50 turn spot with topset

Hi,

obviouslly it sucks to fold, but vs a passive player you have to do it. If he was relatively agressive, i'm just jamming here because it's possible he is shipping a hand like worse set + fd or something like a double fd + gutter + pair.

Regards,

Skod

u need around 27% and with a set u need 22,5%. That said u would need only a tiny bit of implieds, and u might get a free SD if flush hits river and he had a blocker only.

He has a straight pretty much for sure. So it becomes just a math puzzle.

You need 27,5% and with 10 outs you have 24%. If that was all-in, you should fold.

But opponent has 8,33 left, and 46/15 player is going to call your river bet always. To justify the call you would need at least 12,30 in implied odds. As opponent has less than that behind, you can't have it. So fold, even with the implied odds.

So the conclusion might be here too that the turn-c-bet was too big in this situation.
With a smaller bet there are more implied odds, because opponent can only raise potsize.

No, that would be results oriented thinking. We don't know if he has straight or not, and as we bet we estimate that he doesn't have it too often and want to get value from the rest of his range. And against that part of his range we don't want to miss value but betting smaller.

Originally posted by Kyyberi
He has a straight pretty much for sure. So it becomes just a math puzzle.

You need 27,5% and with 10 outs you have 24%. If that was all-in, you should fold.

But opponent has 8,33 left, and 46/15 player is going to call your river bet always. To justify the call you would need at least 12,30 in implied odds. As opponent has less than that behind, you can't have it. So fold, even with the implied odds.

Actually that 24% is a bit too much, because it considers, that the opponent has no blockers in his hand. 2 is known from vilains hand (assuming he has AQ or Q9) the other 2 will have on average 0,5 blockers to our full house draw. (This was from Tom Chambers Advanced PLO theory, kinda good stuff, coz before that I was counting with all outs unblocked)

It doesn't matter as 2 of opponents cards are considered "unknowns". You can do the math and calculate all the possible blocker combinations with probabilities and you will come to the same number: 10/42 = 23,8%.

Sometimes opponent has a blocker, sometimes he doesn't. If he doesnt, then we have 10/40 = 25% equity. If he has one, we have 9/41 = 22%. If he has two, we have 8/40 = 20%.

If you count that average of 0,5 blockers, then we would have 9,5/40 = 23,8%. Same as above.

Unknowns are unknowns, it doesn't matter where they are in terms of probabilities.