Roulette vs Mathematical Betting Strategy. Please Comment and Criticize!

1. Jul 9, 2011

elegysix

Ok so on the roulette table the odds against winning on red or black are 1.111 to 1.
This is about a 47% chance to win. Payout is 1 to 1.

Say minimum bet is 1$and bets are allowed in multiples of 1$. up to some limit.

Suppose I want to win 1$for every spin I play. I bet 1$, and I lose. so my net gain= -1$Now if I want to have won 1$ for each spin to this point, I need to bet 3$and win. (1$ for the net, 1$for the last spin, and 1$ for this spin)

I lose my 3$bet. now my net gain = -4$

So then I bet (4 + 3)$= 7$. winning this bet will cover my -4 net and give me 1$for each spin I've done. If I win at any time, I restart my betting at 1$.

table of bets on a losing streak:
N(streak) | B(bet)
............1|1
............2|3
............3|7
............4|15
............5|31
............6|63
............7|127
and so on.

recurrence relation:
$B_{N}=2*B_{N-1}+1$

Usually there is a maximum bet limit, and if my losing streak is so great that I cannot bet the required amount, I will consider that money lost and restart my betting at 1$. There is some probability that I will have a losing streak too long, and at that point lose a fixed amount of money. But will that loss be greater than what I have gained up to that point? Suppose the maximum bet is a multiple of the minimum bet. At what point does this ratio cause my net gain to become probably negative or probably positive? there must be some minimum multiple which gives me a higher probability of having a net gain than a net loss, and some maximum multiple for which the probability of a negative net exceeds the probability of a positive net. ( the higher the ratio of max/min bets, the less likely I'll have a streak too long, thus meaning I'll be less likely to lose that money and vice versa ) I don't know much about statistics, and I would love to have someone help me solve this before I decide to go to vegas and try this out :) 2. Jul 9, 2011 mathman I won't comment on your particular betting strategy. However I can assert without fear of contradiction that (assuming the wheel is is in perfect condition) there is no winning betting strategy. The only way to beat a roulette wheel is to find one that is slightly imperfect and place bets to take advantage of the imperfection. 3. Jul 9, 2011 micromass 4. Jul 9, 2011 pmsrw3 On average, yes. The ratio doesn't matter. Your expected loss at every spin is positive, and the expected total loss is just the sum of the expected losses for each spin. There is absolutely no system that has an expected positive gain. There is an optimal strategy that minimizes expected loss: bet the minimum every time, regardless of what has happened in the past. And there's one strategy that's even better than that: don't bet at all. It was solved long ago. People have been asking this question for literally hundreds of years. In fact, your exact system has been proposed many times and analyzed to death. But I'm going to bet that you won't believe it :-) 5. Jul 9, 2011 SteveL27 An engineer years ago solved the problem. He started by recognizing that a roulette wheel is a mechanical object, hence imperfect. He built a little wearable computer and stood at the wheel for hours, recording the results. His computer then figured out the bias of the wheel, allowing him to beat the house. In the end they gave up the system after winning only about$10k. One of the members of the team burned a hole in her skin when the computer shorted out.

http://en.wikipedia.org/wiki/Eudaemons

6. Jul 9, 2011

pmsrw3

Cool! I'm surprised there's enough mechanical imperfection in a Roulette wheel to allow the house advantage to be overcome.

7. Jul 9, 2011

elegysix

I reviewed this. There is a section where they calculate expected gain and loss. In the martingale system this is always negative. This is not the case for my system.

The method I suggest is very similar but modified - In the martingale system you only net gain 1*BaseBet after breaking a streak of N losses. In this system you net N*BaseBet for breaking a streak of N losses. This gives the system an edge, and the calculation for expected gain/loss is actually positive. (N being the shortest streak to require a higher bet than the table limit.)

example, expected gain/loss:

martingale system:
(max N = 6)
Prob of 6 streak loss: (20/38)^6=2.1256%,
P of not 6 streak: 1-2.1256%=97.8744%
expected gain/loss = 1*(.978744) - 63*(.021256) = -1.339118

(negative meaning expected loss)

This system:
probabilities are the same,

expected gain/loss (max N=6) = 6*1*(.978744)-120*(.021256)= 3.321774

which is positive for an expected gain.

8. Jul 9, 2011

micromass

This is incorrect. Not having a 6 streak loss does not mean that you won 1$every time. For example, a thing that could have happened was that you won first time and then lost 5 times. In this case you would lose money, you would not win$1 every time. so the factor 6*1*(.978744) is incorrect.

9. Jul 9, 2011

pmsrw3

It's thinking like this that makes running a casino such a profitable business :-)

10. Jul 9, 2011

elegysix

The case of winning once and losing the next five is irrelevant, if that were the case I'd bet again, because I'd be under the max bet limit. Assuming my funds are enough to cover a 6 streak loss, If I lose, I bet again. I do this until I have a streak of 6 losses. Therefore If I lose my 6th bet on the streak, I'm out 120$, but if I win I'll have a net profit of 6$ + 1$/spin for however many spins I'd done before the streak of 6. for any set of N spins ending in a win, it is a gain of N$. If at any point there is a streak of 6 losses, (due to a max bet limit I cannot continue my strategy), I will lose 120$. However, if I lose 5 bets but win my 6th bet, it is still a gain of 6$.

That is why I put 6*(.978744)-120(.021256)

If this is wrong I would appreciate it if you would elaborate - thank you :)

Last edited: Jul 9, 2011
11. Jul 9, 2011

pmsrw3

OK, let's suppose you're in that situation. You've lost 5x in a row, and you place your sixth bet. If you win, you net $6. If you lose (which is slightly more probable), you lose$120. And, umm, you think this means a net GAIN?

12. Jul 9, 2011

elegysix

You're forgetting the 1\$/spin you got for every spin until that point.

Prove me wrong: Show me that the average number of spins until a streak of 6 has likely occurred is less than 120.

13. Jul 9, 2011

micromass

Fine, I calculated the average number of spins until a streak of 6 and it seems to lie around 95.

What I did was to introduce following transition matrix:

$$P=\left(\begin{array}{cccccccc} 0.47 & 0.53 & 0 & 0 & 0 & 0 & 0 & 0\\ 0.47 & 0 & 0.53 & 0 & 0 & 0 & 0 & 0\\ 0.47 & 0 & 0 & 0.53 & 0 & 0 & 0 & 0\\ 0.47 & 0 & 0 & 0 & 0.53 & 0 & 0 & 0\\ 0.47 & 0 & 0 & 0 & 0 & 0.53 & 0 & 0\\ 0.47 & 0 & 0 & 0 & 0 & 0 & 0.53 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ \end{array}\right)$$

The chance on getting 6 consecutive failures on the n'th attempt is given by $P^n(1,7)$ (thus the first row and seventh column of Pn). So the average number of steps needed to get 6 failures is

$$\sum_{n=1}^{+\infty}{nP^n(1,7)}$$

I calculated this with n going to 10000, and I got about 95.

But anway, what you propose is well-studied. You must read up on the optional-skipping theorem of martingale theory. It says that whatever strategy you try, you cannot win a game that is not in your favor.

14. Jul 10, 2011

elegysix

Thank you for the detailed equations & explanation! I don't understand how that matrix works, but I'll take your word for it. I think I should take a course on stats/probabilities lol

15. Jul 10, 2011

pmsrw3

Using

$$\sum_{n=1}^{+\infty}{nP^n}=P(1-P)^{-2}$$

I get 97.207971 (using p=18/38). I get the same answer by solving the hitting time equations:

$$\begin{array}{l} k_6=0 \\ k_0=1+p k_0+(1-p) k_1 \\ k_1=1+p k_0+(1-p) k_2 \\ k_2=1+p k_0+(1-p) k_3 \\ k_3=1+p k_0+(1-p) k_4 \\ k_4=1+p k_0+(1-p) k_5 \\ k_5=1+p k_0 \end{array}$$

$k_i$ is the hitting time from i losses in a row to 6 losses in a row.

I'm kind of surprised it's this high, actually. The probably that six throws are all losses is about 1/47, so I thought it would be around 47.