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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