Luck adjusted winnings in SNGs

[Parket]

I don't see how this can be completely wrong. And I never stated that when you're HU in an SNG, the calculation would be different. My ICM remark was when there are more than 2 players. Then you can use ICM to determine your EV when you bust.

On the contrary, you say that with the correct calculation it is possible to lose more than a BI or to win more than the 1st prize. This is complete bollocks of course. Your Expected Result in a single tournament can NEVER be less than the buy-in or more than the 1st prize. The definition says it already itself, the Expected Result of your tournament, is what you would win 'on average'. How can the average be less than the minimum or more the maximum ???

I don't have a reference to this method. It's pure common sense.

Again let me give the example. Say we play a hyper HU SNG with 500 chips each. 1st prize is $10.
H1: I win a showdown hand for 250 chips. I'm now leading 750-250.
H2: We have a preflop all-in coinflip. I'm now 50% favorite to win the $10. So from the $10 I'm entitled to $5 already. There is a 50% possibility we'll have to keep on playing for the other $5 equity. I do lose and we're back at 250 each.
H3: I again win a showndown hand for 250 chips and leading 750-250
H3: We again have a preflop all-in coinflip. I'm again 50% favorite to win. Of course I win $10 when I do, but the tournament could have ended in H2 already and EV wise there is only $5 to distribute anymore, so I'm entitled to 50% of that $5 instead, i.e. $2.5. I however lose and we continue playing for the remaining $2.5 equity, with 500 chips each again.
H4: We have a preflop all-in coinflip. I'm 50% to win the tournament for that remaining $2.5 equity, opponent has the other 50%.

So the total $EV for me in this tournament is $8.75, and for my opponent it's $1.25.
You could calculate it differently as well. My opponent needs to win 3 consecutive coinflips to win the tournament. If he had lost any of the coinflips, he would have lost the tournament. So he had a .5*.5*.5 = .125 probability of winning the $10 and a .875 probability for losing, which also leads to an $EV for him of $1.25.

If I understand correctly, the new method will simply add the cEV of every individual hand. If you do that, you will end up with a cEV=1250, meaning a $EV of $12.50.

EDIT: and I'm definitely going to read that 2+2 thread as soon as I have time, because I cannot imagine that not a single person would have posted something like that.

[Parket]

PS: let me also explain how it would be different for 3+ players.

The above example does not show you what would happen if a tournament would have continued after the last all-in ended it. In my example, the tournament will end regardless of the outcome (except if there would be a split pot - in fact, this is possible and would need to be taken into account in the $EV calculation). Anyway, if stacks were not equal, then you have no idea how the tournament would have gone on if the outcome of the last all-in was different. As a result, the only way to put a $EV to that outcome is by simply looking at stacksizes.

Let's alter the previous example by changing the outcome of H2:
H1: I win a showdown hand for 250 chips. I'm now leading 750-250.
H2: We have a preflop all-in coinflip. I'm now 50% favorite to win the $10. So from the $10 I'm entitled to $5 already. There is a 50% possibility we'll have to keep on playing for the other $5 equity. Instead of losing, I win, and the tournament is over.

Now what is my $EV ? Well, we would have continued playing for the remaining $5 equity, but we didn't, so the only way to divide that is by looking at stacksizes. We'll be at 500 each again, so the remaining $5 is divided evenly. Result, my $EV is $7.5, opponent's is $2.5.

Now this is exactly what you would do in 3+ player situations. You would use ICM calculation to divide the remaining equity if you would have still been in the tournament.
Example: 4 players left. I have 500 chips, the others have 2500 left.
H1: I have a preflop all-in coinflip and win it. There was a 50% chance that I bust in 4th and won $0. There is a 50% chance that I survive with 1000 chips and opponents have 2500,2500,2000.
H2: I have another preflop all-in coinflip against the 2000 guy, but this time I lose and am out. Again there was a 50% chance that I lost this. The other 50% chance I would have been at 2000 chips, with opponents at 2500,2500,1000.

What is the $EV of these hands ? Well, it's 50% of 50% of my ICM equity with 2000 chips.
This is basically also what my algorithm above is trying to explain (and I didn't proofread so there may be errors in it) : at each all-in, you have a branch of possibilities where your $EV is the sum of the probability of that branch multiplied by the $EV of that branch. The $EV of a branch may be a) $0 where you bust out of the money, b) an actual prize if you bust ITM, c) your ICM equity if you would have won but you actually bust, d) the continuation of the $EV calc algorithm for the remainder of the tournament in the case you won the all-in.

[Parket]

If I understand correctly, the new method will simply add the cEV of every individual hand. If you do that, you will end up with a cEV=1250, meaning a $EV of $12.50.

Sorry, this is incorrect. You will end up with a cEV=+500, or an expected $ profit of $5, which would be equal to an $EV of $10 in my calculations, as I disregard the initial BI.
In an Adjusted Profit graph, you would obviously always have to subtract the BI from the EV that my methodology (which always returns a non-negative $EV) returns. And I know, in poker terms you typically refer to EV as an expected profit, not an expected result.

[Parket]

One final remark before I let you guys fire at it .

Note that in my methodology, you can completely ignore the non all-in hands. This may be surprising, but it is not if you realize that what we're trying to do here is calculate a global expected outcome of a tournament by taking into account each possible outcome of an all-in. The reshuffling of chips in between the all-ins is irrelevant because you implicitly take it into account in the next all-in by looking at the stacksizes at that moment.

This means that calculations should go fairly quickly. Admittedly, if you sum cEV of each hand, it's also a no-brainer for non-all-in hands as you then simply take the actual chips won.

[Zangeeph]

Again let me give the example. Say we play a hyper HU SNG with 500 chips each. 1st prize is $10.
H1: I win a showdown hand for 250 chips. I'm now leading 750-250.
H2: We have a preflop all-in coinflip. I'm now 50% favorite to win the $10. So from the $10 I'm entitled to $5 already. There is a 50% possibility we'll have to keep on playing for the other $5 equity. I do lose and we're back at 250 each.
H3: I again win a showndown hand for 250 chips and leading 750-250
H3: We again have a preflop all-in coinflip. I'm again 50% favorite to win. Of course I win $10 when I do, but the tournament could have ended in H2 already and EV wise there is only $5 to distribute anymore, so I'm entitled to 50% of that $5 instead, i.e. $2.5. I however lose and we continue playing for the remaining $2.5 equity, with 500 chips each again.
H4: We have a preflop all-in coinflip. I'm 50% to win the tournament for that remaining $2.5 equity, opponent has the other 50%.

The reason that doesn't work is because it creates a bias. It favors players who are good at all ins when the blinds are low. It is against those who go all in later when the blinds are high. Because if you usually go all in later in the tournament, it matters less due to your algorithm already awarding a fixed amount of the prizepool to the player if he already went all in earlier on. What I'm saying is, according to your algorithm the order of the all ins matter. It shouldn't - that's not how poker works. If a player went all in with 250 against 750 then later on in the same tournament it was 750 to 250 in an all in, the second one would matter less even though its the exact same situation.

I think your algorithm might actually work in a tournament where the blinds remained fixed. But I'm not sure; I'd have to think about it more. Anyway, it's not up for debate. Your calculations are incorrect. There really aren't multiple 'correct' ways to do it. We already have the best way to do it (excluding calculations that use simulations of future hands). Just as there aren't multiple ways to calculate the EV of going all in $100 With AsAh vs KsKh, there aren't multiple ways to do luck adjusted winnnings.

The only way to really improve luck adjusted winnings would be to account for dead cards by estimating the range that other opponents fold (in non HU situations), but I imagine that's beyond the scope of what PokerTracker devs care about (as it would take a lot of effort to get only a small improvement).

[Parket]

Your calculations are incorrect.

Seriously LOL. I even wonder to what extent you are familiar with probability theory. If there's one thing I'm certain of, it's that the algorithm is a 100% accurate for HU SNGs. EV is simply a sum of (probability of an outcome)*(EV of an outcome) for all possible outcomes. That's exactly what the algorithm is doing.

The essential difference is in what EV you are calculating. In HEM (and supposedly PT4 in next beta), you are simply calculating the $EV of every individual hand. This is correct for each hand, but you introduce an error by then actually summing up all hands and calling that the $EV of the entire tournament. Whereas my approach is only calculating the $EV for the entire tournament, and is a 100% accurate for HU SNGs. If in the very beginning I hit a one-outer for 2% equity then the simple fact is that in 98% of the runs of the same situation I would be out of the tournament, meaning that on average I can impossibly win more than 2% of the 1st prize. If however you calculate it per hand, then you end up with high-equity hands later on in the tournament that were simply not going to happen 98% of the times. In that way, those later hands are enormously overrated/overweighed when you add everything up, because they ignore the fact that in the sampling population, they would simply not occur a majority of times. You call it 'according to your algorithm the order of the all ins matter'. Well yeah, obviously it matters! I don't know how long it's been since you last played a tournament, but the first hand in which you're all-in that you loose means you're out of the tournament and the value of any hypothetical future all-ins that you might have played if you had won the all-in, is of 0 importance !

However, where I realize my algorithm would already fall short is in 3+ SNGs. The problem is in all-in hands where I'm not involved. I could take two approaches.
1) The correct one would be to take that one into account. For simplicity let's assume a pure coinflip. Then 50% of the equity would come from the situation that actually occurred in the tournament. The other 50% however would be a situation in which we can only calculate the EV via an ICM model. Now in my case this would mean that already at least 50% of my $EV would come from the theoretical ICM calculation of that single hand, which completely ignores skill levels and the like. Do 3 coinflips like that and already 87.5% of my $EV would come from ICM calculations. So the final $EV would mostly be based on my ICM equity of the stacks of the hypothetical outcomes and a minority would be from my own all-ins. Clearly this sucks and renders the calculation almost worthless because it completely dwarfs my gain in equity of the hands that I did play. In that respect simply adding up the $EV of each hand is going to be superior.
2) Or I could simply ignore the all-ins of other players. But this is equally bad/incorrect, because we all know how someone else busting can greatly increase our equity and should therefore also be part of the $EV calculations. E.g. 4 players left and I'm the shortstack with 1 chip. 2 big stacks clash in an all-in on the turn where the largest of both hits his one-outer on the river. While I'm not involved the hand, I'm incredibly lucky since I had a 2% chance of not finishing 4th. So it is essential to include all hands, not just the ones where you're all-in yourself.

Too bad I no longer have the time nor programming skills to implement the algorithm for HU. I've played thousands of hyper HU SNGs and would love to see how much this $EV calculation would differ (or not) from HEM's implementation.

[Zangeeph]

In HEM (and supposedly PT4 in next beta), you are simply calculating the $EV of every individual hand. This is correct for each hand, but you introduce an error by then actually summing up all hands and calling that the $EV of the entire tournament.

No, try again. If it's correct for each hand then the summation of those numbers has to be correct. It's like saying "well your profit per hour is $10, but you can't say that your profit per 10 hours is $100 as you can't just add them." I really don't have the time to prove that adding numbers together can't create an error. I don't know where to begin explaining because you seem very lost indeed.

Whereas my approach is only calculating the $EV for the entire tournament, and is a 100% accurate for HU SNGs. If in the very beginning I hit a one-outer for 2% equity then the simple fact is that in 98% of the runs of the same situation I would be out of the tournament, meaning that on average I can impossibly win more than 2% of the 1st prize. If however you calculate it per hand, then you end up with high-equity hands later on in the tournament that were simply not going to happen 98% of the times. In that way, those later hands are enormously overrated/overweighed when you add everything up, because they ignore the fact that in the sampling population, they would simply not occur a majority of times. You call it 'according to your algorithm the order of the all ins matter'. Well yeah, obviously it matters! I don't know how long it's been since you last played a tournament, but the first hand in which you're all-in that you loose means you're out of the tournament and the value of any hypothetical future all-ins that you might have played if you had won the all-in, is of 0 importance!

Your algorithm explicitly favors earlier all ins on top of the fact that their importance is already accounted for without the need for mathematics on top. This is because the number of hands players play decreases on average. Eg its likely that the first 20 hands will be played, less likely that the next 20 will be played, even less likely that more will be played and so on. You're essentially doubling this effect.

Too bad I no longer have the time nor programming skills to implement the algorithm for HU.

That's a shame. I would've loved to see your reaction when you realized you created an abomination of an algorithm.

I regularly play Hyper Turbo tournaments. At one point I went 200 buy ins under Luck Adjusted Winnings (according to HEM). Without the reassurance that I was a winning player, I might have quit poker altogether. Luck Adjusted winnings are incredibly important.

[Parket]

No, try again. If it's correct for each hand then the summation of those numbers has to be correct. It's like saying "well your profit per hour is $10, but you can't say that your profit per 10 hours is $100 as you can't just add them." I really don't have the time to prove that adding numbers together can't create an error. I don't know where to begin explaining because you seem very lost indeed.

It's you who seem lost really. Again, adding all hands' $EV is simply incorrect due to the fact that in a large sample those hands won't occur all the times.

Let me give you a little experiment:
Toss a coin 3 times. If the first one is heads, you win $1 and can toss again. If you then toss heads, you win another $1, and can try a 3rd time. If you toss heads again the third time you win a final $1.
Now you play this game once, and the outcome is heads, heads, tail and you win $2. According to your logic, the $EV of this game is the sum of $EV's of every toss, meaning that the $EV is 3 times 50c, or $1.50.

Well, I'll let you play this game for only $1, so you think it's highly +EV for you. Let's play it 1000 times and see who makes the profit.

That's a shame. I would've loved to see your reaction when you realized you created an abomination of an algorithm.

Now this is a funny one for sure, because you are already making the assumption that HEM has the correct algorithm and that if my result would be different it would mean mine is wrong.

[Zangeeph]

It's you who seem lost really. Again, adding all hands' $EV is simply incorrect due to the fact that in a large sample those hands won't occur all the times.

Thats why it will work.

Let me give you a little experiment:
Toss a coin 3 times. If the first one is heads, you win $1 and can toss again. If you then toss heads, you win another $1, and can try a 3rd time. If you toss heads again the third time you win a final $1.
Now you play this game once, and the outcome is heads, heads, tail and you win $2. According to your logic, the $EV of this game is the sum of $EV's of every toss, meaning that the $EV is 3 times 50c, or $1.50.

In that particular instance when it went to 3 flips, yes my $EV was $1.50. But in many tournaments my $EV would be just $0.50.

If we play 1000 games, in a perfect example I would get HHH 0.5^3*1000=125 times. I would get HHT 125 times, I would get HT 250 times, and I would get T 500 times. In HHH my $EV is $1.5, HHT my $EV is $1, HT my $EV is $0.5 and T my $EV is $0. Let's multiply those number by how often I get them. 125*$1.5+125*$1+250*$0.5+500*$0=$437.5.

So per game, my $EV is $437.5/1000=$0.4375 or, including the buy-in of $1, a net loss of $0.5625. Now do you understand why I am correct?

Now this is a funny one for sure, because you are already making the assumption that HEM has the correct algorithm and that if my result would be different it would mean mine is wrong.

No see, the funny thing is watching you struggle over the most basic EV calculations. But the sad thing is watching you attempt to give advice to the developers of PT4 when they already messed it up the first time around.

[Zangeeph]

If we play 1000 games, in a perfect example I would get HHH 0.5^3*1000=125 times. I would get HHT 125 times, I would get HT 250 times, and I would get T 500 times. In HHH my $EV is $1.5, HHT my $EV is $1, HT my $EV is $0.5 and T my $EV is $0. Let's multiply those number by how often I get them. 125*$1.5+125*$1+250*$0.5+500*$0=$437.5.

I just realized I counted these incorrectly. The correct calculation is:
In HHH my $EV is $1.5, HHT my $EV is $1.5, HT my $EV is $1 and T my $EV is $0.5.
This is because my EV is increased by $0.5 according to how many flips I do. This means my EV is 125*$1.5+125*$1.5+250*$1+500*$0.5=$875. So my net won per game is -$0.125.