The three players can be all in preflop, flop or turn, it doesn't matter. Does such a scenario exist where each player has 33.333...% equity with no ties?

As written your problem has no solution. It is impossible to specify three starting hands that give zero probability of ties. A royal flush board would be a tie regardless of hands. We need four of the six cards in the starting hands to be Broadway cards - one of each suit in order to avoid the possibility of a royal flush board. Further at most one hand can have an ace. Otherwise we get a tie on a board like 77552 (this could be avoided if one player had a 7 or 5, but there are too many similar boards to cover all possible such ties with only two kickers left after assigning the Broadway cards). Only one of the two players without an ace can have a king, or else QJT98 no flush would be a tie. Thus we must have Ax, Kx and Qx with the A, K and Q all different suits. We also must put another Broadway card of the fourth suit with one hand. That leaves two cards to be assigned.

Next consider a board like 23456 all one suit. There are four possible boards, one for each suit. We cannot assign four cards from these boards to the player hands because we have only 2 slots remaining after ensuring that royal flush boards are impossible. We also could use the 7s of each suit - this wouldnÂ’t make such boards impossible but would make them not be ties. However again we have to assign all four sevens to prevent ties and we only have two cards left to assign. Hence we cannot assign three starting hands such that there is no possibility of ties.

This is complete trial and error using an online equity calculator, but the closest I’ve come is

Ac9d - 34.86%
KhQd - 32.74%
JsTs - 32.13%

All hands also have 0.27% ties. I strongly suspect that three hands with equal equity is impossible even if ties are allowed, but I do not know how to prove it.

Why shouldn't ties be figured into equity?

JsTs vs JhTh vs JcTc.


ETA:

I think because that gives a trivial solution. I think OP was looking for a spot with three distinct hands that each have equal equity. Having three players with all the same hand is a trivial solution.

They don't have the same hand; they have distinct hands. Otherwise none of them could scoop, which each of them does 8% of the time.

Ok technically, but they all have JTs. They have symmetrical hands. It is obvious that they all will have the exact same equity because they are functionally the same. Two players with AhAc and AsAd also have distinct hands, but it is blatantly obvious that they both have the same equity. I think OP excluded ties in order to exclude trivial solutions where it was obvious that all three hands have exactly identical equities. The question was whether there were hands that are NOT obviously equal in equity, but when the calculation is done they in fact turn out to be equal. I don’t think such a case exists, but as I mentioned above I cannot prove that statement.

If 0.22% of ties is allowed then

Great find! I calculate it more exactly as 33.36% to 33.23% to 33.52% (all 0.21% ties), which is even more impressive.

Wow that is really close! How did you come up with that? Was it just trial and error or did you have some method that you used? Impressive either way though.

Took three goes, started with the basic idea of a small pair vs A high vs K high. Two iterations and I arrived at this.

All players have substantially different hole cards:

player 1: KQo

player 2: K8o

player 3: K2o

Board on the turn: AAA A

As noted above, there is no situation preflop where ties are impossible, and I'm pretty sure there is no situation postflop where there can be no ties but the equities are equal, or even very close to equal.

Guess you missed pokerpops’ post. While the No ties requirement is impossible, he did find three hands that are very close to equal. Not mathematically exactly equal, but close enough to count as a solution IMO. His hands were AJo, KQo (both cards different suits than the AJo) and 66.