302 Roulette Systems (302 Sistemas de Ruleta)

18-10 Method

The next system that I would like to discuss is the 18-10 method. Now, you will notice, by checking the roulette layout, that we have 18 reds and 18 blacks. We also have 18 odds and 18 evens, as well as 18 highs and 18 lows. If we were to place $50 on the black and any one of the black 18 numbers should come up, we would automatically win. If we were to place $50 on the odd and any one of the odds numbers should come up, we would win also. If we played both black and odd together, placing $50 on each one, and any black and odd number hit for that spin, we would win on both, therefore making $100. But let us say, for example, that number 19 should hit; 19 is red, so we would lose our $50 on the black but we would win on the odd - breaking even. Now by playing black and odd together, using the same amount of money on each one for a bet, we are covering 18 blacks plus 10 other numbers on the red. Because we are playing black, we are covering both the events and the odds of black. That's why we don't have to worry about any of the blacks numbers. And by playing odd, we are covering all the odd numbers in red, which gives us another 10 numbers. Therefore, we are covering 28 numbers out of the 36, plus the 2 greens. Against us are 12 numbers. We have 28 numbers in our favour. Let us go back to the chart Number 1, where we ran off 20 games, and let's just see how we would do by playing the 18-10 method. Say that we start off by playing black and odd. We are playing this combination because it gives us 28 of the 36 numbers. If we were playing black and even we would only be covering 26 numbers against 28 playing black and odd. Therefore the advantage to us is to play black and odd or red and even.

Now let us put $50 on black and $50 on odd and see what happens. First, we have 19 red. We lose on black but win on odd, so we break even. By placing the same amount on black and odd our next number is 27 red. Once again, we break even, 6 black. Now we lose on odd and we win on black. Once again we break even. The next number is 11 black and now we win $100; we win on the black as well as on the odd. The next number is 9 red. Because we have odd completely covered we win on odd and lose on black - one has paid the other. The next number is 4 black - win on back, lose on odd.

Our next figure is 24 black - we break even. The next number is 10 black - we break even. The next number is 12 red - we lose $100. Our next figure is 6 black. By playing black and odd we win on one and lose on the other - we break even. Our next figure is 34 red - a $100 loss. At this point we are out of pocket $200. The next figure being 20 black, we win on one, lose on the other. The next figure is 5 red - breaking even again. The next number is 9 red - same thing. The next 27 red - break even- The next figure is 10 black - break even. The next is 2 black - we break even. The next is 14 red - lose $100.

The next figure is 7 red, so we don't make any money. And at the end of the 20 rolls, we have had 3 losses where we lost on both black and odd at the same time, and one win where we won on both at the same time. So adding our wins and our losses, we find at the end of these 20 spins, we are $200 in the hole.

Now, let us play red and even, placing $50 on each and using the same chart; same number of break-evens; 3 wins of $100 each; one loss of $100.

Now we will switch to playing a combination that covers 26 of the 36 numbers, therefore taking the disadvantage of an additional 2 numbers against us. Let us see how we would come out after the 20 games.

We are now playing black and even. Our first number up was number 19 red, therefore we lose $100. Our next number is 27 red; we lost another $100. 6 black wins $100. 11 black breaks even. 9 red loses $100. 4 black win $100. 24 black wins $100. 10 black wins $100. 12 red breaks even. 6 black wins $100. 34 red breaks even. 20 black wins $100. 5 red loses $100. 9 red loses $100. 27 red loses $100. 10 black wins $100. 2 black wins $100. 9 red loses $100. 14 red breaks even. 7 red loses $100.

We have had 8 losses and 8 wins - breaking even at the end of 20 games.

Now let us play red and odd. Our first number is 19 red and we win $100. 27 red win $100. 6 black loses $100. 11 black breaks even. 9 red win $100. 4 black loses $100. 24 black loses $100. 10 black loses $100. 12 red breaks even. 6 black loses $100. 34 red breaks even. 20 black loses $100. 5 red wins $100. 9 red wins $100. 27 red wins $100. 10 black loses $100. 2 black loses $100. 9 red wins $100. 14 red breaks even. 7 red wins $100.

Again, we have had 8 wins and again we have had 8 losses so therefore, we have nothing.

Now we are going one step further, supposing you wanted to play black and low against red and high. It would make no difference because you are only covering an additional 9 numbers. whether you are playing black and low, or black and high. Low meaning 1-18; High, 19-36. You are only covering the other 9 numbers if you should be playing red and low and red and high, so it isn't advantageous one way or the other. Its purely a guess.

Again, if you were to play even and high against even and low, you'll notice you're going to cover the same quantity of numbers, totaling 27, no matter which way you play it therefore, it's really six of one and half a dozen of the others as to which one you would play. So the advantage that I have found in playing this method however, is, to play black and odd, rather than play black and even, gaining that extra 2 numbers and in turn, playing red and even against black and odd, rather that red and odd against black and even.

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