Odds of three sets on the flop? [Archive] - Two Plus Two Older Archives

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bwana devil

03-07-2005, 01:46 AM

Can someone smarter than me tell me what the odds are of three players flopping a set on a 6 handed game? I got to experience that painful feeling earlier today. Wasn't much fun to be on the short end of that statistical rarity.

Thanks
bwana

Party Poker No-Limit Hold'em, $1 BB (6 max, 6 handed) converter (http://www.selachian.com/tools/bisonconverter/hhconverter.cgi)

MP ($57.15)
CO ($52.3)
Button ($89.23)
SB ($187.45)
BB ($130.92)
Hero ($116.5)

Preflop: Hero is UTG with J/images/graemlins/heart.gif, J/images/graemlins/spade.gif.
Hero raises to $3, 1 fold, CO calls $3, Button calls $3, SB calls $2.50, BB calls $2.

Flop: ($15) K/images/graemlins/diamond.gif, J/images/graemlins/club.gif, 3/images/graemlins/spade.gif (5 players)
SB checks, BB checks, Hero bets $1, CO folds, Button raises to $5, SB calls $5, BB folds, Hero calls $4.

Turn: ($30) 8/images/graemlins/spade.gif (3 players)
SB checks, Hero checks, Button bets $10, SB raises to $40, Hero raises to $70, Button raises to $81.23 (All-In), SB raises to $179.45 (All-In), Hero calls $38.50 (All-In).

River: ($399.18) 6/images/graemlins/heart.gif (3 players, 3 all-in)

Final Pot: $399.18
Main Pot: $273.69, between Button, SB and Hero. > Pot won by Button ($273.69).
Pot 2: $54.54, between SB and Hero. > Pot won by Hero ($54.54).
Pot 3: $70.95, returned to SB.

Results below:
SB has 3d 3h (three of a kind, threes).
Hero has Jh Js (three of a kind, jacks).
Button has Kc Ks (three of a kind, kings).
Outcome: Button wins $273.69. SB wins $70.95. Hero wins $54.54.

cwes

03-10-2005, 07:39 PM

Hi bwana

there can be pairs of 13 different values. To have the chance to flop a set, there may be only one player holding a pair of any value. A pair has to contain two of four colors. So a pair of each value can be constructed in '4 choose 2 = 6' ways.

Hence the first player's propability of getting any pair is '(13 x 6)/(52 choose 2) = 78/1326'. The second player's 'Pr(any other pair) = (12 x 6)/(50 choose 2) = 72/1225', the third player's 'Pr(any left over by #1 and #2 pair) = (11 x 6)/(48 choose 2) = 66/1128'.

Propability for the other players to get any hand is exactly 1. Notice that we allow the other players (#4-#6) to hold a pair of a value #1-#3 are holding a pair of.

Muliply these values and you get the propability that "at least three players are holding a pair each that differ in value". I will call this propability 'A'.

Now you still need the propability to flop three cards that each give one of players #1-#3 a set. This propability obviously is 'B = 2/46 x 2/45 x 2/44 = 8/91080'. Using these propabilities we eliminate the problem of other players eventually having caught cards of the values of #1-#3's pairs.

So your propability is 'A x B'. It says, this happens once every 56,280,000 hands that players #1-#3 all get a pair and then flop a set.

Now multiply this propability by '3 choose 6 = 20' to get the propability of this happening to any three out of six players. So it should happen every 2,814,000 hands in a six handed game and every 470,000 hands in a ten handed game (this happens independently from the pairs seeing the flop or not, propabilities do not really care about players decisions).

I hope I got that right...

bwana devil

03-10-2005, 10:54 PM

wow thanks for all the information. that's a heck of a second post. let's keep you around.

i guess i can stand to get burned once every 2,814,000 hands /images/graemlins/grin.gif

cwes

03-11-2005, 04:58 AM

[ QUOTE ]

"at least three players are holding a pair each that differ in value".

[/ QUOTE ]

Has to be: "at least players #1-#3 are holding...". It becomes "at least three players are holding..." as soon as we multiply it by '(#of hands in game) choose 3'.

JonLines

03-11-2005, 07:41 AM

A 1 in 2.8 million chance? You call that a bad beat? I remember this time I had AA........ /images/graemlins/wink.gif

cwes

03-11-2005, 09:08 AM

[ QUOTE ]

i guess i can stand to get burned once every 2,814,000 hands /images/graemlins/grin.gif

[/ QUOTE ]

Dangerous conclusion. I gave the probability of this happening ex ante. I did not give the propability of you being beat by two other sets if you flop a set and two higher cards appear on board.

So if you have got a pair of threes before the flop there obviously are 60 combinations (higher pairs), that can beat your pair. the probability of two other players having two unique valued higher pairs is '(5 choose 2) x (60/(50 choose 2)) x (54/(48 choose 2))'.

The probability of you all flopping a set then stays the same as above (i.e. 8/901080).

So this happens every 485,549 hands in a six handed game to you if you play a pair of threes. Remember this is only about flopping a three if you play a pair of threes and then being beat by two higher sets flopped. It is not about being beat by two higher sets gained on the flop and/or turn and/or the river nor being beat by any other hand.

Given there are three pairs before the flop the probability they all flop a set is just 8/91080 = 11,385^-1.

Given there are three pairs before the flop, your threes and two higher ones, the probability of you and at least one of them flopping a set (and hence beating you unless you get a fourth three and they do not get a fourth whatever) is '2/46 * 4/45 = 258.75^-1'. So starting with a pair against two higher pairs, flopping a set and already being beat by a higher flopped set will happen every 258.75 times you see the flop against two higher pairs.

Different story, huh?

But, hey! It gets even worse. Given you start against two higher pairs with your pair and you flop a set, the probability of at least one of them also flopping a set is just 4/45 = 11.25^-1. From 2.8 millon hands to 11 hands... creepy.