1-3-2-6 System to win in roulette

Your initial bet is 1 unit, the second 3 units, the third 2 units and the fourth 6 units. Let's assume that each unit is $10 and the odds are 1:1 - even money.

The first bet is $10. When winning, $10 is added to the $20 on the table making the second bet $30. When winning again on the second bet, there would be $60 on the table. Of this you take down $40 and the third bet is now $20. If the third bet wins, you will have $40 on the table to which you add $20 making a total of $60 for the fourth bet.

If the fourth bet wins, there would be a total of $120 on the table, all of which is net profit. Now all the bet is taken down and you start the system all over again at $10. Also, each time you lose, at any level, you start all over again at $10.

If you lose the first bet, your loss is $10. The second level loss is $20 (because you added another $10). At the third level, a loss will give you a net profit of $20 (because you have taken down $40). At the fourth level, a loss leaves you breaking even (because you put back $20 out of the $40 taken down).

The attraction of this system is that you risk $20 at a chance of making $120 net profit. This means you can lose six times at the worst level (second bet), and with one win (a set of four wins in a row) get your money back.

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