Quote:
Originally Posted by Chumbardo
You get a pocket pair about .06 of the time. The chances that someone else has a pocket pair is about .5 at a a 10-handed table. The chances that you hit a set on the flop is ~.12. The chances that someone else hits a set given that you hit a set is ~.08 (due to card removal of the first person hitting a set).
So the chances of set over/under set is ~.06*.5*.12*.08 = 0.0000288 = .0288% ~ 18/625 ~ 1/3500. Which means you should be getting set over setted 1 in 7000 times. Divide that by 30 hands per hour and that means it should happen about one every 230 hours. So in 1500 hours at 30 hands an hour at a 10 handed table assuming all pocket pairs make it to the flop, you should be on the losing end of a flopped set over set about 6.5 times, and you're saying it happened to you 40 to 50 times.
I'm a little rusty with this stuff. Can somebody run me through the calculation for why the probability someone else has a PP is roughly .5? I can calculate the probability a specific player has a PP easily enough.
P(WeHavePP) = (13 choose 1)(4 choose 2)/(52 choose 2) = 1/17
P(WeHavePPAndSpecificOtherPlayerHasPP) = 1/17(12 choose 1)(4 choose 2)/(50 choose 2) = 72/20825 ~ .003457
That's the probability say, we get a PP and the guy to our left also get's one. But that's not very useful in computing the probabilities for the whole table. Also this is only looking at the scenario there are two pocket pairs in play. To be complete we should consider up to nine pocket pairs in play, though to be realistic cutting it off at three or four would probably suffice.
If we treat the dealing of ten players cards as ten independent Bernoulli trials with P(PP) = 1/17, then yeah okay P(someone has PP) = 1 - (16/17)^10 = .4546, which is close to .5. BUT this number is off because they aren't independent Bernoulli trials and gives misleading results because although cases in which three or more players have a pocket pair are rare, the probability of set/set is significantly more in these cases.
Also, .4546 gives the probability anyone at the table has a PP, when what we want is the probability someone else has a PP after accounting for our own PP, right? And we could estimate that in a similar fashion, but yielding P(WeAndSomeoneHavePP) = 1 - (1-72/20825)^9 ~ .0307
Anyway, if I'm following your logic, the equation should be
P(set/set) = P(underset) * P(WeAndSomeoneHavePP) * P(WeFlopSet) * P(WeAndOtherFlopSet)
Ignoring paired flops and quads/set situations
P(set/set) = .5 * .0307 * 1/8.5 * 1/8.5*2/3 = 0.0001416378
Solve(1/x = 0.0001416378,x) yields x = 7060.26 hands for set/set
Given y hands per hour, then we have 7060.26/y hours on average for set/set
Setting y=30 gives 235.34 hours, y = 40 gives 176.50 hours
I agree 40+ times over 1500 hours is probably an exaggeration but I'm not really sold on this calculation. Also, rare events like set/set have a lot of randomness in their occurrence because they are so rare. What I mean is yeah, maybe the magic answer is "set over set once per 250 hours" but results are unlikely to be close to that over even fairly large sample sizes. And we always like to ignore the possibility someone is experiencing extreme negative variance in some way.
Another thing to consider...we should be on the winning end of set/set far more than the losing end. This is because good players are capable of raise/folding hands like 22 - 77 preflop if the odds to setmine are not there, or just folding vs. short stack raises. There are many situations it's not profitable to see a flop with small pocket pairs.
So I got a little off track here, so let me repeat my original question -- where does the calculation that someone else has a pocket pair half the time come from? Because I'm having trouble reproducing that.