yatzr

someone once described a roulette strategy that sounded pretty good. First off, this person studied the black/red occurances and noticed that more than half of the time that there were two black, or two red, there would be another one of the same color after that. I don't know if this was just coincedance (sp?) but he said that's what he saw.

Now, he had a system. He would bet kinda small, but keep a standard bet...lets say $1. He would bet after two of one color hit (to supposedly get better than 50/50 odds). If he won, he would keep the winnings and bet a dollar at the next occurance. If he lost, he would bet a dollar at the next occurance to try to make up for the loss. If he lost again, he would raise his bet to try to break even. Once he was back to even, he would go back to the $1 bet. I may be missing something, but to me (even without the whole 2 or 3 of a color in a row thing) this sounds pretty good as it's unlikely to lose too many times in a row when the odds are close to 50/50 anyway. It would take forever and it would take backup cash, but I think it would work. What do you guys think?

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CoachAlan

Here's a simple math lesson that may help you. There are 38 spots on an American roulette wheel. When you win, you are payed as though there are 36 spots. Those extra two spots are the house's edge. In this case, that edge equals about 5.25%.

No matter what bet you make on the entire board (with the exception of the five-number 0, 00, 1, 2, and 3 bet, which is even worse) the house edge is that same 5.25%.

Thus, no matter how you split, multiply, divide, hedge, or whatever your bets, the house has a 5.25% advantage. In the short term, luck may be on your side. In the long term, the house ALWAYS wins.

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BigGov

It's a strategy based on luck.

You have the same chance of winning every spin of the wheel, it doesn't matter what happens before, or what happens after, each spin is independent of each other.

Roulette is a pure gambling game, there is no skill involved.

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One day an Englishman, a Scotsman, and an Irishman walked into a pub together. They each bought a pint of Guinness. Just as they were about to enjoy their creamy beverage, three flies landed in each of their pints. The Englishman pushed his beer away in disgust. The Scotsman fished the fly out of his beer and continued drinking it, as if nothing had happened. The Irishman, too, picked the fly out of his drink but then held it out over the beer and yelled "SPIT IT OUT, SPIT IT OUT, YOU BASTARD!"

MPower

My strategy: go play craps.

If you insist on Roulette, make it a cheap table with no 00. Sit back play little money and enjoy your free drinks for an hour or so.

yatzr

Yes, I realize the whole thing with what color comes next is flawed. That's not really what I was asking about though. The strategy of betting small and then raising if you need to get back to even is what I wanted to know what everyone though about (I realize this sentence makes it sound like a dumb strategy...go back and read the initial post for in depth). This is only for betting colors, no numbers involved.


edit: I thought I'd give an example to help illustrate it. Lets say I'm going to bet nothing but red the entire time. I bet 1 dollar and win the first one. That's an extra dollar in my pocket, but I'm not going to count it towards my cash. I bet 1 dollar again and lose. I'm now down 1 dollar (since my initial winning isn't counted toward my cash). Now I have to get back to even, so I bet a dollar. If I win, I go back to the beginning and repeat. If I lose again, I now bet $2 to try to get back to even. Lose again and bet $4 etc. Sooner or later, I will win and go back to even. All of this while keeping winnings on the side. A repeated process should end me up with winnings as long as I don't lose so many in a row that my bet would have to break limit to get back to even (which is why my initial bet has to be small). Granted, the winnings are small, but they are winnings.

I thought I'd also put this in a flow chart like description.

Step 1: Bet $1

win? winnings up $1, go to step 1................lose? go to step 2

Step 2: Bet $1

win? you're even so go to step 1...........lose? go to step 3

Step 3: Bet $2

win? you're even so go to step 1...........lose? go to step 4

Step 4: Bet $4

win? you're even so go to step 1...........lose? go to step 5

Step 5: Bet $8

win? you're even so go to step 1...........lose? go to step 6

The steps just go on like this. Sooner or later, I will go back to step 1 (even if the odds are only 47%ish).


I think I'm going to test this at the dumb betting game at our county fair. It's a checkered dart board and you bet which color the thrower will hit (there's also the black lines inbetween and you can bet on those...they pay 2 to 1,... red/white bets are about 40-45% odds). The only problem is min bet is $1 and max bet is $10, so if I lost 5 times in a row I wouldn't be able to bet high enough to go back to even.

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BigGov

It's basically just setting yourself up to lose big.

One bad stretch and you're done.

If your payroll is high enough to support such a long stretch, then it's probably a waste of time to play. It's basically double or nothing every time on less than 50% odds.

In the end, you'd end up about even, or broke.

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One day an Englishman, a Scotsman, and an Irishman walked into a pub together. They each bought a pint of Guinness. Just as they were about to enjoy their creamy beverage, three flies landed in each of their pints. The Englishman pushed his beer away in disgust. The Scotsman fished the fly out of his beer and continued drinking it, as if nothing had happened. The Irishman, too, picked the fly out of his drink but then held it out over the beer and yelled "SPIT IT OUT, SPIT IT OUT, YOU BASTARD!"

yatzr

after some googling, it turns out that I just described what many believe to be one of the best money managing systems for roulette and blackjack called the Martingale II. But like I said before, the flaw is in table limits. If you could always bet enough to go back to even, there would always be garaunteed profit. I think I'll still try it out at the dart thing though. I'll let you all know how that goes (lucky for me I have a very high paying job this summer).

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CoachAlan

You're welcome to try it, but consider that to cover eight losses, you need a bankroll of $255 to win ONE DOLLAR! Below is a chart I got from http://wizardofodds.com/gambling/scam.html of how the martingale system works over a theoretical one million trials.

Sure, the red peak on the right hand side is mighty tempting, but the smaller red rise on the left still indicates a net loss with this system.

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zenmaster10665

just stick to blackjack....it is the only place you have anywhere near equal odds of winning.

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hannukah harry

eh, stick to poker, where you have to beat a player rather than the house.

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mirevolver

My advice with Roulette: don't play it.

But if you are going to play, then the first thing you need to do is find a single zero roulette wheel. That cuts the house's edge. If you are in Vegas, you can find them in downtown vegas or on the strip there's the Stratosphere, and I think I have also seen them in the Tropicana but I'm not certain.

Once you find a single 0 wheel, bet the even money bets (Red - Black, Even - Odd, 1-18 - 19-36) and keep your bets small.

And as with any game it is always wise to have a system of keeping track of your money so you know when it's time to walk.

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corybmcfly

If you need $255 to cover 8 losses, that means you need to win 255 times before you lose 8 consecutive in order to just break even. If you take the amount of spins it would take to get those 255 wins (probably somewhere around 550) and calculate the odds of having 8 straight losses somewhere within those 550 spins, you'd come to the same house advantage as if you were playing spin for spin. Only spin for spin, your loss is a dollar a little more than every other spin. Martingale, your loss is $255 approximately once every 450 to 500 spins......or in other words broken down to roughly a dollar a little more than every other spin.

You could probably win enough in the short run to go buy yourself some dinner, but if you got into that unlucky streak of 8 straight losses right off the bat, it becomes a $255 dinner (plus the price of dinner).

Stompy

I just got back from Atlantic City and I have a nice roulette strategy.

The tables were a $15 minimum there, so this actually helps minimize losses.

First off, you wait until there are 4 consecutive colors in a row: 4 black/4 red.

In this example, let's say 4 blacks. 3 black, 0/00 works too, which is preferable. You place $20 on red, $15 on black.

If you win, great! You got $5. If you lose, you only lose $5.

If you lost, you then bet $25 on red, $15 on black. If you win, you get $10, which cancels out with the 5 you just lost, giving you $5 profit.

If you LOSE, you flat out bet $15 on red. If you win, you break even, if you lose, which is unlikely but still possible, you lose $30 instead of $45.

Now the reason this works is because with each consecutive result, let's say 4/5 blacks, mathematically, the odds are FOR it being red. To give a bit of insight: try flipping a coin and getting heads more than 5 times in a row.

I tried this and actually won $50 over the span of an hour.

Funny thing, and proof that it doesn't ALWAYS work: I lost $100 because of 14 blacks in a row. I did my steps above, took my $30 loss and walked away.

I came back 15 minutes later and the board was lined up with black numbers, so I thought "guaranteed red.." I put $30 on red, lost it. Next time, put all I had on red (kinda stupid, but then again, mathematically, the odds of it being black again are VERY slim to none).. black again.

After that, there were 3 more black balls.. so people were losing left and right. It was crazy.

In any case, my method really does get some nice results

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corybmcfly

That's cool you made money. Although the odds of rolling red or black are the same going into a spin with 14 previous black spins as they are on any other spin. The only way the odds become slim to none is if someone had predicted the 14 consecutive black spins before they had happened.

BigGov

The odds of it being red or black are perfectly equal. Previous spins have absolutely zero affect on the next spin. It is ALWAYS a set percentage. It doesn't matter if there's 15 blacks in a row, there could be 50 more. There's about a 50% chance it's going to be a certain color, and that percentage doesn't change.

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One day an Englishman, a Scotsman, and an Irishman walked into a pub together. They each bought a pint of Guinness. Just as they were about to enjoy their creamy beverage, three flies landed in each of their pints. The Englishman pushed his beer away in disgust. The Scotsman fished the fly out of his beer and continued drinking it, as if nothing had happened. The Irishman, too, picked the fly out of his drink but then held it out over the beer and yelled "SPIT IT OUT, SPIT IT OUT, YOU BASTARD!"

Stompy

Not exactly true.. like I said, try flipping a coin and getting 5-6 consecutive results. It's not very likely. Yes, a flip is still 50%, but it's not exactly that black and white when dealing with the statistics of the outcome.

It CAN happen, it's just after so many times, the odds are very against it being the same again.

If that mathematics of it all doesn't convince ya, just watch the boards for 30 minutes or so. It's not common for it to go above 5.

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mirevolver

The odds of red or black ar the same on any spin. But the statistical odds vary based on past results. If there are 4 blacks in a row, statistically chances are higher for a red on the next spin.

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Stompy

Unless I have my math wrong, the odds (probability) that you can flip a coin and have it land heads 10 times in a row is 1/1024. Not very probable. If someone flipped a coin 9 times and got heads each time, you could safely bet (with 1/1024 odds) that the next flip will be tails.

There are only 2 possible outcomes when it comes to flipping a coin, and you must multiply that by the amount of occurances you're looking to calculate.

2*2*2*2*2*2*2*2*2*2 = 1024

I think the casino might have rigged their roulette because the chances of getting 14 blacks in a row (and it WAS 14 blacks in a row), is 1/16384.

Of course, that doesn't account for the 0/00 results, but even if you count those slots, it won't skew results THAT much. Either way, odds of getting 14 black in a row are just... not very likely at all and for that to have happened was just utterly insane. You should've seen the people at that table with their jaws dropped. So much money was placed on "red" after the 10th black (and was lost) it wasn't even funny.

For it to have hit 14 black in a row... at the 10th time it was a 1/1024 shot. It passed. The 11th time, odds were 2048. 12th time: 4096.

I have a hard time believing that table wasn't rigged after doing the statistics on these results (again, 0/00 not counted, but won't skew results by that much)

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BigGov

It's not a safe bet, it's still a 50% chance. Previous tendencies have absolutely zero affect on what will happen next. It doesn't matter what happens before, or after, each spin is 100% *independent* of the next. If you walk into a casino any roulette table you go to will give you the exact same odds of winning even if one has had 100 reds in a row and another has had 100 blacks in a row.

Yes, it is very unlikely to hit 14 blacks in a row, meanwhile, on that 15th spin, there is still a 50% chance that the spin will be black. You're trying to find tendencies in a completely random game.

As for odds being utterly insane, I've been dealt bullets in back-to-back hands in hold 'em, *THOSE* odds are insane. Anything is possible.

__________________
One day an Englishman, a Scotsman, and an Irishman walked into a pub together. They each bought a pint of Guinness. Just as they were about to enjoy their creamy beverage, three flies landed in each of their pints. The Englishman pushed his beer away in disgust. The Scotsman fished the fly out of his beer and continued drinking it, as if nothing had happened. The Irishman, too, picked the fly out of his drink but then held it out over the beer and yelled "SPIT IT OUT, SPIT IT OUT, YOU BASTARD!"

mirevolver

With each individual spin there is a 50% chance that it will be black (assuming no 0/00).

However, the stastical odds of it being black consecutive times is reduced exponentially the more times it is consecutively black.

(again, assuming no 0/00)
On a single spin:
Black: 50%
Red: 50%

On two spins:
Both black: 25%
Both red: 25%
1Red/1Black: 50%

On three spins:
3 Black: 12.5%
3 Red: 12.5%
1 Black 2 Red: 37.5%
2 Black 1 Red: 37.5%

On four spins:
4 Black: 6.25%
4 Red: 6.25%
1 Black 3 Red: 25%
2 Black 2 Red: 37.5%
3 Black 1 Red: 25%

The furthur down the chain you go the likelyhood of consecutive colors is reduced.

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Stompy

If that was the case, you'd have the same chance (in this case, you say 50/50) of getting 10 blacks in a row as you do getting any arbitrary combination, such as 6 blacks and 4 red.

As mirevolver said, each individual spin of the roulette wheel (not counting 0/00) yields a 50/50 chance of getting red or black since those are the only two results that can come out of it.

If you were to actually take a coin right now and flip it 5 times, you have a 3% chance of getting all heads/tails. The 50/50 only comes into play when you say "ok, what are the chances of me getting heads vs. tails". That's 50/50. When you take a look at the bigger picture and take into account past results, you getting consecutive results is no longer 50/50.

The exact equation to figure out the statistics of the result is, for lack of a better explanation, the amount of results the object can yield times the number of "turns" you want to calculate. For example, a dice has 6 sides. The chances of you rolling two consecutive numbers (six then six, or one then one) is 1/36, or 2.7%.

The same applies to roulette. I'm tellin ya, it works! Of course, it isn't 100% fool proof, but if you follow what I explained above, the odds are in YOUR favor of winning. It's rare that you'll lose that $30, but it can happen. Chances are, you'll either break even, or better yet, always have a $5 win.

When I first started my method, I won $10, then happened upon a scenario where I lost $30. I got all that back and then some within 45 minutes. The thing is.. you gotta watch the tables and it is pretty monotonous process.

Find a cheap copy of Hoyle Casino games, or better yet, find an online roulette wheel and try it out. You'll see what I mean.

[edit]
Check out this site for a better explanation: http://www.blarg.net/~math/probability2.html

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CoachAlan

Mind you I live in Las Vegas, and am very familiar with what goes on inside casinos...

On most new roulette tables, there is a board that tells you the last 14 or 20 numbers that came up. Stompy, if your system worked, this board would be a great boon to the roulette player. If it really aided the player to know what numbers have come up previously, the casinos would not spend the extra money to install those boards on new tables to let you know about it.

The fact is, those boards PREY on players like you: those who are unaware of the law of independent trials. Your example of the increasing unlikelihood of hitting the same color is flawed. Let me demonstrate the correct formulation for a four-roll series (not accounting for zeroes) :

Color Permutation..........Chance of hitting
R,R,R,R....................................6.25%
R,R,R,B....................................6.25%
R,R,B,B....................................6.25%
R,B,B,B....................................6.25%
B,B,B,B....................................6.25%
B,B,B,R....................................6.25%
B,B,R,R....................................6.25%
B,R,R,R....................................6.25%
B,R,B,R....................................6.25%
R,B,R,B....................................6.25%
R,B,B,R....................................6.25%
B,R,R,B....................................6.25%
R,R,B,R....................................6.25%
B,B,R,B....................................6.25%
B,R,B,B....................................6.25%
R,B,R,R....................................6.25%

Total possible permutations = 16
Total combined likelihood = 100%

You see, it's easy to say that the likelihood of rolling four blacks in a row is only 6.25%. What people fail to realize is that the odds of rolling red, then black, then black, then red, is also 6.25%.

And whatever the first three colors in the order are, the likelihood of the fourth number being black or red is still 50%.

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Stompy

That's why there's 3 iterations of it. You don't just toss all your $$ down on red if there are 4 blacks. Of course the odds of getting 4 in a row is 6.25%. That's when you START it.

Since you start at 4, you're betting that it won't be 5 in a row (3%). If you happen to lose, you're then betting against it being 6 in a row (1.5%), and then finally if all else fails, the breaking even point which is 7 in a row (.7% failure). Meaning... if it happens to land on the SAME color 6 times in a row, you have a 99.3% chance that the next spin will be red.

Of course, since red/black are the only possible results that may come up. No one is refuting this. Keep in mind we aren't counting the 0 and 00 results that may come up. You're talking about the likelihood of the next spin, I'm talking about the odds (statistically) of it being 5, 6, or 7 in a row.

The likelihood that it's black or red is obvious, it can only BE red or black, but calculating future results if you have past results in front of your face is most certainly based on past results. We're not talking about the likelihood of the outcome, we're talking about the odds of getting x in a row.

You're right though, if Casinos knew it helped you, they wouldn't do it... however, aside from pretending that 0 and 00 won't come up, I can't find a flaw in my plan.

Take an adjusted deck of cards, remove a few so that it matches the roulette wheel's total black/red slots, put 2 jokers in (let them be 0 and 00), and to humor me, just try it for about an hour. Flip a card and bet yourself if it'll be red/black. Re-shuffle the deck to keep it even. Flip another card.

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Stompy

Hm, I'm totally aware of the law of independent trials considering I explained the exact same thing you did above.

You're calculating the likelihood of hitting any combination of 4 spins. Since only 2 results can come out of a spin, the formula is:

1/(2*2*2*2) = 1/16 or 6.25%. Read what I wrote above regarding the 2 die rolls. The chances of you rolling two sixes vs. rolling snake eyes are exactly the same: 1/36.

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hannukah harry

stompy, here's the flaw in your system as i see it... you and CA are both looking at it from different terms. he's looking at it from turn to turn, and you're looking at it long term.

the thing is is that the past doesn't matter. there may have been six black rolls in a row, the next roll, because it is independent, has a a 50% chance of being black and a 50% chance of being red. there isn't a 99.3% chance of it being red because the previous pattern doesn't matter to the ball and wheel.

hell, you saw 14 black spins in a row at the casino, doesn't that tell you something? you can't make long term bets because even if you look at the statistics of what it the odds are x number of black spins, you're not betting htat far down. only the next one matters, not the previous or future rolls. to think that the roulette wheel works that way you've built your method is a misapplication of statistics. it may work some of the time, but whenever it hits enough in times in a row on one color, it'll bust you.

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bookerV

Here is the best roulette strategy to save your money...

Don't play. Take your money and walk over to blackjack.

There is so much luck involved in roulette, I don't trust it at all.

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Stompy

hannukah harry, then the same would apply to flipping a coin since each flip is independent.

That doesn't make any sense because you're saying you have a 50% chance of flipping heads 5 times in a row, and that's not true.

[edit]
I'll write a program later on to simulate the method. I'll make an array for the 36 numbers along with 0 and 00 and I'll have the computer randomly pick between them.

Every 5th one, I'll have it perform the betting method I explained above and I'll run it through 100,000 times or so and see how often it comes out ahead.

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CoachAlan

No, what he's saying is that the fifth flip, regardless of the four flips before it, has a 50% chance of flipping heads. Don't you see that saying, "It went three reds in a row, so there's only a 6.25% chance of it being red again," is the same things as saying, "It went red, then black, then black, so there's only a 6.25% chance of it being red."

Black, Black, Black, Black, Red; and Black, Black, Black, Black, Black, are equally likely. As is Black, Red, Red, Black, Red (in that order).

If you're still convinced that your betting system will work over the long term, take on the "Wizard of Odds Challenge." It's here: http://wizardofodds.com/gambling/scam.html. He will put up $20K against your $2K that your system will lose over the long term (1 billion bets).

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Stompy

I just wrote something pretty basic, but it spins 1000 times and counts how many times the results go over 7 consecutive colors in a row.

I took that, looped it 20 times and took the average. So 1000 spins, 20 times, the average that black or red showed up over 7 times was 1.42, which is accurate because I increased it to 100,000 and it came out around 142 times. (I don't mean to throw it off track by using "average", but it rarely ever goes above 4 times per 1000 rolls)

Results are completely random each time, so the computer isn't spitting out the same data.

I'll finish the auto-betting and all that later on (gotta run out for a bit), but so far.. odds are very slim that any color will go over 7 in a row.

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CoachAlan

I have just come up with an idea that I think will definitively demonstrate what most of us seem to have grasped.

Check the results of your simulation to see what happens after four consecutive red spins. I bet that the results will go R, R, R, R... R just as often as they go R, R, R, R... B. That' because the fifth spin is still a 50/50 spin.

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Stompy

The probability of both happening is 1/32, or 3%. So yes, they will happen equally as often. The result of the fifth spin is 50/50 because it can only be red or black, but as mentioned above, we're calculating probability of a sequential set of results, not the probability of the result of the fifth spin itself.

I'm not saying it's fool proof, because I'll probably run this thing through a billion times and find out that the player ends up with a loss somehow, but I think a lot of people are confusing chance of an individual spin's result with the sequential probability, which is what my lil "plan" was based on.

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CoachAlan

Aren't you betting on "an individual spin's result"? And isn't the sequential probability of B,R,R,B,R,B,R,B,B,R identical to R,R,R,R,R,R,R,R,R,B?

The chance of rolling Red then Red is 25%. The chance of rolling Red then Black is 25%. The chance of rolling Black then Red is 25%. And the chance of rolling Black then Black is 25%. In fact, all possible sequential outcomes are equally likely when their order is determined.

I think you are confusing the mathematical concept of combination with that of permutation.

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