Bluff Odds

Okay, I know the correct manner for calculating bluff odds, but does anyone else have a problem with it?

For example, I think a lot of people know that betting 1/2 pot needs to work 1/3 of the time to break-even.


Let's say pot is $100 and we're betting $50, the calculation to find the break-even point is as follows:

Your Bet / (Current Pot + Your Bet)

so in this example: $50/($100+$50) = .333 or 33%

We need our opponent to fold 33% of the time for us to break-even. This is because 66% of the time we lose our $50 bet, 33% of the time we win the $100 pot, so we calculate: .666(-$50) + .333($100) = $0

Now I fully understand the whole concept that once you put money in the pot, it's no longer yours. However, it seems the concept of "break-even" is a bit misleading.

Let's say our opponents do in fact call us 66% of the time, we're all in, and whenever called we cannot win.

When we win the pot, which we will 33% of the time, we net an actual profit of $50.
When we lose, 67% of the time, we lose $100.

.33($50) + .67(-$100) = $16.65 - $67 = -$50.35

So even though it's "Break-even" since the money in the pot is no longer ours, our actual dollar value that we earn on this play is negative. This has been bothering me for awhile, and I'd like to hear some thoughts.

I havfe a terrrible time with it persojnally I think mthat it is just one of those truly maddening things that you cant do alot about but either you got it or you dont and unfortunatelty I dont lol. Good luck to yiou though maybe you can get it figured out.

The $100 in the pot belongs to no one and is up for grabs. So when you bet $50 and the opponent folds, your win is $100.

If you get called and lose, you only lose the $50 that you are going to bet. The $50 you put in before doesn't count because it ceased to be yours when you put it into the pot.

That is just negative for the plays that you lose, but it is break even because essentially, you will lose 2 times for every 1 times that you win, so -50.35(loss#1)-50.35(loss#2)+100(win)=-0.70. I could be wrong, but I am almost positive that the only reason that it isn't completely break even in the math is because of the unexact(word?) numbers, ie. .33 instead of .3333333~ and .67 instead of .6666666~. So basically in order for you to break even you have to win one time for every two losses with this bet. As far as your $50 that is already in the pot, like Arjonius said, that isn't counted into your odds anymore because there is no decision on that anymore because it is already in the pot. Hope that helps, not 100% sure what you're looking for.

I'm aware that the logic goes that once money is in the pot it is no longer yours. However this does not change the fact that winning the hand earns a net profit of $50.

Nor that losing the hand results in a net loss of $100

This is what I'm getting at with this thread, not trying to debate the correct manner in which to make the calculation.

Edit: Also, the 3/1000 isn't going to make a big difference, .333($50) + .667(-$100) = $16.65 - $66.70 = $-50.05

absoluthamm said:

not 100% sure what you're looking for.

Click to expand...

I am proposing that perhaps there are some issues in the current structure for mathematically investigating a particular play's EV value. In this instance, what % of the time we truly need an opponent to fold for a given bet size in order to make money on the hand.

absoluthamm said:

you will lose 2 times for every 1 times that you win, so -50.35(loss#1)-50.35(loss#2)+100(win)=-0.70.

Click to expand...

This isn't right either. Yes you will lose 2 times for every 1 time you win.
However, those times you lose, the $100 from your stack goes into the pot and to your opponent.

When you win, you end up with $50 of your opponent's money.

Those 2 losses, you lost $200. That one win gave you $50. You ended up down $150, which is where the -$50.05 (when done to the thousandth) comes from. That is the EV of the play. On average you will lose $50.05 if your opponent folds 33% of the time.

Mase31683 said:

This isn't right either. Yes you will lose 2 times for every 1 time you win.
However, those times you lose, the $100 from your stack goes into the pot and to your opponent.

When you win, you end up with $50 of your opponent's money.

Click to expand...

Ok, the big point is that you have to forget about the fact that you previously put $50 into the pot, because that is no longer your money, so it doesn't have to do with the EV of your how much you lose on a single bet, so it is ONLY that $50 that you are concerned about when it comes to how much you are investing.

Also, to reword the first part of my phrase.... Not that you will lose 2 and win 1, but you will break even WHEN you win 1 and lose 2... there are always times when your opponents will auto-fold/auto-call anything you bet. Along with that, there are going to be times when you try this bet as a bluff and you will still end up beating an opponent(think two Ace highs with yours having the best kicker), that will add some variance to the formula as well. It seems like you're getting pissed about what I posted the first time, I was just trying to help you figure it out...

No, I'm not mad or anything like that. I could see how it looks that way, but I was just trying to be more clear about what I was trying to say. I know that once you bet $x it's not yours anymore. But what I'm looking at is the net result of a hand being played.

When the hand is done you've either won $x or lost $y, which is what I'm talking about. I get that at that particular point, determining a bet's EV, any money in the pot isn't yours. It doesn't matter how the $100 got in there. If you bet $50 at it, then getting a win 1/3 of the time will in fact break even from that point.

I'll probably just discuss and work on this in person with some people in all honesty. I'm aware this goes against all current thinking, but I always want to know why, and figure things out for myself; it's just how I work.

Good point of view, i didn't had think about this break even yet.