[AaronBrown]

Call the amount in the pot before the turn betting P, and the amount of the current bet B. I assume the flush draw guy either has a flush draw or an Ace-high straight, but not both, and that he is equally likely to have either. I assume there are nine cards flush cards among the 44 in the deck if there is a flush draw, and eleven if there are not. If you change these assumptions, the answer changes.

If you fold, you get zero.

If you call now and there is no flush card on the river, you win (P + B)/2 if only one other player has an Ace, and P/3 if both other players have Aces.

If you call now and there is a flush card on the river, so you fold, you get –B.

If you call now and there is a flush card on the river and everyone calls the 2*(P + 3*B) bet, you win P/3 if the raiser had the straight, and lose 2*P + 7*B if he has the flush. If you call and the other straight person doesn’t, you still lose 2*P + 7*B if the third player has the flush, but now you win (P + B)/2 if he doesn’t. However, in both cases, you lose more if you lose than you win if you win, and 55% of the time you will lose. So you can’t call on the river.

Now we can figure the EV’s of calling on the turn. 10 times out of 44 there is a flush card on the river, so if you call and fold for a flush card, out of 44 deals you lose B 10 times, win (P+B)/2 17 times and win P/3 17 times. Your overall EV is (85*P – 9*B)/264. This is greater than zero (so the strategy is better than folding) as long as B < 85*P/9. So unless the bet is more than nine times the pot, you should at least call.

Finally, we should consider raising on the river. I assume all-in is large enough that the flush draw has to fold. Therefore you win (P+B)/2 if there is a flush draw, and P/3 if not. Your expected gain is (5*P + 3*B)/6. This is always larger than the EV of calling.