# The hidden evil of the EV-

Playing positive EV situations is the key to becoming successful at poker. According to Sklansky, every time one plays EV+ he/she wins something (regardless of the outcome of any particular bet) and every time he/she plays on negative EV, he/she loses something. This is pretty straightforward and it doesn’t take particularly deep thinking to comprehend.

What I’ve just discovered the other day though, is that EV- is a far bigger evil than the above statement implies.

To make a long story short: **every time you play a negative EV hand in poker, your loss on it becomes worse than on a similar EV- hand played before.** That means every time you play a negative EV situation, you lose a little bit more on it than you have on the exact same situation before.

Sure, I understand that EV- is not the same on every EV – hand you play, but we can suppose that it is because we can always calculate the average EV-/hand, for a certain range of hands, and the above theorem remains valid. Let’s consider 10 consecutive EV- situations played by John. John is a beginner, and he has no idea how to spot EV+ so he plays all his starting hands, and he plays them all bad.

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X1 is the negative mathematical expectation on his 1st hand, X2 on his second and so forth up to X10. If we calculate an arithmetic average on all his X-es we get X1+X2+…….+X10/10 = Xm.

Xm describes the average negative EV on a hand he plays (out of the 10 considered).

Let’s say he comes into the game on a $10 bankroll, and on all the 10 hands that he plays, he wagers exactly $1 (this is not a real life situation, of course, but we’ll use $1 wagers to make the math easier – we can calculate an average bet/hand the same way we did with the EV-).

On his first hand he loses Xm%, on his second hand he loses Xm% again, and so forth, all the way to the 10th hand. Because playing negative EV value hands doesn’t mean he’ll lose every time, it’ll be fair to assume that John wins a couple of pots.

He won’t stop of course, and he’ll re-invest his winnings in the form of a few more negative EV wagers. Let’s say he makes 5 more $1 bets, with the same negative EV (Xm%). That means that after his 15th hand, he’ll have lost not Xm% of $10, but rather Xm% of $15, which is obviously more.

With that in mind, we can safely assume, that after having made his first 10 bets (the equivalent of his bankroll), every new bet he makes with the same Xm% negative EV, will actually carry bigger and bigger losses for him more than Xm%.

This is what the casinos call the house drop or the house hold. John will be giving that ‘hold’ up to his opponents. Since John could’ve come into the session on a bankroll of $1, it is safe to suppose that he’ll be paying more and more and more on every negative EV $1 hand he plays, after each negative EV hand. The longer he plays, the more he’ll be giving up on each individual hand.

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If you weren’t sure why you were supposed to do everything in your power to avoid EV- I hope this article has made it clear for you that it will nibble away at your bankroll in ways you would never have imagined.

This is the ‘house drop’ phenomenon in poker, only John won’t be ‘dropping’ it to the house but to the entity formed by the opponents who ‘milk’ him.

I know what you’ll say… you’ll say that wagers vary wildly on different hands and the EV- does too. All you need to do in that case is use the formula I used above for calculating an arithmetic average for the bets placed/hand, and the EV-/hand, and you’ll be able to apply the whole algo based on those values.

The EV- doesn’t just rob you of the money that you can mathematically track in a direct manner. It robs you of much more over time.

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