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  #1  
Old 10-18-2005, 07:55 PM
Vincent Lepore Vincent Lepore is offline
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Default Can one overcome a -EV game?

I made a post on another forum in which I claimed that without betting caps on a game like Craps a double up betting system could be used to eventually ruin a Casino that has a finite bankroll.

A poster in response, called me an idiot (might be right), and said that no huge bankroll could overcome a -EV game? I believe he is wrong and I am right. I would like some help from you probability guys to decide this issue.

Vince
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  #2  
Old 10-18-2005, 08:00 PM
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Default Re: Can one overcome a -EV game?

Define 'overcome'

If your goal is to bust the casino, you can make the chances of this anywhere up to (but not including) 1 with a finite bankroll

Without involving infinities it will still always be -EV

edit: assuming no table limit
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  #3  
Old 10-18-2005, 11:56 PM
Vincent Lepore Vincent Lepore is offline
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Default Re: Can one overcome a -EV game?

[ QUOTE ]
Without involving infinities it will still always be -EV

[/ QUOTE ]

Well the point is not to change a negative EV to a positive EV. The point is that involving infinites on the side of the -EV player one can take all of the Casino's finite bankroll. I do not know the math or the xact amount necessary but I would bet that one could come up with two reasonable finite numbers (one for the Casino and one for the opponents) would yield a > %99.999999 probability that the -EV player would win all of the Casino's money using a double up system

Vince
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Old 10-19-2005, 12:50 AM
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Default Re: Can one overcome a -EV game?

[ QUOTE ]
[ QUOTE ]
Without involving infinities it will still always be -EV

[/ QUOTE ]

Well the point is not to change a negative EV to a positive EV. The point is that involving infinites on the side of the -EV player one can take all of the Casino's finite bankroll. I do not know the math or the xact amount necessary but I would bet that one could come up with two reasonable finite numbers (one for the Casino and one for the opponents) would yield a > %99.999999 probability that the -EV player would win all of the Casino's money using a double up system

Vince

[/ QUOTE ]

See my post
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  #5  
Old 10-18-2005, 08:28 PM
yellowjack yellowjack is offline
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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.

[ QUOTE ]
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


[/ QUOTE ]
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  #6  
Old 10-19-2005, 12:21 AM
Vincent Lepore Vincent Lepore is offline
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Default Re: Can one overcome a -EV game?

Suppose the Casino has a $1 billion dollar Bankroll. Suppose I have a 500 trillion dollar bank roll. Both a lot less than infinite. I bet my bankroll on the pass line of their dice table until one of us is broke. What is the probability that I break the Casino before they break me? I bet that is close enough to %100 in my favor as to be accepted as the expected result. For a bunch of mathematicians I don't understand how you guys cannot accept the power of large numbers. Don't forget the Casino has no Max on their bets so I just pick a number large enough in comparison to their bankroll that makes it a near certainty that I win. It doesn't have to be an infinite number. The casinos also know that there are people in this wolrd that have bankrolls that can cause extreme damage and even destroy them if they do not cap their bets.

Vince
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  #7  
Old 10-19-2005, 12:56 AM
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Default Re: Can one overcome a -EV game?

Hi Vincent-

I'm going to make a nice assumption about you and think you aren't a troll, despite a flood of silly posts today by you. Here's my take on the matter:

If Casinos didn't cap their bets, yes, theoretically a gambler with a giant bankroll could ruin them. Let's make this easy: Assume the game gives you a 40 % chance of winning, so it's -EV. If you have a bankroll of $1.023 trillion and the casino has $1 billion, you start by wagering $1 billion. If the casino wins, you wager $2 bil, then $4, etc, until they are broke or you are. The casino would have to beat you 10 times in a row to take your $1.023 trillion bankroll. (2^10) With a 60 % chance of beating you in any given game, the casino will succeed at this venture .6 % of the time, or 6.05 times out of 1,000. (.6^10) So, 6 times out of 1,000, the casino takes $1 trillion from you. 994 times out of 1,000, the casino gives you $1 billion. This is a net loss of $5.05 million for you, per hand of blackjack or whatever it is we're playing.

This seems like bad odds for you, and I'm sure billionaires would be lining up to bankroll the casino. This is why this theorem doesn't work. Casinos cap their bets not because it's profitable, but because they don't have a line of billionaires waiting to back them. But if someone set up an explicit deal such as yours, I'm sure there'd be a long line [img]/images/graemlins/smile.gif[/img]

This, by the way, is why Negreanu's strategy of a massive NL buyin isn't fundamentally a bad thing for the players. Whether one should be playing against Negreanu is another story, but if he's making poor plays, then it's +EV for a good player to take him on, as long as the player keeps a good amount of their bankroll off the table at any given time.
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  #8  
Old 10-19-2005, 01:12 AM
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Default Re: Can one overcome a -EV game?

Exactly, the casinos have a BR in respect to the bets they allow the people to wager so there is hardly and ROR, just like us poker players. But Vincent was talking about the Martingale system not just trying to crack them in one shot.
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  #9  
Old 10-19-2005, 01:31 AM
Vincent Lepore Vincent Lepore is offline
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Default Re: Can one overcome a -EV game?

[ QUOTE ]
But Vincent was talking about the Martingale system not just trying to crack them in one shot.

[/ QUOTE ]

Yes, I did mention Martingdale but I also said a double up system. I believe that if you allow an inifinite bankroll then the martingdale system will also work to bust a finite casino bankroll.

Vince
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  #10  
Old 10-19-2005, 01:26 AM
Vincent Lepore Vincent Lepore is offline
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Default Re: Can one overcome a -EV game?

Thank you for the post it was very well done. I accept your conclusion especially the one I was looking for:

[ QUOTE ]
If Casinos didn't cap their bets, yes, theoretically a gambler with a giant bankroll could ruin them.

[/ QUOTE ]

You may call my posts silly but anyone reading them should have gotten the point that my comments were based on "theory". But instead I got a rash of insults. You at least understood that theoretically large numbers are very powerful.

I reread Negreanu's comments on massive bankroll. I will agree that there is no inherent mathematical advantage to a massive comparative buy given the conditions he outlines But there are advantages that a skilled player can develop through proper use of his massive bankroll. As long as the skilled player is making good plays a skilled opponent will not gain anything but the massive stack will likely prevent a him from making marginal +EV calls. Mainly because when making a decision the more likely the call is close the less likely a good player will make it. That is an advantage but not really quantifiable.

Another factor that will give the massive stack an advantage is that there are players that rarely, if ever, quit when they double up once or even twice if they havent been playing for a long period. Now the massive stack will always have them covered and be able to turn an earlier loss into a win. Even the greatest players make mistakes and also lose close decision. The massive stack will be able to always be in a position to take all of there chipsvalmost without regard to how much they have built their stack when this happens. Negreanu does not give enough credit to the psychological part of poker and the thought of losing ones buy in in a single swoop. Negreanu claims that poker is a people game yet points to mathematics when it is to his advantage. Yes there is no way to quantify the psychological effect that facing losing everything one has in front of them has on their game but I bet it is a substantial consideration to most solid poker players.

Vince
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