 times, the probability of you getting KK at all is about 1/5, so just getting KK on the button twice in a row is about 1/6125.
Remember, this is just the probability of it happening the second time (when you were the button after it happened the previous time you were on the button, if I read it correctly) with the same person getting the AA.
Assuming the probability of you having KK and any other player having AA is six times that, then even if you were the button on 50 hands, the probability of it happening at all is about 1/150, so the probability of it happening while you were on the button and then it happening with the same player the next time is about 1 in 6.768 million. Note that if you are the button 50 times, the probability of you getting KK at all is about 1/5, so just getting KK on the button twice in a row is about 1/6125.
July 3rd, 2015 at 4:34:13 PM
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For the exact calculation it's best to use inclusion-exclusion (credit to BruceZ, another new member of this forum, for teaching me advanced uses of this technique).
P(hero KK & at least one villain AA when 7-handed) = 1/221 * 6*C(4,2)/C(50,2) - C(6,2)/C(50,4) = x
P(hero KK & specific villain AA) = 1/221 * C(4,2)/C(50,2) = y
It sounds like OP wants to know the chance of the first happening this hand and then the second happening within 5 hands from now.
That's x*(1-(1-y)^5) =~ 1 in 68,016,741
If it doesn't have to start from this hand, but can be any 6-hand window within a session of N hands, that can be calculated too. Or another thing one could ask is what's the probability of the two events occurring at any times within N hands (without the limit on their separation).
P(hero KK & at least one villain AA when 7-handed) = 1/221 * 6*C(4,2)/C(50,2) - C(6,2)/C(50,4) = x
P(hero KK & specific villain AA) = 1/221 * C(4,2)/C(50,2) = y
It sounds like OP wants to know the chance of the first happening this hand and then the second happening within 5 hands from now.
That's x*(1-(1-y)^5) =~ 1 in 68,016,741
If it doesn't have to start from this hand, but can be any 6-hand window within a session of N hands, that can be calculated too. Or another thing one could ask is what's the probability of the two events occurring at any times within N hands (without the limit on their separation).
