View Full Version : Odds of 3 sets being flopped and quads being hit on the turn?

05-11-2005, 08:27 PM
I've seen set over set as I'm sure everyone has. However I've never seen three sets flopped. I was curious what the odds of three out of six players having pocket pairs and ALL of them flopping sets. On top of that what are the odds of that happening AND one of the sets turning into quads.

Party Poker 1/2 Hold'em (6 max, 6 handed) converter (http://www.selachian.com/tools/bisonconverter/hhconverter.cgi)

Preflop: Hero is UTG with 8/images/graemlins/diamond.gif, 8/images/graemlins/spade.gif.
Hero calls, <font color="#CC3333">MP raises</font>, CO calls, <font color="#666666">1 fold</font>, SB calls, BB calls, Hero calls.

Flop: (10 SB) T/images/graemlins/heart.gif, 8/images/graemlins/heart.gif, 6/images/graemlins/heart.gif <font color="#0000FF">(5 players)</font>
SB checks, BB checks, <font color="#CC3333">Hero bets</font>, MP calls, CO calls, SB calls $1.86 (All-In), <font color="#CC3333">BB raises</font>, <font color="#CC3333">Hero 3-bets</font>, MP calls, CO calls, BB calls.

Turn: (13.65 BB) 6/images/graemlins/club.gif <font color="#0000FF">(5 players, 1 all-in)</font>
<font color="#CC3333">BB bets</font>, Hero calls, <font color="#CC3333">MP raises</font>, CO calls, BB calls, <font color="#CC3333">Hero 3-bets</font>, <font color="#CC3333">MP caps</font>, CO calls, BB calls, Hero calls.

River: (29.65 BB) A/images/graemlins/diamond.gif <font color="#0000FF">(5 players, 1 all-in)</font>
BB checks, <font color="#CC3333">Hero bets</font>, <font color="#CC3333">MP raises</font>, <font color="#CC3333">CO 3-bets</font>, BB calls, Hero calls, <font color="#CC3333">MP caps</font>, CO calls, BB calls, Hero calls.

Final Pot: 45.65 BB

Results in white below: <font color="#FFFFFF">
SB has 3d As (two pair, aces and sixes).
BB has 9c 7d (straight, ten high).
Hero has 8d 8s (full house, eights full of sixes).
MP has Tc Ts (full house, tens full of sixes).
CO has 6s 6d (four of a kind, sixes).
Outcome: CO wins 45.65 BB. </font>

05-11-2005, 08:44 PM
I posted this question not long ago after it happened to me.
Set over set was somewhere between 1 in 176 and 207 ?
Set over set over set was like 1 in 1700. /images/graemlins/confused.gif

05-12-2005, 12:24 AM
I posted this question not long ago after it happened to me.
Set over set was somewhere between 1 in 176 and 207 ?
Set over set over set was like 1 in 1700. /images/graemlins/confused.gif

[/ QUOTE ]
I assume that's 1 in 1700 when there are three players with pocket pairs. So I would would need to know how often 3 players have pocket pairs in the same hand to figure out how many times set over set over set happens out of all hands. Unfortunately I'm not sure how to calculate that.

05-12-2005, 12:22 PM
Well lets see.

To get the Number of at least 3 players having PP &gt; PP &gt; PP in 6 max is:

C(6,3)*(13(C(4,2))*12(C(4,2))*11(C(4,2)))/C(52,12) =
20 * ((13*6) * (12*6) * (11*6))/C(52,12) = .000035919862891074924 or 1:~27838

To get set &gt; set &gt; set on the Flop:
(6*4*2)/(46*45*44) = 5.270092226613965E-4 or 1:~1896

So to catch a quad from the set &gt; set &gt; set:
3/43 or 1:~13.3

So of this ever happening is:
(1:~27838) * (1:~1896) * (1:~13.3) = 1:703,744,640

And I know I've done something wrong.


05-12-2005, 12:48 PM
Actually, I think I did get it wrong:

The odds that you will have PP &gt; PP &gt; PP preflop in 6max is:
(C(6,3)*(C(13,3)*C(4,2)^3)) / C(52,12)
(20 * (286 * 216)) / C(52,12) = 1:~167037

So the chance of this happening to the turn:
(1:~167037) * (1:~1896) * (1:~13.3) = 1:~4.2 Billion

I probably still did something wrong...


05-12-2005, 03:17 PM
In order to answer this type of question you need to make certain assumptions and start at a given point. I will find a slightly different number. I am going to solve the problem that given you start with a pair.

1. You find your self in a set/set/set situation after the flop.

2. One or more of you end up with four of a kind.

Obviously you must assume all people dealt a pocket pair stay to see the flop and all people with a set on the flop stay to the end.

Given that you have a pocket pair!

The flop gives you sepecifically a set

=2*(12c2)*4*4/(50c3) = 10.7755% or 1 in 9.28 flops

Prob two of your five opponents also hold a set

=18/(50c2)/(48c2)*(5c2) = .0139% or 1 in 7,676.67 times

Prob one or more of your opponents ends up with four of a kind

=(3*40+3)/(43c2) = 13.6213% or 1 in 7.34 times

So given you flop a pocket pair the probability of seeing a set/set/set flop is:

=10.7755%*.0130% = .0014037% or 1 in 71,242 times

So given that you have a pair the probability that three of you flop a set and one or more end up with a four of a kind is:

=.0014037%*13.6213% = .0001911% or 1 in 523,019 times

Now if you wanted to start from scratch and say what is the probability that on the next deal I get a pair and the above happens you will be dealt a pair 1 in 17 times so multiply the above numbers by 17.

This does not answer the posters question of how often this will happen at a six person table. It answers how ofter will I be involved in the above in a six person table.


05-12-2005, 04:01 PM
Darn it. I think I still made a mistake.

Its the first term thats the hardest.

I give up. At least for now.