I am perplexed because it seems that I have come up with a system for beating the roulette which has a positive mathematical expectancy/hope... but only in the calculations, which I think that I've made wrong.(adsbygoogle = window.adsbygoogle || []).push({});

So I want to know in what I have failed doing the calculus.

Well basically the method consists in betting on the three dozens... at the same time!. We start with 0'5 on each. Then we multiply by 1,5 the dozens that didn't turn out.

(1'5 is the minimum by which we have to multiply the initial bet so that it keeps giving the same benefit as the first one, while covering the losses.) Well. It is a type of martingale as you can see.

What we do next, is to stop the sequence, the progression of a determined dozen, once it reaches 3 consecutive spins without turning out, so that it goes back to the initial bet (0'5), and yeah, we accept the losses of that dozen.

The thing is that if we calculate the mathematical expectancy/hope, we see that we're supposed to have benefits, that is, that it is positive. At least it is so in the way I calculated it by, which I suspect that is mistaken

We will lose money with this system in 2 cases:

-a determined dozen is not turning out (at least during 3 spins), whilst the other two including the zero are. Chances of this are: (25/37)^3=0.3085. (30.85%).

-a determined dozen plus the zero turn out 3 consecutive times. Chances of this are (13/37)^3=0.043 (4'3%).

With this results we have that, each 100 spins, in average, 30'85 of them we will have a 3 consecutive dozen without turning out, so we would have the following losses: 30.85*2,375=73.27€. (2,375 is the acumulated amount from the 1st spin to the last in which we accept losses and reinitiate) (0'5+0'75+1,125= 2'375).

The same happens in the second case, but now, as only one dozen is turning out, we would lose that amount in the other 2 remaining dozens! So we would lose:

4'3*(2*2'375) = 20.4€.

There is a third case that I have not taken into account as its losses are really unsignificant. That case is when the 0 turns out 3 consecutive times.

Well, On the other hand, we are wining 1€ each spin, whatever the outcome (well.. we don't when the 0 is the outcome though). So in 100 spins, we will win roughly 97€. (As the 0 is expected to turn out 2'7 times).

If we add the losses and subtract it to the earnings, we clearly see that the expectancy is positive.. so we should win with this system in the long term.

However I computed it with a sample of 5000 roulette spins and...... the result is that I lose.

What did I calculate wrong or I didn't take into account?

Thanks beforehand

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# I Beat the roulette wheel with this?

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