## Why is this working? My system is beating the Roulette wheel and dont know why...

Hey everyone, new member with a question.

I am terrible at any type of math or science problem. I am a big poker player and me and my buddy were in Atlantic City waiting for a poker tournament and think we have found a way to beat roulette. I need to be proven that we are wrong.

The system is based on the martingale system but with a twist. Standard martingale system just bets a specific color as many times needed until won, doubling each bet to cover loses on previous bets. Some say that in the long run the martingale system may work, if it wasnt for the max bets keeping you from doubling if reached that high.

Well my friend and I started playing $10 bets. What we thought was that we would go to the roulette area and wait for a table to hit a certain color AT LEAST 4 times in a row. So once we saw a table with 4 reds we would immediately bet$10 on black. We would then double the bet until we won. As soon as we won our bet, whether on the first roll or the 8th roll, we would take our money and get up, starting the system all over again.

If at any point we hit a green, we would get half our bet back (standard in Atlantic City) and cut out loses and stop the system and move to another table.

We would then wait until another table had a streak of 4 and do it all over again. In about 2 hours, moving from table to table each time, we made $400 a piece. I did this online a few nights ago with 6 tables online, only betting when a streak of 4 hit and made$900 in about 2 1/2 hours.

This is my thought, please explain to me why I am wrong...

As soon as one color hits, there is a 1 in like 64 chance that it is going to hit the same color 8 times in a row. So, if I am waiting until the 4th color in a row to hit and only bet on the 5th roll and above, are the odds not HUGELY in my favor to be a winner in the long run??

What am I missing? Did we only get ridiculously lucky not only live but online as well?

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age

Recognitions:
Homework Help
 Quote by TheStone Did we only get ridiculously lucky not only live but online as well?
Yes.

 What am I missing? Why isnt my theory correct?

Recognitions:
Gold Member

## Why is this working? My system is beating the Roulette wheel and dont know why...

Whether the ball lands in a red or black pocket is not dependent on previous spins. Every spin is an independent event. You got really lucky.

 The OP never mentions what sized bankroll he was using. It is uncontroversial that the martingale works, provided you have a big enough bankroll. This has nothing to do with predicting roullette outcomes (which are of course independent), and everything to do with selecting tables with bets that are small compared to your bankroll. Had the OP tried this scheme with a $50 bankroll at a$20 table, odds are he would have been cleaned out rather quickly. But if you play at $10 tables using a$500 bankroll, your odds of success become appreciable.
 Casinos have table max and min limits to prevent the unlimited doubling of bets.
 My bank roll was over the max bet. I had about $5k in my pocket and only making$10 bets. Although I had $5k in my pocket, I was only willing to lose about$1k though. I understand that each bet is independent of each other. But my bet is not against red or black... my bet is against red running 8+ times in a row. Im looking at it like this. The chance that its going to run red 8 times in a row is 1 in 64. Wether the wheel knows it or not, my bet started on the first time it ran red. It has now run red 4 times in a row. Im now making my bet and willing to continue my bet until it hits black. The chance that the next 4 rolls are going to be red is 1 in 64 correct? If thats not the case, then how can the equation of heads running 8 times being 1 in 64 be true? So Im willing to bet $10, and doubling it, up to 5 times.$10 $20$40 $80$160. Even after making the $160 bet I am offering the casino 16 to 1 odds ($160 to win $10) but getting 64 to 1 odds on my chance. Is this not right? I am partially retarded and could be WAYYYYYY wrong here.  Recognitions: Gold Member If you have HUGE pockets and the casino did not have betting limits, the martingale system could work for you if you had a 50;50 chance. But the martingale system will only work for you if you have a 50:50 payout:loss. Roulette is not set up this way because the house has a set number of "house wins" pockets to get their cut. You got lucky. Be happy and don't try it again unless you want to lose your shirt. Recognitions: Homework Help Science Advisor  Quote by quadraphonics The OP never mentions what sized bankroll he was using. It is uncontroversial that the martingale works, provided you have a big enough bankroll. This has nothing to do with predicting roullette outcomes (which are of course independent), and everything to do with selecting tables with bets that are small compared to your bankroll. Had the OP tried this scheme with a$50 bankroll at a $20 table, odds are he would have been cleaned out rather quickly. But if you play at$10 tables using a $500 bankroll, your odds of success become appreciable. (I'm assuming 2/38 chance of losing half, 18/38 of winning, and 18/38 of losing.) On the first spin, bet$10:
Win: 18/38 (stop with +$10) Green: 2/38 (stop with -$5)
Lose: 18/38 (go on)

Second spin, best $20, assuming a loss on the first: Win: 81/361 (stop with +$10)
Green: 9/361 (stop with -$10) Lose: 81/361 (go on) Third spin, bet$40, assuming a loss on the first two:
Win: 729/6859 (stop with +$10) Green: 81/6859 (stop with -$20)
Lose: 729/6859 (go on)

Forth spin, bet $80, assuming a loss on the first three: Win: 6561/130321 (stop with +$10)
Green: 729/130321 (stop with -$40) Lose: 6561/130321 (go on) Fifth spin, bet$160, assuming a loss on the first four:
Win: 59049/2476099 (stop with +$10) Green: 6561/2476099 (stop with -$80)
Lose: 59049/2476099 (stop with -$310) Anyone want to add these up and find the expected winnings? Recognitions: Homework Help Science Advisor  Quote by TheStone I understand that each bet is independent of each other. But my bet is not against red or black... my bet is against red running 8+ times in a row. Once you've seen red four times in a row, the chance that it will be red five times in a row (those four plus one more) is 18/38, not (18/38)^5.  Quote by TheStone Im looking at it like this. The chance that its going to run red 8 times in a row is 1 in 64. Wether the wheel knows it or not, my bet started on the first time it ran red. It has now run red 4 times in a row. Im now making my bet and willing to continue my bet until it hits black. The chance that the next 4 rolls are going to be red is 1 in 64 correct? If thats not the case, then how can the equation of heads running 8 times being 1 in 64 be true? The chance that the next four are red is ((# of red) / (total # of spaces))^4, not ((# of red) / (total # of spaces))^8.  Quote by TheStone So Im willing to bet$10, and doubling it, up to 5 times. $10$20 $40$80 $160. Even after making the$160 bet I am offering the casino 16 to 1 odds ($160 to win$10) but getting 64 to 1 odds on my chance.
If there were no green spaces, you'd have a 31/32 chance of winning $10 and a 1/32 chance of losing$310. Because of the green your expected winnings are lower.

 Quote by CRGreathouse Once you've seen red four times in a row, the chance that it will be red five times in a row (those four plus one more) is 18/38, not (18/38)^5. The chance that the next four are red is ((# of red) / (total # of spaces))^4, not ((# of red) / (total # of spaces))^8. If there were no green spaces, you'd have a 31/32 chance of winning $10 and a 1/32 chance of losing$310. Because of the green your expected winnings are lower.

Can you say everything that you said, but pretend you are talking to a 10 year old kid. Seriously man, I barely passed algebra in high school.

I see what your doing, but dont understand it.

Recognitions:
Homework Help
 Quote by TheStone Can you say everything that you said, but pretend you are talking to a 10 year old kid. Seriously man, I barely passed algebra in high school.
When you see things happen, the chance that they will have happened is not small, it's 100%. If you've already seen four reds in a row, the chance you'll see four reds is not 1/16 but 16/16. The chance that there will be eight or more reds in a row is not (slightly below) 1/256, but (slightly below) 1/16.

It's slightly below, not exact, because the ball may land on a green space instead of a black or red.

 Yes, you got lucky. Basically, what you're saying is this: Flip a coin. Observe what it comes up... say, for the sake of argument, it's heads. By your method, you would bet that a tails would come up next. That is, P(heads on second) = 1/4 and P(tails on second) = 3/4. The problem with that is this: imagine somebody else shows up after you flip the coin the first time, and has no idea what you're doing. If you ask him what the probabilities of heads vs tails is, he would likely respond: P(heads) = 1/2 and P(tails) = 1/2. Of course, he's right, because it doesn't matter what happened in the past... only what is still going to happen. Does that clarify it any?
 Then how can anyone come up with an equation for the chance of heads flipping 8 times in a row?
 Blog Entries: 3 Bets,2,4,8,16,.... Expected Losses:1E[0.5]+2E[0.5*0.5]+4E[0.5^3]+... Expected Winnings:1E[0.5]+2E[0.5*0.5]+4E[0.5^3]+... It doesn't look like it converges. Fraction of People that win after one play 1-0.5 Fraction of People that win after two plays: 1-0.5^2 Fraction of people that win after three Plays 1-0.05^3 It looks like the system should eventually work for many more people then it doesn't but for the people it doesn't work for they'll lose big keeping the average winnings zero. After about 20 plays starting at 1$the bet to win has went up to 1 million dollars. About one in a million people will still be playing. For this one in a millionth person that had over a million dollars to bet, I can't help wonder if the first 10 bets really mattered that much to them. However, if they started at something meaningful say$1000 then they are much more likely to end up betting one million dollars. In fact if they started at 1000 dollars they'd have about a one in 1000 chance of still playing by the time the bet got to 1 million dollars. The moral is, if you start out betting something meaningful there is a one in 1000 chance you'll lose big.
 TheStone: If all eight flips are still in the future, then it's easy to calculate the probability of getting 8 in a row. All that you need to do is this: There are 2 ways to flip a coin... get heads, or get tails. There are 4 ways to flip two coins... heads heads, heads tails, tails heads, and tails tails. There are 8 ways to flip three coins: {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}. There are 16 ways to flip four coins: {HHHH, HHHT, HHTH, ..., TTTT}. There are 32 ways to flip five coins: {HHHHH, HHHHT, HHHTH, ..., TTTTT}. ... There are 256 ways to flip eight coins: {HHHHHHHH, HHHHHHHT, ..., TTTTTTTT}. To find the probability of getting 8 tails in a row, just count the number of ways to do it (in this case, just one) and divide by the total number of ways 8 coins can be tossed (256). This yields a probability of 1/256, which is only like 4 parts per thousand, or under 1%. HOWEVER: if you want the probability of getting tails on the last toss (so the last character in the string is T, not H) then you do the same thing: count and divide by 256. It should be clear, with a little thought, that exactly 128 of them end with a T. Dividing 128 by 256 yields 1/2. The thing is that the events are independent... the outcome of one does not affect the other. In mathematics, we right this as P(A|B) = P(A) - the probability of event A given that event B has occurred is still just the probability of A. Think of it this way: what if I toss a die three times and get 6 each time. Then I flip a coin. Does the die affect what the coin will do? No. So why should previous flips of the same coin matter? The coin doesn't remember what it came up last time.

 Quote by TheStone The chance that its going to run red 8 times in a row is 1 in 64.
The key to getting this is to notice that the probability of getting *any* specific string of 8 outcomes is the the same. RRRRRRRR is exactly as likely as RRRRBBBB or RRRRBRBR or BBBBRBRB.

This is a common situation in probability where intuition gets you into trouble. Because the string RRRRRRRR looks very "ordered," we have an intuitive sense that it should be very unlikely to arise randomly. But that's not really the case: the intuitive sense of "order" here is not applicable. There is no less likelihood of RRRRRRRR than any other string of 8 outcomes.