Don't really understand how the "$EV won" in HEM can be greater than the 1st prize money..
Like in $12 9 player sng.. The 1st prize is $49.50.. I have one tourney where my $EV won is greater than $49.50 (exactly $52.05).. Doesn't really make much sense to me...
You can also lose more than 1 buy-in in the AIEV line ♦
Just imagine you get it in tons of times in a tournament and you're always ahead, but you lose many of the showdowns. Your EV will keep going up forever as long as you always get it in good (according to ICM obviously) and don't bust.
On the other hand you can also keep getting it in badly and winning and therefore lose more than 1 buy-in in your AIEV.
but my expected value for a $12 tourney can never be less than -$12??? that makes no sense at all...
I probably sound dumb as fuck... but well I gotta ask to learn something lol 
Your $EV can be -10 buyins in theory, there is no boundary -.-
If you bust in first hand while getting it in on the river HEM will show your $EV at 0. (which is -12$, because you payed the buyin)
shouldn't the boundary be the entire prize-pool?
If you bust every opponent in just 8 hands (9 handed tourney) and your opponents are always drawing dead when they are all in. Would your EV be 100% of the prize pool?
Originally posted by elhh82
shouldn't the boundary be the entire prize-pool?If you bust every opponent in just 8 hands (9 handed tourney) and your opponents are always drawing dead when they are all in. Would your EV be 100% of the prize pool?
yea that makes sense...
Originally posted by malosanmaka
Your $EV can be -10 buyins in theory, there is no boundary -.-If you bust in first hand while getting it in on the river HEM will show your $EV at 0. (which is -12$, because you payed the buyin)
on the other hand this makes no sense...
The only time your chips are worth the buyin is when everyone has an untouched starting stack no?
I'll let pzhon explain this better, but some of you are confusing HEM's $EV (luck adjusted EV) with current equity in a sng (according to ICM). Those 2 things aren't the same.
Originally posted by elhh82
shouldn't the boundary be the entire prize-pool?If you bust every opponent in just 8 hands (9 handed tourney) and your opponents are always drawing dead when they are all in. Would your EV be 100% of the prize pool?
HEM's $EV would be equal to first prize (4.5 buyins). It is calculated as "prize won" + "luck adjust". Because there weren't any all ins <100% "luck" would be 0.
Originally posted by 8979687
The only time your chips are worth the buyin is when everyone has an untouched starting stack no?
In first hand your 1500 chips are worth exactly the buyin. (minus the fee)
But you can easily have a situation later on where your chips are worth aprox. 1/9 of the prize pool = 1 buyin.
For example chip counts of 4450, 4000, 4000, 1150(HERO). Here your 1150 chips are worth close to 1/9 of the pool or ~1 buyin.
I wrote an article on this phenomenon in backgammon. The title was "Unbiased Nonsense." Basically, if you are adding up the results of 100 tournaments, it is ok for the value given to a few tournaments to be greater than first prize, or less than losing the buy-in, even though those seem not to make sense. If you "correct" these, then you actually introduce a bias so that your adjusted results might no longer converge to your skill advantage. This is a worse problem than getting a silly-seeming result for one tournament, so it is reasonable to accept the luck adjustment.
Imagine that you play a freezeout starting with 2 chips each, and take a coin flip for 1 chip at a time. The all-in luck-adjustment lets you factor out some of the luck, when you lead 3:1 or trail 1:3, but ignores the luck when you are tied 2:2. So, the sequence win-win gets adjusted from 100% wins to 75% wins, since the second win is visible and the luck from winning is factored out. The sequence win-lose-win-lose-win-win gets adjusted from 100% to 125% wins because the luck adjustment sees that you had 2 losses and 1 win while you were all-in, which appears unlucky by 25%. The luck adjustment cancels this bad luck by adding 25% to the result of 100%. The all-in luck adjustment overlooks the 75% luck you had while not all-in.
Over all, adjusting for part of the luck in an unbiased fashion is better than not adjusting, even if it sometimes produces some odd results. Even though the adjusted results for a particular tournament can be outside the reasonable range, the overall variation in the adjusted results should be significantly smaller than in the unadjusted results. It takes fewer tournaments to estimate your ROI within 5% if you use the adjusted results.
Originally posted by malosanmaka
I'll let pzhon explain this better, but some of you are confusing HEM's $EV (luck adjusted EV) with current equity in a sng (according to ICM). Those 2 things aren't the same.Originally posted by elhh82
shouldn't the boundary be the entire prize-pool?If you bust every opponent in just 8 hands (9 handed tourney) and your opponents are always drawing dead when they are all in. Would your EV be 100% of the prize pool?
HEM's $EV would be equal to first prize (4.5 buyins). It is calculated as "prize won" + "luck adjust". Because there weren't any all ins <100% "luck" would be 0.
you're right sir. so if you actually run bad and still ship the tourney. For example, if you lose a few flips (so you're under ev) but manage to win it all in the end with you opponent putting in money when drawing dead (even ev), then your EV prize would be more than first prize..
I hope this is correct now.
all makes sense now... so It's pretty much that having the amount you can win or loose "uncapped" gives a better view of your real roi... than the more "Real" approach of having it capped at loosing 1bi.. and winning what the 1st prize is worth..
ty for your reply phzon and the other dudes 
