How to calculate profit

[chief444]

Right but you don't just subtract 3 SB's for each from the total EV. AK's overall EV for the hand will be just under 4 and KQ's will be 4-AK's EV and just barely positive using the 10% and neglecting redraws. In actuality KQ is calling incorrectly so AK's EV must be just greater than 4 and KQ's is a little less than 0.

[SeaEagle]

Ack. You are right. I miscalculated the implied odds. I was wondering how the implied odds could be so big that they made the flop call correct (I rationalized this to myself by saying the implied odds had AK making 3 huge mistakes on the turn and river). But, in fact, on the turn KQ gives up his .6sb equity in the pot by folding 90% of the time so the implied odds needs to be 6sb*10% - .6sbs*90%, or .06sbs. i.e. KQs expected gain after the flop betting is .06sbs.

This makes the overall EV 3.6+1.8-.06 or 5.34 for AK. When you subtract the 3sbs he invested into the hand (you have to do this) his overall EV is 2.34sbs.

In other words, if I played you with this hand 100 times (and I got AK), I'd expect to profit, on average, 234sbs. If you folded to the flop bet like you should, I'd expect to make 200sbs.

Aaaand, if I just checked the flop for some reason, I'd expect to make 154sbs.

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In actuality KQ is calling incorrectly so AK's EV must be just greater than 4 and KQ's is a little less than 0.

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Right. If you want to look at it from the flop onward, then AK will make 4.34 on average and KQ will lose .34sbs.

Edit: added the check=154 number.

[SeaEagle]

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EV of AK = .91*4 + .82 + .91*.09*6 + -.09(6) = 4.46 + .4914 - .54 = 4.4114 SB

Oops! Different answer, and quite wrong I believe.


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Heh. Just rereading this thread. I'd guess 4.41sbs is darn close to AKs EV if the J only turns 9% (instead of 10%).

[elindauer]

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What about the times when a jack hits on the turn and the river is a queen?

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You're correct Harv, but for simplicity sake, let's ignore this. It only complicates the math without changing any of the conclussions.

Thanks,
Eric

[elindauer]

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Are you saying that the EV for a gutshot draw is positive in a 4sb pot?

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With the stated assumption that AK will lose 3 BB every time hit, yes. He's just barely getting odds to call at 11:1.

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Or are you saying that the gutshot's "piece" of the pot is .08 and that AK's "piece" is 3.92?

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Yes, I'm also saying this. Given the assumptions stated about how these players are going to play their hands, the 4 SB in the pot get divided up 3.92 for the AK, .08 for the gutshot.

[elindauer]

SeaEagle, you are proving my point.


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AK will win, on average... 4.8sbs

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Ack. You are right... This makes the overall EV... 5.34 for AK.

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You've now taken several cracks at this very simple situation and come up with two different wrong answers. Not only are your answers wrong, they are WAY wrong. Not only are the WAY wrong, they are getting WORSE as you continue to "fix" your estimate.

This just shows how much more confusing your view is. It's not easy to calculate things your way. Here's how to do it, with your request that we make a J hit 10% of the time:

When KQ misses (90%), he loses 1 SB. When he hits (10%), he gains 11 SB. Therefore:

ev (KQ) = .90 (-1) + .1 (11) = .2 SB
ev (AK) = pot - ev (kq) = 3.8 SB

This is the answer. Wasn't that simple?

Good luck.
Eric

[elindauer]

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Heh. Just rereading this thread. I'd guess 4.41sbs is darn close to AKs EV if the J only turns 9% (instead of 10%).

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You could guess that, but you'd be wrong.

[elindauer]

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This should be .09(11) - 1 = -.01 right? He's getting a 9% chance to win 11 SB's for the cost of 1 SB.


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No. You're ignoring the fact that when he wins, he gets his SB back, so he only loses his flop call 91% of the time.

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AK's EV if KQ calls is just 4+.01 = 4.01. If KQ is calling incorrectly then AK's EV certainly can't be less than if KQ folds right?

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KQ is calling correctly. This is just a function of your missed math above though.

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91% of 1 SB = .91 SB's. I'm not sure where the .82 comes from...

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No. Here's an obvious counter-argument. If you bet and are called and you have only a 50% chance of winning, you don't make any money. The thing you are ignoring here is that, while the AK wins .91 SB of his opponent's bet, he loses .09 of the bet he put in himself.

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...or why you're comparing it to checking.

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I'm comparing it to checking because this is the method of calculating profit that is "typical" and was defended in the other thread.

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I'm not sure what this is... the EV of betting is the pot...


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Your change increases this term, which makes the calculation even MORE wrong than before. You'll have to defend this.

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Basically, I'm just not making any sense of the second half of your example...

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Yes, isn't it confusing? That's the point. Looking at things the way I did in the 2nd half is very difficult. Looking at things the first way is very easy. We should use the first way.

Good luck.
Eric

[SeaEagle]

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When KQ misses (90%), he loses 1 SB. When he hits (10%), he gains 11 SB. Therefore:

ev (KQ) = .90 (-1) + .1 (11) = .2 SB
ev (AK) = pot - ev (kq) = 3.8 SB

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Heh. I must stop drinking and posting.

You are correct. And if you look at my original post, it's correct as well (AK will average a profit of 1.8sbs on this hand). The implied odds are so huge that they overcome the "mistake" of KQ calling the flop bet.

FWIW, my "fix" introduced the same problem you had in your "wrong" answer in your original post: I double-dipped KQ losing his equity (or you would probably say that I didn't account for KQ winning the entire pot 10% of the time).

When I sober up in the morning, I'll have to think on this further.

p.s. Did you know when you made up this example that KQ was in fact correct in calling a 10-1 draw only getting 5-1? This is pretty unintuitive and I almost didn't post my original (correct) post because I couldn't believe this was true.

[SeaEagle]

Ok, now that I'm able to think coherently about this.
Either we are doing the same thing in almost the same way, or you have something cool as hell.

First let me paraphrase your approach to make sure we're communicating on the same wavelength.
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ev (KQ) = .90 (-1) + .1 (11) = .2 SB

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In English: EV is the equity you have in the final pot minus any immediate investment (I switched the order of the terms since I think it flows easier). Yes?

So the part that's different from my approach is that I calculate immediate EV but don't have a way to add in the implied odds and usually just fudge them (or screw them up as I did last night). Your approach has the implied odds baked in - if we can only figure out what the final pot will be and how often you'll win it, and if there's any further investment required to win the pot. If you have that part down, then your approach has the benefit of being far more accurate than the traditional approach. In the OP hand, your calculation correctly showed a +EV for calling while mine would normally only look at immediate odds and show a solidly -EV (-.4sbs assuming no implied odds). Obviously, the improvement in accuracy you can achieve would make a huge difference over time.

So how do you do that? Let's add just a smidgen of complexity to this hand. Let's say there's a 3rd player (P3) who acts last and has unknown cards. We can assume that KQ is still drawing to the nuts with no redraws so that he'll still win the pot 10% of the time, but now AK is sharing the other 90% with P3. So, if P3 folds to the flop bet, all the calculations are the same as above. But P3 may or may not raise, and if he does AK may or may not reraise. How do you calculate the EV if we introduce this small complication?