Roulette Strategies

Probably the most popular of all is the Martingale. It's most likely as old as gambling is itself. This system has you increasing your wager after a loss… the old "double or nothing" routine.

You continue to increase your bets in an up-as-you-lose fashion until you finally do win. At that point, you begin the progression all over again. Your objective is to win 1 unit, usually on the even-money wagers. If it were a fair game, you wouldn't get hurt in the long run. Let's assume that the casino was feeling generous, so they removed the zeroes from the rotor and left the pay outs the same. You are there and observe that several "black" numbers appeared, so you decide to wager on "red" up to five times in a row, to try and hit at least once. Your betting progression would look like this:

1. Bet $5 on red. If you win, repeat step 1. If you lose, go to step 2 (50% of the time).
2. Now bet $10 on red. A win takes you back to step 1. A loss takes you up to step 3 (25% chance of happening).
3. Bet $20 on red. If you hit, go back to step 1. If you lose, go to step 4 (12.5% chance).
4. Bet $40 on red. A win has you start a new progression. A loss has you go on to 5 (1 in 16 or a 6.25% chance of occurring).
5. Now bet $80 on red. A win puts you back at step 1. A loss also puts you back at step 1 (or to the ATM machine). At this point, you have lost the whole progression. The chance of losing the whole series is 1 in 32, which is 3.125% of the time.

You'll note that if you win at any time during the progression, you will be ahead $5 or 1 basic betting unit. You will have won that progression and will attempt a new one. Because the zeroes were removed (a fair game), your chances of winning $5 in the first step are exactly ½. Your chances of losing $5 are therefore ½ also.

The chances that you will lose steps 1 and 2 are [1/2]^2, or [1/2] x [1/2] = ¼. The chances of losing the first 3 steps are [1/2]^3 = 1/8, for a 1 in 8 chance. Following this down through step 4 (1 in 16 chance), and finally at step 5, the probability of losing the whole series is [1/2]^5 = 1/32, or 3.125% of the time. That means you will win the progression 96.875% of the time! "That's virtually a lock!" you say. But let's take a closer look…

If I played 32 cycles of the progression, I will win my 1 unit 31 out of 32 attempts or 96.875% of the time. So, 31 x $5 = $155, not bad.

But, I will lose the whole series 1 time in 32, or 3.125% of the time. This means 1 x ($5 +$10 +$20 +$40 +$80) = <$155>.

That works out dead even! $155-$155 = $0. I guess that's why they would call it a fair game. By the way, the length of the betting progression has no bearing here on the consequences… longer or shorter, it still breaks even. Over the long haul, the person betting on this game will not get hurt.

Let's say that the casino wasn't making any money offering a fair game, so they added the 2 zeroes back in, but kept the payouts the same. The progression will look the same, but the chances of winning are reduced. We now have 20 ways to lose out of 38 numbers:

1) $5 on red. You will now lose [20/38]^1, or 52.63% instead of 50% of the time.
2) $10 on red. Now loses you [20/38]^2, or 27.70% instead of 25% of the time.
3) $20 on red. You lose [20/38]^3, or 14.58% of the time instead of 12.5%.
4) $40 on red. Will have you losing [20/38]^4, or 7.67% instead of 6.25% of the time.
5) $80 on red. You now lose [20/38]^5, or 4.04% instead of 3.125% of the time.

"So I lose a little less than 1% more often," you say. "What's the big deal?" Let's look at the unfair game more closely:

In 32 cycles, you will now win only 95.96% instead of 96.875% of your wagers. So, (0.9596) x 32 x $5 = 30.707 x $5 = $153.54. "Not too far off from the $155, previously," you're quick to point out.

But, you will see we are losing much more… (0.0404) x 32 x $155 = 1.2928 x $155 = <$200.38>!!

By the way, you now know how a system can win 96% of the time (as claimed in some sales ads) and still lose money overall. The systems sellers forget to mention that the one loss will, over the long run, wipe out all of your winnings… plus! The calculation above shows our net loss is $153.54-$200.38 = <$46.84>. This is the average you will lose over 32 cycles of playing this progression. "Wow, I'm losing 1% more often and I'm averaging -$46.84 over 32 attempts… I know, I'll just increase the progression to 6 bets! That should more than pick up that extra 1%," you reason.

Well, let's see…

6) Now raise bet to $160 on red after 5 straight losses. This bet will lose [20/38]^6, or only 2.13% of the time! Great!

Hold on… the judges are conferring…

In 32 cycles, you will win 97.87% of your wagers. So, (0.9787) x 32 x $5 = 31.318 x $5 = $156.59.

But, you are losing… (0.0213) x 32 x ($5+ $10+ $20+ $40+ $80+ $160) = 0.6816 x $315 = <$214.70>!!

$156.59-$214.70 = <$58.11>, for a net loss of $58.11 every 32 cycles on average. "I'm losing 24% more money by extending the progression from 5 to 6! Let's shorten it down to 4 instead," you correctly conclude. Now you are on the right track. The longer your progression, the more you will lose with that progression. Remember, you are exposing larger amounts to that house edge on the back end of your progression. It's simple really, the more that you bet, the more you will lose.

Using our calculations from above we can determine what our expected loss will be on a 4-bet progression. We notice that we will lose $40+ $20+ $10+ $5 = $75. This will happen 7.67% of the time. So, (.9233) x 32 x $5 = $147.73 in average winnings and (.0767) x 32 x $75 = <$184.08> in loses. Our resulting net loss has been reduced to <$36.35>! If you follow the mathematics all the way down to a progression of 0 bets, then you will reduce the losses down to $0.