If everyone at a 8-handed game just limps and checks any two cards all the way to the river, and we deal 10000 hands.
On average:
- How many times at least 1 player hits a set on the flop (with pocket pairs)?
- How many times at least 1 player hits a set by the river?
- How many times a player hits a set on the flop (set over set over set counts as 3)?
- How many times a player hits a set by the river?

Thanks!

Assuming 8-handed NLHE where every player goes to showdown on every deal (and including the relevant quads and full houses in the tallies), here are the results of a simulation of 100 Million deals:

5,433,063 times at least one person hits a set on the flop

8,713,397 times at least one person hits a set by the river

5,533,107 players hit a set on the flop

9,028,862 players hit a set by the river.

As I always say, it's a good idea to await some sort of confirmation of these results (mistakes have been known to occur).

Also, it is undoubtedly rather straightforward to derive good approximations of these percentages (which I have not done) but I am not sure how easy it would be to derive exact answers.

I confirm whosnext's flop stats:

N(unpaired flop) = C(13,3)*4³
N(pair flop) = 13*12*4*C(4,2)

P(set | unpaired flop) = 8*3*3/C(49,2) - C(8,2)*3*3²/3/C(49,4) + C(8,3)*3³/(5*3)/C(49,6) = 6.00423232 %
P(boat+ | paired flop) = 8*(3+1)/C(49,2) - C(8,2)/C(49,4) = 2.70787316 %

P(flopped set+ using pocket pair) = [C(13,3)*64*.0600423232 + 13*12*4*6*.0270787316] / C(52,3) = 5.43166269 %

For the expected number of sets, we need the chance of exactly 1, exactly 2 and exactly 3. Since I used inclusion-exclusion, you can just take my calculations above and insert a couple coefficients.

E(flopped sets per hand) =
[C(13,3)*64(.058867365 + 2*.00116775 + 3/138728) + 13*12*4*6(.0269465788 + 2*.000132152769)] / C(52,3)
= .055318275

In 10k hands that translates to 553 flopped sets.

Since whosnext's sim agrees with those, I would bet his full-board stats are right too.

Thanks everyone! Very useful.