It’s not so easy to choose the most suitable strategy for the Roulette, as this game is almost unbeatable and after Check the probability of winning the game is always -2,7%. Still, there is a number of systems that help a gambler beat the Roulette.

Nevertheless, these systems may bring you loss as well as winning.

The main principle of any roulette game system is raising bets in case you lose. Below you can see gamblers’ most preferable system – Donald’s – Nathanson’s system, which is considered the most reasonable and simplest one.

**DONALD’S – NATHANSON’S Roulette Strategy**

In accordance with this system, the bet is fixed. You cannot change it as there are no zero and negative bets in the Roulette. Most players prefer making bets on Red. For example, let’s bet 1$. If Black occurs, you should bet **1$ more**; if Red occurs, you should lower the bet by one.

Providing that you have bet 1$ and Red has occurred to you, you win. Then as I have already said, you should lower the bet. Thus, the next bet will be zero that means that you miss the next turn. In this case, I always advise to make a bet on Red and watch what will occur further on. Let’s assume that Red occurs once more. Then you win again, thus you must lower your bet. Consequently, the following bet will be -1.

By the way, the negative bet on Red is the bet on Black. Whatever the course of the game is, you should always stick to the rule set above. If Red occurs, lower the bet; if Black occurs, raise it.

Let’s suppose that the wheel turns for three times, and the result of each turn is occurrence on Red. After the first turn you win 1$, then you bet zero, and after that your bet is -1$. Afterwards I advise you to reduce the bet to -2$ and to bet 2$ on Black.

I can’t but mention that the property of invariance is actively used in the Roulette. The size of the winning doesn’t depend on the sequence of Red and Black.

For instance, the original bet is 1$. The Roulette has made 36 turns. Providing that Red has occurred for **20 times, the winning is 14$.** In case Red has occurred for 17 times, the winning doesn’t change; it is also **14$.**Donald’s system was counted on this fact, while Nathanson just intensified it. For example, by Donald’s system, you should raise the bet if zero occurs. By Nathanson’s system, you should raise the bet according to a certain module, that is you raise the bet by one providing that it is positive; and you lower the bet by one in case it is negative. However, if zero occurs, even once out of 36 turns, it is quite difficult to make out a real benefit.

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If the bet is positive, and zero corresponding to Black occurs to you, the winning is paid in accordance with the system. Thus, in case Red occurs for 20 times, Black for 15 times, zero for 1 time, your winning will be 14$ again. By the way, zero plays an important role here: due to zero, Red occurs less frequently.

Now let’s suppose that the bet is negative and it corresponds to Red. In case Red occurs for 20 times, its actual number is 21 together with zero. Thus, by Donald’s – Nathanson’s system, your winning is only 6$ instead of 14$. However, if Red occurs under 18 times, the winning is higher.

Moreover, zero may occur to you when you’ve made a zero bet. Keep in mind the following: **if you raise your bet, zero corresponds to Black;** in case you lower it, zero corresponds to Red.

I also advise you to pay attention to all the previous games. It may happen that Red occurred more frequently than Black did. In this case, raise the bet. And vice versa, if Black occurred more often, lower the bet.