Likely ahead, outs if you're behind

[PoBoy321]

KHere's a quick question, and it's likely been brought up before, but I can't find anything in the search.

Let's say you have K /images/graemlins/heart.gifK /images/graemlins/diamond.gif in Hold 'Em and the flop comes K /images/graemlins/spade.gifQ /images/graemlins/spade.gif3 /images/graemlins/spade.gif. Hypothetically, an early position bettor moves all-in for an a 2x pot-sized bet. It's on you to call and you're getting 3:2 on the call. If you're ahead, great. If you're behind, you're still on a 2:1 draw.

How often do you have to have the best hand for this to be a +EV call?

[AaronBrown]

You have to win 40% of the time, 2 times out of 5. If there is X in the pot preflop, when you win, you win 3X and when you lose you lose 2X. If you win twice and lose three times, you win and lose 6X to break even.

If your opponent has a flush, she cannot have any of your outs (K, Q, 3). So you have a 29% chance of winning. So you must think there is at least a 16% chance she does not have the flush. How much more than 16% is break-even depends on the probabilities of her having a four flush, AJ or AT or JT or QQ or A3 or some other hand.

[Siegmund]

If he moves in with a 2x-pot-sized bet, you need to win 2/5ths of the time to break even on a call.

If we assume you are winning either always or 1/3 of the time, we just need to solve 1 * X + (1/3)(1-X) = 2/5 -> X = 1/10.

Being ahead 10% of the time is enough.

Edit: Just to avoid any confusion between mine and AaronBrown's post... he has done the same calculation but taken 29% instead of 1/3 as his estimate of your chance of winning if behind, and gotten (40-29)/(100-29) ~ 16% as his answer.

[ihaveapigyo]

[ QUOTE ]
1 * X + (1/3)(1-X) = 2/5 -> X = 1/10.


[/ QUOTE ]

Do u mind explaining how u determine this?

[Siegmund]

[ QUOTE ]
[ QUOTE ]
1 * X + (1/3)(1-X) = 2/5 -> X = 1/10.


[/ QUOTE ]

Do u mind explaining how u determine this?

[/ QUOTE ]

In words:

You want to win 2/5s of the time overall to show a profit. (One side of equation = 2/5.)

Some proportion of the time (x) you always win (1).
The rest of the time (1-x) you win only one-third of the time (1/3).

More generally, on one side we have a list of (the chance we are in a given situation) * (the chance we win if we are in this situation), which when added up gives us our overall chance of winning, and then we compare this to our pot odds to see if we are showing a profit or not.

The same method applies (*) to deciding whether a semibluff is profitable or not.

You bet, he folds: you always win.

You bet, he calls: you win if your draw comes in, lose if it doesn't.

You are risking 1 bet to win the N bets already in the pot. If (probability he folds) + (probability he doesnt fold) * (probability your draw comes in ) > 1/N, your semibluff shows a profit.

(*) actually, you win N bets when he folds, and at least N+1 bets when he calls and your draw comes in, and lose 1 bet when he calls and your draw doesn't come in, so a slightly more complicated version, with (probability of situation) * (probability of winning or losing in this situation) * (amount you will win or lose) on the left, and size of bet on the right, is needed. In the original question, we were considering calling an all-in bet, and the amount of money involved was the same for every case.

[jogsxyz]

Doesn't that make it a mandatory call? He will raise all in with QQ and 33. 6 ways. 45 ways for the flush.