Roulette?

[Pat]

I have a question about playing Roulette. Assuming you have an unlimited bankroll (I know this is impossible, but bear with me.), is it statistically possible to win while playing Roulette?

Assume that you have an initial bet of $10 and are betting on "red" every time.
If you win then bet $10 the next time.
If you lose double your bet until you win.
After you win go back to the original bet of $10.

Assuming you have unlimited bankroll, will this work?
Just curious,
Pat

[Teddy]

Although I am by no means an expert on casino gambling, just two quick points.

First, your odds of winning aren't 50-50 (since there is a double zero and a zero on most tables)

Second, tables have max bets for this exact reason. A typical table where I'm from may have a min of $20, and a max of $1000.

Just quickly..
20, 40, 80, 160, 320, 640, over..
That's 6 spins in a row red and you are screwed.. The probability of that happening is approx (.53)^6.. or about 2%..

I've heard it said that Roulette simply can't be beat, and I think it's true.. That being said, there's just about no easier way to have a good time at a table without having to pay attention at all.. ie: Take the wife. /forums/images/icons/smile.gif

[Wildbill]

Roulette is beatable if the wheel is biased, but not for any other reason. In your theory, well I suppose you would never lose because you would eventually get a win in the millions of dollars bet size. Once again, bet limits would get you in any reasonable example. If you played something like you said, at some point you would hit the nastiest losing streak and the owner of the casino would suddenly say "ok, game over" as you hit the worst run imaginable and he would be happy with your thousands of dollars.

[Terry]

This is known as the Martingale system... double up until you win... or go broke.

Much has been written about it. Try a web search.

The definitive answer to your question is<font color="red"> NO </font color> , Roulette cannot be beaten without a sophisticated (almost certainly electronic) method of detecting a biased wheel. Such systems are very expensive and hard to come by. They are also illegal in some places.

[Nick B.]

Yes you could win if you doubled up after every loss. Only if you had an absolutely unlimited bankroll and no table max. Unfortunately in the real world nobody has an unlimited bankroll and table maxs do exist, so you will lose.

[youtalkfunny]

[lorinda]

You would also require infinite time, which unfortunately is just as unlikely!!

Lori

[offTopic]

Others have (correctly) answered the question with regard to the Martingale system and casino roulette.
If you'd like an interesting read on a group's attempt to "Beat the House", check out Thomas Bass' book "The Eudaemonic Pie". Don't try this at home!!!

The Eudaemonic Pie (http://www.thomasbass.com/work13.htm)

[Trefo]

High stakes you would have to play and even then there's still one zero on the table, and again there's usually a max on the bets......though i hear Binion's Horshoe in Vegas will honor any bets.

[lorinda]

If you'd like an interesting read on a group's attempt to "Beat the House", check out Thomas Bass' book "The Eudaemonic Pie". Don't try this at home!!!

Originally published under title "The Newtonian Casino"

Lori

[Zoe's Echo]

Pat,

I - ur - unfortunately spent the better part of the six months ending in September to devise and implement just such a theory.

Utilizing a random number generator, I calculated 45,000 spins to identify streaks.

The results indicated that if you staggered your initial bet until at least a string of 5 consecutive outcomes had appeared - odd/even, high/low or black/red and then started small that you could print money like an ATM. These five spins could not include any Zeros.

The methodolgy that I developed included being compensated for each spin that I played so for instance: if the limit was $10 then the betting would be as follows: 6th spin $10, 7th spin $30 ($20 to win back my two $10 bets and an incremental $10 of profit), 8th spin $70, 9th spin $150, 10th spin $310; 11th spin $630, 12th spin $1,270. This is where it got tricky.

Most tables in AC have a $2,000 "even bet" limit. Therefore, profitability stops at the 12th spin. However, my probabilites showed that it was mathmatically correct to play the 13th and 14th spins at $2,000 to limit the downside. (If you quit after the 12th you were out $2,470).

The Zeros that everyone keeps speaking of above are non-issues as long as you select a 0/00 table. These tables give you 1/2 of your bet back on the even bets and tend to extend your profitability to 13 and even 14 spins (the earlier in the streak the better).

The model indicated that 99.71% of streaks would solve within the 12th spin. Though there would inevitably be losses it appeared that the long term would be quite profitable.

Armed with this knowlege and $10,000 I headed to AC. I played at the Taj ($2,000 table limit) with 16 tables spinning in one pit. In my first 8 hours of play I cleared $2,200. Only once did I get to the 12th spin. /forums/images/icons/smile.gif

When I compared the probablities I was surprised that though there was much more profit expected from the number of "plays" experienced. The reasoning was that the probability on the actual casino wheel was skewed to the bottom. In other words, many more 9, 10 and 11 spin durations then under random normal probability.

Also contributing was the fact that you can't be at every table on the sixth spin - so you might not get to until the 11th spin. In this case, I would bet the expected profit for an 11 spin sequence ($60) and be profitable much later into spins. There was no way to account for these opportunities in the random number generator or my expected profit model.

On my second trip I experienced a 13 spin streak in which I had my first loss, after rebounding to $200 ahead I had the misfortune of being in a 14 spin streak. However, I only lost a minimal amount of cash as I entered after 7 spins.

I finished the night up $800.

However, after looking at the numbers what was a 99.7% proability of solving in 12 spins had fallen to 97.3%. /forums/images/icons/confused.gif Though the 2.4% adjustment might not seem like much it had a tremendous impact on profitability as the absolute losses are much higher once the streak goes past 12 spins.

Over Labor Day, the wheels came off. In one night I saw a 13 and a 14 at Bally's. After shaving off, $4,000 from my stack I revisited my options.

With the new probablities it would take a higher table limit to make a long term go of the system. I knew that Vegas has $10,000 limits at the Bellagio but 1) I live on the east coast and 2) you need a $10,000 seed on a $2,000 limit table on the $10,000 you need at least $40,000 of walking aroung money. It needed to be cash because if you got stuck in a bad run you couldn't ask them to hold the wheel while you go to the cage.

I found a compromise at Ceasar's in AC - though they only have nine tables in the pit they have a $5,000 table maximum. I did not up my stake but kept the same $10,000 level. It was better as spin number 13 was now profitable but the downside of going to the 14th and 15th spins was much darker. My entire bankroll could be lost in one sequence.

And on Labor Day, one fine streak of 15 consecutive blacks did me in! /forums/images/icons/frown.gif

Now I am not prone to conspiracy theories - but the streaks in the actual casinos are much longer than statisical evidence backed by 45,000 spins. I will say that whether the human element of the dealers cause the streaks or when there is big money out they can "call the ball home" to a specific place - it doesn't matter.

Without a huge bankroll (and the balls to carry it around), a $10,000 table maximum (still only profitable to 15 spins) and some luck - you will suffer losses and they will eat away all of your profit and then some. The wheel people talk of spins of at least 20 consective results. Say good bye to your Vegas bankroll there.

A new theory came out of the research but remains untested at this point.

Good Luck!

[Ed Miller]

No. That is a losing strategy... as is placing any bet at Roulette.

[Ed Miller]

Geez... I'm sorry... you've clearly spent a lot of time thinking about this, but you should know that you are dead wrong. Every bet you place at a Roulette wheel is -EV. It doesn't matter how you place your bets or in what order or magnitude... they are all -EV. It doesn't matter if you "miss a spin", if you double your bets, if you half your bets, if you flat bet. They are all independent trials and you are a favorite to lose each one of them. Your system is destined to lose your entire roll if you play long enough.

This Martingale system is akin to laying odds. You can look at it like this... you are betting $1050 to win $1 that something that is 1000-1 against won't happen. If you do this for a short period of time, you are by far the favorite to come out a winner... as most likely the 1000-1 shot won't come in and you will simply collect $1 on each bet. When the 1000-1 does come in, though, you lose everything you've gained and more. You will lose your entire roll if you do this longterm.

[Ed Miller]

Yes you could win if you doubled up after every loss. Only if you had an absolutely unlimited bankroll and no table max.

This is not correct. Even with an unlimited bankroll and no table max, you are still destined to be a loser. While the chance that you will have an infinite streak of losers is infinitesimal, if you multiply that chance by your infinite losses, you will come out behind at exactly 5.26% of your total action.

Think of it this way... say you bet $105 to win $1 that a 100-1 shot won't come in. Clearly a bad bet. Now make the bet $1050 and the odds 1000-1. Still a bad bet. Now keep adding zeros until the numbers go to infinity. It is still a bad bet, no matter how many zeros you add (even though, in the limit, the chance of losing your bet becomes infinitely small).

[Terry]

Hmmm. It sounds as if you have a brain and some patience. I suggest you use both of them to read a book or two on probability as it relates to gambling.

If, when you say random number generator, you mean you programmed a computer to test your system, I suggest that there is a flaw in your program. If the program were written correctly, any betting progression system for roulette would always fail. You will have streaks of 16, 17, 22 losers. You will.

Testing systems for games which have a negative EV on all bets can be entertaining and/or enlightening, but it can never be profitable. After some study, when you are firmly convinced of this fact, time spent studying beatable games becomes even more entertaining and profitable. /forums/images/icons/smirk.gif