The system roulette-Donald Nathanson

Nowadays old Thomas Donald system underwent a critical revision on the part of a serious mathematician Lev Natanson. He reasoned as follows:

I always bet on red. Suppose the initial rate - $ 1. After falling out of black I increase the bet by one unit, and after the loss of red - is reduced by one. But what do I do if I put a dollar on red, and won? According T.Donaldu, the rate should remain unchanged since either zero or negative rates does not happen. And actually, why? - Mathematician thought. And I tried: It turned out very interesting.

In order not to deviate from the canons of the system, after betting on red and win, you need to reduce the rate per unit. If you bet $ 1, the next bet should be zero. What is a zero rate, it is clear: the next launch of the roulette you just miss. But it is zero on the red and watch carefully what will fall - to know how to put the next time. Suppose again fallen red. You have won and should reduce the rate again. Next rate (in the system) must be equal to -1 (minus one).

And what is a negative bet on red? This is - a bet on black! Whatever happens in the future, only one rule: on a roll of black rate increases, the red on a roll - is reduced.

Suppose, for example, when the first three launches roulette all the time drops red. After the first run, we won $ 1, for the second time "set zero", and the third - minus $ 1 (dollar on the black). Before starting the fourth, we must lower the rate to minus $ 2. Put $ 2 on black.

We can prove that if 2 N roulette runs red and black for fall N times, the prize will be exactly N initial rates. Regardless of the number of deletions of red (and thus, black) performed "invariance": the sequence in which the red alternates with black, on the size of a prize is not affected. Suppose Roulette started 36 times. Your income (positive or negative) is shown below.

The number of red deposition 14 15 16 17 18 19 20 21 22 23
Profit -22 -6 +14 +18 +14 +6 -6 -22

For example, if the red fell 20 times, then at the initial rate of $ 1 gain is $ 14. If the red fell only 17 times, and you will win $ 14. Curiously, the income distribution is symmetric about the middle of the table.

The table reflects only those cases where the frequency of loss of red and black are slightly different (in other "hands," you are the biggest loser). It is the closeness of these frequencies and expected T.Donald. Nathanson just went in his footsteps and "aggravated" system. To complete the picture, recall of zero.

According to Donald T., a roll of zero next bet should be increased. In a modification Nathanson it is necessary to increase in absolute value. In other words, if the rate is positive, it should be raised by one, if negative - lower. Unfortunately, the appearance of zero breaks beautiful invariance property, and your income to determine unequivocally fails. We restrict ourselves to the case when the 36 launches roulette zero falls only once.

Let a roll of zero rate has been positive. Then zero is completely equivalent to black, so the income is determined on the same table as before. For example, at 20 deposition of red, 15 black and one zero payoff will be $ 14. Just do not think that zero no matter what no effect: it reduces the expected number of deletions of red.

Zero may fall and a negative rate. Now, it is equivalent to the red. If the red fell 20 times, due to zero the number of his appearances is actually equal to 21. Instead of $ 14 (according to the table), we actually win only $ 6. But if the red fell at least 18 times your income increases.

And finally, zero may appear at zero rate. You can do whatever you like: the rise of zero rate will be equivalent to black, with a decrease - red. Yet look at the background: if the red fell out more often than black, is to increase the rate if at least - on the contrary. So you would like to converge the frequency of loss of both colors. Mr. Donald would have been pleased.