Can one overcome a -EV game?

[Vincent Lepore]

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

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

[yellowjack]

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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[SheetWise]

You'll find no end of posters who literally believe (from The Doctrine of Chances) that "anything that can happen, will happen" -- and will use abstractions such as infinity to "prove" it. Since your question states "eventually ruin a Casino that has a finite bankroll", it could be proven with 99.999999... certainty.

[pzhon]

[ QUOTE ]
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.

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Your bankroll is also finite.

You have a probability of winning. That probability is less than 1. Your expected value is not positive.

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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.


[/ QUOTE ]
Your friend is right that you can't produce a positive expectation by a betting system. You need to have a +EV game. This has been covered many times before.

Imagine your task is to write down numbers that add up to 100. You can be as creative as you want, you can even use a random number generator, but you can only write down negative numbers. This is impossible for essentially the same reason you can't beat a -EV game by a betting system. I hope that makes it more intuitive to you.

Infinite (dream world) Yes, because it's never ending. All the money in the world today, NO! So it doesn't really matter but;

Consider this, there are world records for winning and losing streaks recorded in the history of gambling. For example, games like craps, roulette, BJ, etc. Think about how many times the games have been played. "IF" one could figure out how many hands of BJ have been played vs. the longest losing steak recorded should reflect what the odds say should happen. 1,000 hands= (1) 10 hand losing streak, 1,000,000 hands= (1) 20 hand losing streak, etc. Not exact but you get the point. Sooner or later the streak will happen if you play enough and will eventually be more than all the money in the world (not infinite) and the world will go broke except for one person. Sooner or later gambling winning streaks and losing streak records will be broken once enough events have happened which is never ending. Though it will take many life times to see this.

Goodluck

[Vincent Lepore]

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Without involving infinities it will still always be -EV

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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

[Vincent Lepore]

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

[ 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

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See my post

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.