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#2
10-13-2006, 12:13 PM
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Re: Floped full house Vs floped quads 6 handed
100%
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#3
10-13-2006, 01:01 PM
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Test yourself here
[ QUOTE ]
Just wondering what the probability of this happening is? I had my full house cracked by quads in only a small amount of hands and got stacked both times. [/ QUOTE ] While this problem may sound difficult, it should actually be very easy for anyone with a solid grasp of basic probability. Rather than spitting out the exact answer as I usually do, I want to give others a chance to think about it first, as this problem is useful for exposing deficiencies in understanding and technique. Anyone who cannot get the exact answer in under a minute has something important that they can learn from this problem. To restate the problem, we want the probability that any given hand will have full house vs. quads on the flop in a 6-handed game. Assume for simplicity that any player will see the flop with any pair. It is not necessary to use the inclusion-exclusion principle. Basic probability techniques are all this requires. EDIT: Changed assumption to read "any player will see the flop with any pair".
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#4
10-13-2006, 01:22 PM
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Re: Test yourself here
Alright Bruce, just had a go at it. I studied stats in Uni a long time ago and haven't done much since. With your assumptions it seems a bit easier however i'm probably wrong here.
Is it 1 in 19760??? I'm more interested in knowing without the assumptions though...
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#5
10-13-2006, 01:27 PM
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Re: Test yourself here
[ QUOTE ]
Alright Bruce, just had a go at it. I studied stats in Uni a long time ago and haven't done much since. With your assumptions it seems a bit easier however i'm probably wrong here. Is it 1 in 19760??? [/ QUOTE ] No. [ QUOTE ] I'm more interested in knowing without the assumptions though... [/ QUOTE ] The problem cannot be solved without assuming something about what pairs the players will see the flop with, or at least how many. If you want a different assumption you need to state it.
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#7
10-13-2006, 01:33 PM
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Re: Test yourself here
i like the new bruce!
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#9
10-14-2006, 07:09 AM
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Re: Test yourself here
Ok, i'm going to have another go here and show you some reasoning. Before i thought your assumption was that everyone was playing with a poket pair, i should wear my glasses more often.
So it's two events, firstly two people dealt any unique pocket pair, the odds of this would be 52/52 * 48/51 * 3/50 * 3/49 = 1 in 289.23611 Next, i'll say it AA v KK here so the probability of flopping a full house vs quads is the combined probability of it coming AAK, AKA, KAA, KKA, KAK, AKK.. Six handed their are 40 cards left so the probabilities are AKK 2/40 * 2/39 * 1/38 = 1 in 14820 AKA 2/40 * 2/39 * 1/38 = 1 in 14820 KKA 2/40 * 1/39 * 1/38 = 1 in 14820 KAA 2/40 * 2/39 * 1/38 = 1 in 14820 AKA 2/40 * 2/39 * 1/38 = 1 in 14820 KAA 2/40 * 1/39 * 1/38 = 1 in 14820 so 6 in 14820 = 1 in 2470 So the combined probability of these two events is 1 in 714,413.1944 recurring. Hows that? any better?
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#10
10-14-2006, 11:25 AM
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Re: Test yourself here
Ok- i'm going to take a stab as well
[ QUOTE ] So it's two events, firstly two people dealt any unique pocket pair, the odds of this would be 52/52 * 48/51 * 3/50 * 3/49 = 1 in 289.23611 [/ QUOTE ] I think this should be (52/52 * 3/51) * (48/50 * 3/49) = p(player A receives pocket pair) * p(player B is dealt a first card that does not match player A's pair) * p(player B pairs this first card) =~ 0.003457 then, of 48C3 possible flops, 4 will make quads and a full house. 4/(48C3) = ~0.000231267 therefore, the probabilty of this happening at a 2 person table is 0.003457 * 0.00023126 = ~ 8e-7 or roughly 1 in 1.25 million I am anxious to see how to extend this result to a 6 person table. I thought of using a binomial distribution here- however I am fairly certain it does not consider the times when 3 or more people have been dealt pocket pairs- and is therefore not applicable?!? Many thanks to Bruce for all his help.
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