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young nut
10-19-2004, 03:28 PM
Recently I had a discussion with one of my buddies about gambling in casinos. We got into the argument of what games are best to play, as in which games have the highest EV. I kept arguing that poker was had the highest +EV for me because I felt I had an edge because of my poker knowledge and strategy. He said that craps or black jack would be the best bet, simply because they have the least house advantage.

He then began to tell me a story about one of his friends father who took trips to vegas constantly. According to my buddy, he was told by his friends father about a roulette strategy that was a garunteed winner every time. I told him that this was statistically impossible, but he explained anyways:

He said the strategy goes like this:

Place the minimum bet that you want to gamble with on RED.
if you lose, double that bet and play RED again.
Keep doubling your bet until red hits, and then go back down to your minimum bet.

He explained that no matter how often you won loss, each time the wheel struck red, you would be exactly +1 more minimum bet.

I thought about this strategy and began to think he was right, but my instincts kept telling me that there had to be a flaw somewhere in this strategy. But the only two flaws I could come up with are these:

1. You would need a very very large bankroll to play this strategy to its fullest. Because given that your bets are doubling each failure, after 10 or so failures you are placing a very large bet on the table.

2. There will be a time where the wheel hits BLACK thirty times straight (although statiscially very rare, it will happen eventually) And by the 30th losing bet, you will be placing an enourmous amount of money on the table just to win back your initial small bet.

does anyone see other flaws in this strategy. I keep thinking to myself, if I had an enourmous bankroll to play with, this strategy would work every time. Because it seems to me that the only constraint on continued success would be to run out of money to bet with. I wish I could write a simple program to test this theory, but I am not very good with computers. Any thoughts / comments appreciated.

jason1990
10-19-2004, 04:20 PM
This is a classic mathematical problem, sometimes referred to as the "Gambler's Ruin". However, it is often difficult to dissuade people of their belief once they get set on believing that it works.

First, here's the mathematics: call the minimum bet 1 unit (dollar, euro, chip, whatever). Suppose you have a bankroll of 2^n - 1, exactly enough for n bets. You then apply the strategy until one of two things happens: (A) you lose all your money, (B) RED eventually appears and you are up 1 unit. The probability of RED appearing is 18/38 or 9/19. So the probability of A is (10/19)^n. The probability of B is, therefore, 1 - (10/19)^n. If A occurs, you lose 2^n - 1. If B occurs, you win 1. So you're EV is

-(2^n - 1)*(10/19)^n + 1*(1 - (10/19)^n)
= 1 - (2^n)*(10/19)^n
= 1 - (20/19)^n.

It's easy to see from here that your EV is negative, no matter what n is. For example, suppose you enter the casino with 1,048,575 units. This corresponds to n = 20, giving you an EV of about -1.8.

But looking only at the EV is misleading. What's really happening here (when n = 20) is that you have a 99.999734% chance of winning 1 unit and a 0.000266% chance of losing 1,048,575 units. For some people, when they look at it this way, the strategy looks tempting.

But you should tell your friend's father that, when he uses this strategy, he's effectively playing a "reverse lottery". For example, what if someone offered you a free lottery ticket under the agreement that, if the numbers on the ticket do NOT come up, you win \$1. But if they DO come up, you lose everything you own (including the winning lottery ticket). This particular example is actually (for most people and most lotteries) +EV, yet I doubt very many people would be willing to gamble everything they own for a chance to win \$1, no matter how good that chance may be.

Side note: I doubt very many people would have the guts to follow this strategy through. If you lose 12 times in a row (which WILL happen if you use this strategy on a regular basis) and you're on a \$5 roulette table, you must wager over \$40,000 on your next bet! That's going to be a lot of money to anybody who's trying to win \$5 using this strategy. Besides which, most casinos have a house limit which determines the maximum size of any particular wager.

magiluke
10-19-2004, 05:18 PM
The very reason there is a maximum bet in casinos... Me and a bunch of my friends have been trying to figure out a way to abuse roulette, and the best we can do is just reduce the amount we lose, never actually getting a positive Expected Value (except through bad math).

MicroBob
10-19-2004, 07:34 PM
[ QUOTE ]
Me and a bunch of my friends have been trying to figure out a way to abuse roulette, and the best we can do is just reduce the amount we lose

[/ QUOTE ]

ummmm....the only way you can 'reduce the amount you lose' is by betting less (or not betting at all). Anything else that you THINK might be reducing the amount you lose (by varying bet size and what-not) is evidence that you are not doing your math correctly.

If you are playing inside-numbers then just look at each bet as -5.26%.
For each \$100 you bet you lose \$5.26. It doesn't matter how you vary your bet-size back and forth, etc etc. You just lose.

If you are betting red-black or evens-odds or whatever I think the percentage mihgt be 2.8% (not sure...but it really doesn't matter....you just need to know that it's -EV).

If you are determining through your calculations that you can sometimes lose more than 2.8% and other times lose less than 2.8% then try-again. It's ALWAYS the same no matter what you do.

To the original question -
some people who subscribe to the doubling-up +1 theory will poo-poo anyone who says 'yeah, but what happens when you lose 20 bets in a row?' by saying "the odds of that happening are so astronmical it doesn't even matter."

well, it can happen and it does matter.

IF you had an inifinite bank-roll and IF the casino allowed any size bet up to infinity then eventually you would have to hit a winner and you would get your precious roll plus \$1 back.
But since NOBODY owns an infinite amount of money and no casino will allow an infinite bet the idea that this strategy is a guaranteed winner is incorrect.

1st bet = 5
2nd bet = 11
3rd bet = 21
4th bet = 41
5th bet = 81
6th bet = 161
7th bet = 321
8th bet = 641
9th bet = 1281
10th bet = 2561

Now aren't you going to be feeling a bit silly betting \$2561 on a single-bet just to finish up \$1?
What if you lose that \$2561....now you're over \$5k in the hole?

For every \$1k you win in single-dollars using this system you will lose another \$1200 or so (don't know exact number and don't care) when you hit a bad run and get clobbered for a ton.

You can have fun playing this way if you like. A lot of people do.
But the casinos WELCOME people who use this type of system (it's a VERY WELL KNOWN system) because they know they are going to be net-losers....just like anyone else playing a -EV game.
I would strongly recommend not trying to make your living on this ridiculous system.

young nut
10-19-2004, 07:46 PM
I agree with both of your posts (bob and jason) and I appreciate you explaining the reasons mathematically why this strategy is -EV. I argued for about an hour with my buddy that the strategy had to be flawed, and my main reasoning was because casino's ALWAYS have a statistical edge in every game that is played against the house. He basically just used the infinite bankroll argument to counter.

But another interesting question about the infinite bankroll. Lets assume that one did have an infinite, or near infinite bankroll and that there was no max bet on a roulette table, would this strategy ever work out to be +EV? It seems to me that statistically the house would still have the edge, but on face the problem looks like with an infinite bankroll you can be a garunteed winner.

Nottom
10-19-2004, 08:19 PM
[ QUOTE ]
But another interesting question about the infinite bankroll. Lets assume that one did have an infinite, or near infinite bankroll and that there was no max bet on a roulette table, would this strategy ever work out to be +EV? It seems to me that statistically the house would still have the edge, but on face the problem looks like with an infinite bankroll you can be a garunteed winner.

[/ QUOTE ]

This was discussed in depth in a thread a few months ago. If you had an infinate bankroll and the casino would accept any bet, than the martingale system would work.

Of course if you have an infinite BR winning a bet at roulette isn't going to change anything.

MicroBob
10-19-2004, 08:50 PM
With an infinite bankroll it would work.

With a 'near-infinite' bankroll it would technically not work.

However, if you had a bankroll of somewhere around \$\$100,000,000,000,000,000,000 and were just starting with \$5 bets on red/black and trying to win \$1 extra after doubling-up AND the casino allowed for the possibility of placing any size bet you like then I would be VERY comfortable with your low risk-of-ruin on this.
But, as Nottom pointed out, you certainly wouldn't care about winning \$1 at a shot if you had a bankroll this size.

Your friend has fallen for a common gambler's fallacy.
As I said before, this is a VERY WELL KNOWN system and there are MANY PEOPLE who claim that it is a guaranteed winner. In fact, it is included in SEVERAL books on gambling.
If this system was a guaranteed winner than everyone would do it and the casinos would all be broke.

If your friend can't understand the math then perhaps the ideas that this system is included in books, lots of people know about this, and casino-personnel are trained to WELCOME players who use these systems and comp them generously might start to hel phim understand that it might not be quite as sure-fire a system as he wants to believe.

FWIW - I had heard of this double-up Martingale system in 1995 during my first EVER trip to a casino ("what's a double-down?").

Some of the less-idiotic books on Blackjack (like World's Greatest BJ Book) and others go into detail as to why these vary-your-bets type strategies won't work....but they are essentially saying the same stuff as has already been said in this thread.

goldcowboy
10-19-2004, 11:16 PM
The "opportunity cost" associated with keeping that near-infinite bankroll liquid and available for betting as opposed to having it invested and earning a return would likely overwhelm any potential "gain" associated with this betting system.

ILL34GL3
10-19-2004, 11:36 PM
Anyone who plays roulette is a moron, plain and simple. How many pro roulette players do you know? 'Nuff said.

jason1990
10-20-2004, 01:41 AM
I think I saw one on the "Vegas Challenge". He was up against a professional 3-card poker player. Boy, that guy sure knew how to bet that Pair Plus!

magiluke
10-20-2004, 03:40 AM
I mean, reduce how much we lose according to the expected value. We tried systems that used the thirds, systems that used the 50/50s, streets, different incrementing systems; we couldn't get a positive expected value, just increase it depending on where we stopped and how we bet...

jason1990
10-20-2004, 09:46 AM
I think what MicroBob was getting at is this: (1) the net EV of any two bets is always the sum of the EV's of the individual bets, and (2) the EV of any bet on the roulette table is -1/19 times the amount bet. (Well, actually the quint bet has a worse EV, but it's the only exception. Let's just assume you never play the quint bet, which is a bet on 0-00-1-2-3.) So, no matter how you place your chips on the table, your EV is -1/19 times the amount you have on the table. You cannot change this no matter what you do.

As for stopping strategies, the situation is similar. The total amount you bet in a session, call it N, is determined by your strategy and is based on the outcome of the spins of the wheel. So it's random and it has an EV, call it E[N]. Then the EV of your net winnings in the session is simply (-1/19)*E[N].

What you can change, though, is your variance. And there are many ways to do this by playing various different combinations of bets.

young nut
10-20-2004, 10:36 AM
So, in actuality, the max bet on a roulette table is what gives the house the statistical odds? (Would that be a correct assumption)

MicroBob
10-20-2004, 11:58 AM
[ QUOTE ]
different incrementing systems; we couldn't get a positive expected value, just increase it depending on where we stopped and how we bet...

[/ QUOTE ]

Different Incrementing Systems will not change ANYTHING.

Stopping where and when you bet doesn't change ANYTHING.

The more you bet, the more you lose. Varying the size of your bets does not change this.

If you think you were changing the amount that you could win or lose by doubling-up on one bet and cutting it in half the next bet, etc etc then you are very mistaken and your calculations were wrong.

MicroBob
10-20-2004, 12:09 PM
The max-bet and your inability to ever hold an infinite-bankroll makes your particular system ineffective.

But the house's statistical odds just comes from paying out less than the odds of the occurance happening.
If you bet an inside-number on double-0 roulette it will have a 1 in 38 chance of hitting yet it only pays 35:1.
If this 1 in 38 occurance paid-out 40:1 then the advantage would be for the player.

Some people look at the 0 and 00 as the reason the house has such an edge. Without the green zero's then many of the roulette bets are essentially even-money.
But really it's not the fault of the green zeros. You could keep the zero's and remove number 17 and 25 and accomplish the same result.

If we were flipping a coin and you paid me \$2 every time it landed heads and I paid you \$1.50 every time it landed tails then eventually I would do pretty well because your paying more and than you are going to take in on this 50-50 occurance.
Roulette is little more than the same coin-flipping bit just using a few more numbers.

TomCollins
10-20-2004, 12:46 PM
There was a professional roulette player on the board here a few months back. But he could watch the wheel and figure out where the ball would stop. I think hes selling the NL Party Poker strategy now.

MrBlini
10-20-2004, 05:00 PM
No, the house has excellent odds on every single bet. It makes money statistically no matter what the wager. That there are 38 possible outcomes and that the house pays as if there were 36 is what gives the house the advantage in roulette.

The maximum bet is a tool to manage variance, not expectation. It is designed to minimize the house's day-to-day variance and to protect it from ruin. (With no maximum bet, a doubling system could be used in reverse to try to break the house, and most gambling houses want a 0% risk of ruin, even if they would make more money statistically by taking a greater risk.)