After being stacked when my set ran into a bigger set I decided to do a little math and see what are the chances of being that 'lucky'.
For the sake of simplicity we'll assume that every PP will pay to see the flop.
Also note that when I say Full Ring (FR), I mean 9 opponents and accordingly 6max means 5 OPPs.
1) Chances to get PP:
there are (52*51)/2=1326 theoretically possible holdings, 13 PPs and 6 ways to be dealt each which leads to 13*6/1326=0.058823 or 5.88% or (1-to-17) 1/17.
2) Chances that at least one of your opponents has different PP himself:
first we must take 2 (your hole) cards out of the deck so there are (50*49)/2=1225 possible holdings and remaining 12 PPs, 6 ways to be dealt each -> 12*6/1225=0.058775 or 1/17.01 AGAINST 1 OPPONENT (HU).
With 9 OPPs (FR): (1 - (16/17)^9) = 1/2.38
With 5 OPPs (6m): (1 - (16/17)^5) = 1/3.82
By now we have two different PPs preflop. Since we want to know how often will our OPP have the better one (OBV half of the time), we need to divide these with 2 ->
HU: (1/2)*(1/17.01)= 1/34.02
6m: (1/2)*(1/3.82)= 1/7.64
FR: (1/2)*(1/2.38)= 1/4.76
3) Chances of you flopping a set:
first we must take 4 known (hole) cards out of the deck -> 1-((46/48)*(45/47)*(44/46))=0.12234 or 1/8.17
4) Chances of your OPP flopping a set:
now we must take 5 known cards out of the deck: 4 hole cards and one flop card that brought us a set which also means there are 2 remaining flop cards that OPP might hit set with -> 1-((45/47)*(44/46))=0.084181 or 1/11.88
5) Putting it all together:
now we just have to multiply all of the above:
(HU) (1/17)*(1/34.02)*(1/8.17)*(1/11.88)= 1/56133
(6m) (1/17)*(1/7.64)*(1/8.17)*(1/11.88)= 1/12606
(FR) (1/17)*(1/4.76)*(1/8.17)*(1/11.88)= 1/7854
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SYNOPSIS:
On average, how often will it happen that you flop a set AND your opponent flops a bigger set?
Roughly about ...
HU: once in 56.1K hands
6m: once in 12.6K hands
FR: once in 7.8K hands
