[cwes]

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...