The oldest roulette systems Part 2

Some folks will try to snag you with the idea that you can somehow decrease this $.48 loss by only betting the red numbers in the third column. The problem is that, while this phony Three-To-Two offers a whopping 22-percent chance of hitting a 15:1 payoff, you’re only robbing Peter to pay Paul. Really, the amount you lose per spin remains constant regardless of whether you play the real or phony version. For the real one, your total bet each time was $25 on a $10 table and your loss each spin averaged out to roughly .2 percent of that. The expected-value loss for the phony Three-To-Two is roughly the same (.3 percent), but your total exposure goes up slightly to deal with the multiples of eight generated by having to bet eight separate props evenly. Assuming again that you’ve found the only French wheel in Europe to accept dollars and have a $10 minimum, you would have to bet at least $24 on Black for every $16 you bet on the red numbers. So, even though your average expected-value loss remains the about same, you’ll wind up losing more money over time. Consider again the expected-value formula:

(-$40 x 10.4995/37) + (-$16 x .5005/37) + ($8 x 18/37) + ($30 x 8/37)=?
(-$11.35) + (-$.22) + ($3.89) + ($6.49)=?
$10.38 – $11.57= -$1.19

So the phony Three-To-Two is out, but it does give us our second piece of advice: Always bet at the lowest denomination possible because, at some point, the expected-value loss becomes so miniscule that it’s negligible. Fore example, on a hypothetical $1-minimum French table where you play $3 on the color and $2 on the column, the expected-value loss is only $.10 for each spin.

Now, once you’ve gotten your loss over time as small as you can get it, the next thing you have to do is overcome it. The best way to do this is to add a little technique that’s vaguely similar to the D’Alembert. We are stressing the word “vaguely” because both practices require you to raise and lower your bets. However, this concept, a true “series system,” has its grounding in statistical theory, not everyone’s lame-brain idea that magical fairies control the wheel. Essentially, it works because, as we know, the probability of the ball landing on a number you bet is very high. It is so high, in fact, that it is actually more probable for the ball to land on two covered numbers in a row (about 49 percent) than it is that you will lose everything once (about 30 percent) or lose everything after a win (about 20 percent). So, if you double both bets after one on them wins, you stand a good chance of doubling what you would otherwise win.

The problem, of course, is that two wins in a row should happen about once every two spins, while a win-loss combination (found by multiplying the probability of a loss and a win) should occur about once every five spins. And, doubling after every win, you will, over a long time, lose about $20 for every $7 you make. The shorthand solution is, once again, to use null betting and this time to wait out after every second win until the next string of two losses occurs; at the beginning of a session, you always assume you’ve won twice already and wait for the first two-loss string before betting. If you win, you’ll double your bet, and if you win again, you’ll stop until the next set of two losses shows up. If three losses in a row do occur (i.e. you get hit with one of them), bet again, and if a four-loss string hits, leave the table. This will decrease your number of regular losses because three wins in a row will only happen about as often as a single loss (both should occur about once every three spins) and, after that second win, your bets become fair game. Meanwhile, it pits the probabilities of winning once (again, 70 percent) and winning after a loss (about 20 percent) against the probability of three losses in a row (about 3 percent), making the casino work extra hard to keep up with you.

Lastly, the Three-To-Two is a grinding system. Even at its best, you won’t walk away with the big hull, but rather, with a tiny bit of extra spending money. Occasionally, too, you’ll have a bad day just like with the Martingale. In this case, however, the Law of Big Numbers won’t give you much protection because you will average a tiny loss for each spin. Our advice, then, is that you set your bankroll at about eight times your initial bet and only allow yourself two losses before quitting. Also, if you play the system and everything is working plus-perfectly, set a win goal of 100 percent your bankroll for that session. (If you bring $8 to a $1 table and are up $8, leave). It may not be very much money. But at least you’ll be in the black.

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The oldest roulette systems

The D’Alembert

Ringing in as the second oldest roulette system, the “D’Alembert” does a lot to correct for the Martingale’s shortcomings. Unlike its forerunner, you don’t need deep pockets to play it and will probably never hit a table limit with it. This is because it is a “series betting system.” That is, it isn’t a true progressive like the Martingale (which calls for only increased bets) and instead requires you to both lower your bet by one unit when you win and raise your bet by one unit when you lose.

The problem is that, although the D’Alembert is named for a famous French mathematician, the theory behind it (that the more often an unselected outcome happens, the more likely a selected outcome becomes) is plain-out hogwash. A number or color is never “hot” or “due,” and if you let yourself think so, you’re bound to wind up in a reality-T.V. version of that movie “Lost in America.” For this reason, we’ve decided not to describe the D’Alembert and strongly advise you against using it. If you do decide to try it for experiment’s sake, however, simply treat it as a Martingale with your initial bet set at twice the table minimum; if you win, decrease your bet by half, and if you lose double it. Otherwise use all the corrective strategies we explained for the Martingale.

The “Three-To-Two”

The Three-To-Two is kind of like the James Bond of roulette systems: it has gone by many names and everyone who realizes it’s there mistakenly thinks he’s a super-genius. Basically, it requires you to bet on the Black and the third column or Red and the second column at a ratio of 3 to 2 (hence its proper name). That is, for every three units you place on the color, you have to place two units on its corresponding column.

At first glance, the Three-To-Two looks promising because, technically, it covers about 70 percent of the wheel. But the truth is, its expected value shows that it’s still a negative-sum bet (i.e. you will lose a small amount of money to the house every spin). Consider: Every time you win on only a color bet (about 38 percent of the time on an American wheel), you will lose your whole column bet, making your real payoff about 1-3. Likewise, if you win on only the column bet (21 percent of the time), you will lose all your color bet, and your real payoff will be 1-2. On the rare occasion that you win both (about 11 percent of the time), yes, you will win a much larger amount, but even this is only a 7-5 payoff. To put it in terms of sheer numbers though, here’s the expected value for a Three-To-Two when it’s played with unvaried bet amounts on an American wheel with a $10 minimum:

(-$25 x 12/38) + ($5 x 14/38) + ($5 x 8/38) + ($35 x 4/38)=?
(-$7.89) + ($1.84) + ($1.05) + ($3.68)=?
$6.57-$7.89= -$1.32

So you see, the Three-To-Two isn’t half as ingenious as the casinos and their lackeys (i.e., non-GP gambling-news sites) would have you believe. We would, however, like to point out that it’s not entirely impossible to use because, like the Martingale, it can be tweaked to some extent. Foremost of these corrections is that, if you’re going to use it, you must play it on a French Wheel that offers En Prison, and you must always take the held-over option. This will up your probability of winning overall and sometimes give you 3/5 of your original wager back when a zero hits. Plugging the Single-Zero numbers into the expected-value equation, we find the loss rate per spin is a little more in our favor:

(-$25 x 10.4995/37)* + (-$10 x .5005/37) + ($5 x 22/37) + ($35 x 4/37)=?
(-$7.09) + (-$.14) + ($2.97) + ($3.78)=?
$6.75 – $7.23= -$.48


Note: .4995/37 is the equivalent to the 1.35-percent house advantage expressed as a fraction; because En Prison only partially protects you from losing your total bet, the possible loss for this fractional value is still $25; the “-$10” represents the amount lost on the column if you win your imprisoned bet back.

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Martingale betting system Part 2

“But that can’t possibly happen,” you say, “because the probability of winning on an even-money prop is 47 percent!”—And believe us, man, we feel you.

But as we explained in Section IV, probability is hellova complex. For instance, it is relatively probable that a long string of losses will occur—which we know because we can calculate the likelihood of it happening by multiplying the probabilities for all the events together. So, going back to our even-money bet, all we have to do to find the probability of a two-loss streak is multiply 20/38 by 20/38. The two twenties equal 400, and the two thirty-eights equal 1444, so the fractional probability is 400/1444. Now, if we simplify that by dividing 1444 by 400, we find that the probability of a double loss is 1/3.61 or about once every four spins.

Really, deciding if you can play a system, then, is as easy as accounting for the amount of money you will probably lose and seeing if, over time, you can still come out ahead. No doubt, we don’t want you committing the “gambler’s fallacy” or falling into a “law of averages” pit trap, but it is a mathematical truth according to the “Law of Large Numbers” that, the more times a wheel is spun, the closer the average number for each occurrence will come to the ideal probability. Again, this doesn’t mean 10 spins or 100 spins. It means 1,000, 1 million, 1 billion spins over time.

Armed with this little bit of information, let’s go back to our example and see how it works out: For one thing, we know that Black is an even-money bet, and for an even-money bet, the probability of losing seven times in a row on an American wheel is 1/89 or once every 89 spins. As we mentioned in Section IV, there are about 105 spins per hour. So on a table with a $500 limit, you will probably lose $635 once every hour. Meanwhile, 105 spins aren’t nearly enough of a sample to get your average number of wins close to the ideal probability. And even if they were, each time you’d win, you’d only net $5 for 49 wins (i.e., a loss of about $400)!

Yet, this isn’t to say winning with the Martingale is entirely impossible. To get the odds in your favor, all you have to do is play a table with a 200-1 limit and “null bet.” Null betting—the practice of waiting out a couple spins until several consecutive losses have occurred—forces the house to produce an improbable string of losses over such a large number of spins that you can’t help but win. Also, because you aren’t actually betting anything on the first couple losses, null betting allows you to lose about the same number of times you would have for a more probable losing streak. In the case of our Black bet, probability states that it will have an 11-loss streak only once in about 1,164 spins. This is closing in on that large sampling we need to ensure that our ratio of wins to losses will average out to about what their ideal probability is. What’s more, we only need to wait out for three losses in a row, which happens every seven spins or so.

Over time, if we continue to wait until a three-loss streak occurs before betting, we should win more than we lose. Because the average number of spins per hour is 105, we can reasonably guestimate that an 11-streak will occur once in about eleven hours, with the house netting a total of $1,275. Meanwhile, because we’d be playing enough spins to get close to the ideal probability and because nothing but an 11-loss streak can hurt us, we’d rack up somewhere in the ball park of $2,500—that’s a profit of about $1,200, not counting the times you’ll lose only half your bet if you’re playing a wheel with Surrender.

So, yes, Virginia, you can use a Martingale to win, but probably not in the way you imagined. You can’t, after all, walk into a casino with a nickel and walk out a millionaire. That’s simply the stuff of fairy tales. In reality, you’ll need a large enough bankroll to hold you through that one big loss, plus play out the rest of a 1,164-spin series.

Finally, we can’t promise that you won’t have two or three—heck, even ten—eleven-loss streaks in one 1,164-spin session. Again probability is at best only a guestimate of what is likely to happen, not what will. The wheel itself has no clue that you’re playing it, let alone what numbers it should choose, and you can almost assuredly bet that you’ll eventually have a couple bad days. For this reason, we’re advising you to set a loss limit: if you reach 600 spins and are still in the hole by a 60 percent margin, scram.

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Martingale betting system

Once upon a time, roulette was simple. One guy would spin a ball around a wheel and a few others would bet on it. Nowadays, things aren’t nearly as cut-and-dry. But some of the time-tested strategies thought up during the game’s early years can still offer today’s player a winning edge when played correctly.

As we teach you to use these classic systems, we’ll also show you how and why each of them originally worked. This might seem like drudgery, but understanding what makes a system tick is the only sure way to use it successfully. Moreover, because there’s very little you can do to get a mechanical edge in online roulette, knowing how to play these systems and correct them on the fly may be your only hope when the information superhighway finally snuffs out brick-and-mortar casinos.

The Martingale

The oldest system on record is known as the “Martingale.” This baby came to the fore back before casinos began setting table limits, and it’s also the main reason they did so. The theory behind it couldn’t get more straightforward: If you bet a proposition long enough, it’s bound to win, and if you increase your bet every time you lose, you’ll make profit when that win finally comes through. The original Martingale was played exclusively on outside bets. A player would, for instance, bet Black at the table minimum, and if he lost, he would double the amount he risked on the next spin. He would continue doubling the previous bet for every loss, and when he eventually won, he would rake in enough at a 1-1 payoff to cover everything he’d lost to that point, plus gain whatever he would have won on the initial bet.

To give you a better idea of how this works, though, let’s look at a betting-sequence chart. The following numbers reflect how the Martingale would play out, assuming we were at a $5-minimum table with no ceiling on how much we could bet at one time.

Seems to work perfectly, right? Well… sort of. You see, as we said before, there are ceilings to how much you can bet on a proposition at one time (i.e., “table limits”). In the past, casinos would set these in a ballpark range, but in the 20th century, gambling wizard John Scarne put a real edge to their sword by advising them to set the their limits at either a 100-1 or 200-1 ratio. What this means is that the table limit will usually be 100 or 200 times the table minimum. Looking back at our betting-sequence chart, then, you can see why this shuts down a classic Martingale; seven losses in a row would make it impossible for you to double your bet again, and you’d simply be out $635.

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