Flopping Set & Two Pair

TheOneWizard · #1 10-08-2006, 06:02 AM

Hey everyone,

I just had a quick question. If one player has a pocket pair and another has two cards different from the pair what are the odds one player will flop a set and the other will flop two-pair? Thanks for your help.

BruceZ · #2 10-08-2006, 10:00 AM

[ QUOTE ]
Hey everyone,

I just had a quick question. If one player has a pocket pair and another has two cards different from the pair what are the odds one player will flop a set and the other will flop two-pair? Thanks for your help.

[/ QUOTE ]

2*3*3/C(48,3) =~ 1439-to-1.

Brizza · #3 09-03-2007, 10:08 AM

They seem like big odds for this event. Do people find that someone flopping two pair and someone flopping a set holds true to these odds?

Furthermore, what are the odds of one player flopping two pair and another player flopping two pair?

What are the odds of one player flopping a set and another player flopping a set?

jay_shark · #4 09-03-2007, 11:38 AM

Brizza , these odds change when there are more players seeing the flop .

Also , when a player sees the flop , he does not have a random hand so there is a higher chance that one may flop a set and another may flop two pair . The more players at your table , the more likely this is to be the case .

2)Player 1 has AB and player 2 has CD and assume there are only two players for now and that A,B,C,D are different :

If this were to be the case then the probability is 0 since there are only 3 cards to a flop .

This means there must be one card in common or possibly both ;ie , AB,AC or AB,BC or AB,AB .

P(AB,AC or AB,BC) =[1248/1326*(3*44/1225)*2]*3*2*3/48c3 =0.000211

P(AB,AB)= [1248/1326*9/1225]*2*2*44/48c3 = 0.0000703

Add those two results and you should get 0.000281...

3)[78/1326*72/1225]*2*2*44/48c3 = 0.0000351...

Brizza · #5 09-04-2007, 12:16 AM

Thanks for your response. And point taken about people seeing the flop not having a "random" hand as such.

From memory I think the chances of set over set on the flop was 167/1

Does that sound right?

insyder19 · #6 09-04-2007, 09:45 AM

I hate maths.

jay_shark · #7 09-04-2007, 09:54 AM

I worked out the probability of set over set .

Here it is again .

[78/1326*72/1225]*2*2*44/48c3 = 0.0000351 . This is the answer for two players at your table . If there are n players then it will be a bit less than n(n-1)/2*0.0000351. For n=10 , the answer is slightly less than 0.0015795 which is about 632.11:1 .