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#1
12-30-2005, 05:14 PM
 Guest Posts: n/a
probability of hands clashing....

So I was playing 1-2 NL at the Taj yesterday, and I held 99 in the small blind. To make a long story short and to spare you all the intricacies of how we got there, on a flop of 988, I was all in and got called by a guy with poket 8's for quads. What are the chances of this happening? It's got to be astronomical. The chances of flopping quads is like 500 to 1 at least, and the chances of another guy flopping top boat on the same hand has got to be astronimical. Anyone?

Ryan
#2
12-31-2005, 01:31 AM
 Guest Posts: n/a
Re: probability of hands clashing....

At the risk of BruceZ taking me to school for bad calculations [img]/images/graemlins/smile.gif[/img] I'll give this a shot.

The chances of two players with pocket pairs (any pocket pairs not of the same rank as opponent):

C(13,1)*(4,2)/C(52,2) = 78/1326 = 5.9%

times

C(12,1)*(4,2)/C(50,2) = 72/1225 = 5.9%

times chances that flop will give one player full house and the other quads:

C(2,1)*(2,2)/C(48,3) = 2/17296 = 0.01%

Total = 5.9% * 5.9% * 0.01% = 0.000003%, or 3.3 million to one.

This assumes only two players, so if other non-eights or non-nines were dealt to other players, the odds get better of this happening.
#3
12-31-2005, 02:57 AM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
Re: probability of hands clashing....

[ QUOTE ]
At the risk of BruceZ taking me to school for bad calculations [img]/images/graemlins/smile.gif[/img] I'll give this a shot.

The chances of two players with pocket pairs (any pocket pairs not of the same rank as opponent):

C(13,1)*(4,2)/C(52,2) = 78/1326 = 5.9%

times

C(12,1)*(4,2)/C(50,2) = 72/1225 = 5.9%

times chances that flop will give one player full house and the other quads:

C(2,1)*(2,2)/C(48,3) = 2/17296 = 0.01%

Total = 5.9% * 5.9% * 0.01% = 0.000003%, or 3.3 million to one.

This assumes only two players, so if other non-eights or non-nines were dealt to other players, the odds get better of this happening.

[/ QUOTE ]

All right except for final the arithmetic and round-off error. It's 2.5 million-to-1.
#4
12-31-2005, 03:39 AM
 Guest Posts: n/a
Re: probability of hands clashing....

Sheesh, thanks guys. I thought it was astronimcal, but I didn't realize that it was THAT astronomical lol.

Ryan
#5
12-31-2005, 04:06 AM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
Re: probability of hands clashing....

[ QUOTE ]
Sheesh, thanks guys. I thought it was astronimcal, but I didn't realize that it was THAT astronomical lol.

[/ QUOTE ]

Keep in mind that's heads-up. At a 10-player table it will be about 9 times more likely that it happens to you(*), or about 278,000-to-1, and if you already have exactly 2 different pairs seeing the flop, then it is 8647-to-1.

(*)Note that it is less than 9 times more likely that an opponent has a different pair since 9 would count the times that N opponents have one N times, but note that when this happens it is also N times more likely to flop quads over full with some opponent assuming that all pairs see the flop, so the over counting is what we want. This would also include cases where 2 opponents share the same pair which we don't want, but these cases are rare, so it will be only slightly less than 9 times.
#6
12-31-2005, 01:15 PM
 Guest Posts: n/a
Re: probability of hands clashing....

Thanks Bruce, I figured it was 9times more likely at a 10 man table. I do have some stat and calculus background, but it is way too removed for me to have figured this out lol. Thanks!

Ryan

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