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Old 10-18-2005, 08:28 PM
yellowjack yellowjack is offline
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Join Date: Nov 2004
Location: Vancouver, Canada
Posts: 263
Default Re: Can one overcome a -EV game?

You can't. Below is a response I made in response to blackjack, but the same trend occurs.

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WWhen you play for $1, it's -EV
When you play for $2, it's -EV

Even though you're doubling your bet each time, it's still -EV

I don't know any strategy for BJ, but I do know that surrendering, doubling down, etc. are only moderately +EV in comparison to the overall -EV in playing. To simplify this, I'll put the probability of winning a hand at 40%, even though I think it is a bit generous (esp. online)

Suppose you're playing a max of 4 hands, where you go from $1 to $2 to $4 to $8 if you lose, and you're stopping at $8 (we'll see later that the max amount doesn't matter, it's still -EV). Let's say you have a 40% chance of winning a hand. If you win at $1 you just play for $1 again.

Note that the win is always just +$1 since when you're playing for say $4, you have already lost $3. So your net gain is +$1.

EV = 0.40*(+$1) + 0.6 * [0.4*(+$1) + 0.6 * [0.4*(+$1) + 0.6 * [0.4*($1) + 0.6(-$15)]]]

EV = $0.40 + 0.6 * [$0.40 + 0.6 * [$0.40 + 0.6 * [$0.40 + (-$9)]]]

EV = -$1.0736 (OP should check this)

So in words, even though the chances of losing 4 bets in a row is small, when you do lose 4 bets in a row it costs a lot. It's worth noting that the more bets you're prepared to make, the worse your EV is.

Take for example just making $1 and $2 bets.

EV = 0.4(+$1) + 0.6[(0.4)(+$1) + 0.6(-$3)]
EV = -$0.44


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