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# The Roulette Wheel (probability) question

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### #1 of 76 OFFLINEAnthony Moore

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Posted February 05 2005 - 11:02 AM

This has probably been discussed before, but I just have a probability question. So Im sure everyone knows this roulette strategy: Start with \$10 on red or black. Lose? put down 20 on same color..lose? put down 40..80..160...320..640.. Win just one of those 7 spins in a row, and you pocket \$10, and start the cycle over. So lets say you pick red. You put all that money on red each time, having enough backup money to cover 6 losing spins.. As long as red comes up ONE time out of seven rolls, you win \$10. If it doesnt, you lose everything As long as red comes up once out of every spins, for 10 cycles in a row, thats \$100. To cover yourself for 7 spins it would cost you \$2550. Its a lot of an investment, but how likely is it you will lose? Whats the probability that RED WILL or WONT come up just ONCE out of every seven spins? So the downside is, even if you win, it will only be \$10 on each cycle. But whats the odds it wont come up? Just curious
Anthony Moore

### #2 of 76 OFFLINEJames T

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Posted February 05 2005 - 12:17 PM

Red, Black, Odds, Even, 1 to 18, and 19 to 36 all have a 47.37% chance of winning. If you want to play the odds game, play craps. If I remember my notes from highschool, it has a 51-52% winning chance.

### #3 of 76 OFFLINEAnthony Moore

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Posted February 05 2005 - 12:38 PM

but the odds have to move in your favor the more spins you get. its not just a 51% chance overall. How would I calculate the probability of red coming up at least once in 7 spins?
Anthony Moore

### #4 of 76 OFFLINEJason GT

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Posted February 05 2005 - 12:54 PM

I thought that someone was gonna reply with a factual post that would have obviated mine (which I didn't end up posting). Anyways... All casino games are tilted in favour of the house. For your original question, Anthony, of
I'll answer the "red won't come up just once" part. From some brief research on the web, there are typically 37 spots on a roulette wheel with one of them colorless. Therefore in a given spin, the chances of you LOSING are 19/37, or about 51.3 per cent. The chances of you losing all seven spins (ie coming up BLACK or NO COLOR) is (19/37)^7 = 0.00941592820025755468171924197107512 Or just under one per cent.

### #5 of 76 OFFLINEJoseph DeMartino

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Posted February 05 2005 - 01:11 PM

No, they don't. The table has no memory. The odds of red coming up on any given spin are exactly the same as on every other spin. It doesn't matter if black has come up on the last 1,000 spins, the odds of red coming up on the next spin are no better or worse than before. Just as with flipping a coin. The "odds" (probablities) are 50/50 every time. In a series of 100 flips you would expect to end up with roughly 50 heads and 50 tails, but if you do the experiment you will almost never end up with exactly that result. And if you've flipped the coin 90 times with a 45/45 split there is no reason you can't get 10 "heads" in a row at the end, and the chances of that last flip producing the "tails" result that it "should" are exactly the same as its coming up heads again. The odds don't change. Folks who believe otherwise are the ones who pay the bill for all that neon down in Vegas.

Regards,

Joe

### #6 of 76 OFFLINEJon_Are

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Posted February 05 2005 - 02:27 PM

This is one of those systems that looks good on paper but falls apart relatively easily in practice. Five, six, seven and more consecutive blacks or reds is not at all uncommon, and that's likely all it would take to bust you. Look at it this way, Anthony...suppose your color misses for six spins in a row. Do you really thing you have the cajones to plop down \$640 for a single 50/50 (more or less) wager? Remember, by this point you're already down \$630. And, even if you hit this huge bet, what are you up overall after sweating this one out? That's right: \$10. Good luck, Jon

### #7 of 76 OFFLINEChris Derby

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Posted February 05 2005 - 02:30 PM

I thought that bill was paid for by the Hoover Dam.
-derby

### #8 of 76 OFFLINEAl.Anderson

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Posted February 05 2005 - 02:31 PM

Joseph laid out the statistical error with the "double" approach. The practical problem is that casinos have table min/max limits; at some point you wouldn't be able to get down your next bet to cover.

### #9 of 76 OFFLINEAnthony Moore

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Posted February 05 2005 - 03:40 PM

does everyone believe this to be true? Because this is the factor. According to JasonGT the formula puts the odds in your favor with a 99.1% chance of victory withe a 7 spin cycle. I do understand that this does not work..otherwise why isnt everyone doing it, right? I just want to know where it falls apart.. So 7 in a row without hitting red one time isnt that uncommon? ( dont know, i havent played much)
Anthony Moore

### #10 of 76 OFFLINEChu Gai

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Posted February 05 2005 - 03:45 PM

### #11 of 76 OFFLINEAnthony Moore

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Posted February 05 2005 - 04:20 PM

There are a lot more scenarios involved with horses. 8 horses race, right? well, this one really only has two different outcomes ( 3 if you count the two greens)
Anthony Moore

### #12 of 76 OFFLINEJoseph DeMartino

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Posted February 05 2005 - 05:48 PM

It falls apart because of the unjustified assumption that past results affect future outcomes. You're still not getting the point that the odds don't change based on "x" number of spins. Each spin is an isolated event, independent of all previous and subsequent spins, and any "system" that assumes that certain outcomes have to obtain within a given number of spins is simply wrong. Regards, Joe

### #13 of 76 OFFLINEKirk Tsai

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Posted February 05 2005 - 05:50 PM

The way it's talked about here, I would think in one run, seven in a row isn't that big a deal. In the long run, however, it seems like could work theoretically. But, what is large enough a sample is hard to determine.

### #14 of 76 OFFLINEJason GT

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Posted February 05 2005 - 08:55 PM

Thanks, Joseph, for bringing up "memory" (or the lack thereof). It falls apart because on any given spin the odds are against you winning. Over a sufficiently large number of spins, you'll lose 51% of the time. Over a sufficiently large number of spins, you will trend towards losing. Spinning seven times in a row without hitting red is highly IMPROBABLE but it is very POSSIBLE. It's that possibility of losing where your wallet grows wings.

### #15 of 76 OFFLINEJerry Almeida

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Posted February 06 2005 - 12:25 AM

Well I'm still not getting it either.

No seriously. I understand that even if I flip a coin 9 times and get 9 heads, it doesn't change the odds on the 10th flip. It's still 50/50. However, can't a series of flips be looked at as a whole different animal together?
What I'm saying is what are the odds of 10 flips all being heads or tails? Is that still 50/50? For some reason that doesn't seem accurate. Help me understand.

Besides, everyone knows that everything in life is 50/50. Either it will happen or it won't. What are the odds of me winning the lottery this weekend, well I either will or won't, so 50/50.
Everybody relax, I'm here. -Jack Burton

### #16 of 76 OFFLINEMike_Stuewe

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Posted February 06 2005 - 12:36 AM

Agreed about the probability of a series. It is very low. A good place to try this would be the Lady Luck casino downtown. I think the minimum spin is 50 cents. However on your other question. If you have already flipped a coin 9 times and all 9 times it comes up heads, then the probability of the 10th time being heads is still 50/50. But, if you have yet to flip the coin, the probability of coming up with heads 10 times in a row is extremely low.

### #17 of 76 OFFLINEBryan X

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Posted February 06 2005 - 02:22 AM

Exactly. The farther you are into your 10 flips of the same side, the higher the odds become that you will be successful. At the beginning, before you've flipped the coin at all, the odds of flipping the same side 10 times in a row is less than one-tenth of a percent ( about .098%). You can calculate this by: .50 x .50 x .50 x .50 x .50 x .50 x .50 x .50 x .50 x .50 = .00098 which can also be written as .098%. After you've flipped it once, the odds rise to just under two-tenths of a percent (about .195%). .50 x .50 x .50 x .50 x .50 x .50 x .50 x .50 x .50 = .00195 or .195%. The odds keep going higer the closer your streak of 'same sides' gets to 10. After 8 consecutive flips of the same side, your odds of making it to 10 are up to 25%. And after 9 flips of the same side, your odds are now at 50%. These calculations make the assumption that you are choosing heads or tails BEFORE your first flip of the coin. The odds start out at .195% if you decide to go with whatever side comes up after your first flip.

### #18 of 76 OFFLINEKen Chan

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Posted February 06 2005 - 03:55 AM

Or you simply don't have that much to chase the double, whichever is more limiting.

Try this simulator. With the default settings, you often win a little, but occasionally lose a lot. (Let me know if there are any errors. The results run straight down the page if you're using IE, but with a more modern browser, they fill the page horizontally.)

### #19 of 76 OFFLINEMike_Stuewe

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Posted February 06 2005 - 06:30 AM

I've been playing a free online game of Roulette. I went from 2000 to 4700 in about 20 minutes with this theory. Then I hit a string of 8 reds in a row and I lost it all.

### #20 of 76 OFFLINED. Scott MacDonald

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Posted February 06 2005 - 08:02 AM

I had this same brilliant idea while in college, but since I was a CS major I thought that I'd write a simulation and run it overnight on the computers in our lab. While it sounds like a rare probability to get a long string of the same color, in practice you will find that it happens all of the time.

As Mike discovered, the only way you can make this work for you is if you do it for a short period of time. If you play this system for any length of time, you will definitely hit the worst case scenario and lose it all. Of course that doesn't mean that you won't hit the worst case scenario on your first attempt.

Trust me, this is not a new idea, but for some reason the Casinos still manage to stay in business
Scott

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