Mathematical Systems - Part 1Part 2 Part 3
Systems, systems, systems… everyone wants a "good" mathematical system to employ at the tables! Some of you have given me the details of systems that you are currently playing and have asked me to analyze them, which I have done. Unfortunately, there are no "good" mathematical betting systems in existence… only bad, very bad and worse yet, systems.
But, before I get too far into the discussion, let me help define what a betting or mathematical system is. A mathematical system is a scheme by which the amount of your next wager is determined by what happened on your last bet. Raise after a loss, raise after a win, or some convoluted method of tracking sequences of losses or wins to formulate subsequent wagers is a system. No matter how great a system you believe that you have, one thing's for certain. There are still 38 pockets on the American double-zero wheel and the casino only pays you 35 to 1. What does this mean? That means that you are taxed on your winnings and not your losses. Losing is free, so to speak. If the game were fair, you would receive 37 to 1 on a straight up win and 18 to 1 on a split, for example. You are always paid as if there were 36 pockets instead of 38. This is how the house is able to run a thriving business! Please realize that all mathematical systems will lose over a sustained period of time.
The House's Edge
Let's calculate the house edge for any roulette wager. First you will take the actual payoff minus the correct payoff. Now multiply that by the probability of hitting your number. Multiply that by 100 to convert to a percentage and you come out with the house edge that the casinos bank on:
[35/1 - 37/1] x 1/38 x 100 = -5.263%, a negative player's edge.
Incidentally, the edge on the single zero wheel is similarly calculated, but remember that we are only shorted 1 unit instead of two. Also, the probability of hitting is 1 chance in 37. So, we have:
[35/1 - 36/1] x 1/37 x 100 = -2.703%, against the player.
"Yeah, but I have this new betting system that I just bought," you say. "The seller unconditionally guarantees that I will win 96% (or some impressive percentage) of the time!" I'll show you that although technically correct, these claims are seriously misleading!!
The "Due" Theory
Most staking systems would have you bet on some number or group of numbers. A group of numbers could be the "red" or "odd" numbers, or the second column numbers, or the 1 through 6 line, etc., etc. Usually, you are instructed to increase the amount of your next bet based on something impressive called "The Maturity of Chances" or "The Law of Large Numbers." Those who know better refer to this as "The Gambler's Fallacy." The endorsers of these "due" systems will have you believe that sooner or later, your group has to come in. Well, that's true over the long run; it could be sooner but it could be later… much, much later at times. Over the course of 50,000 spins, the chances of, say, the number 8 not occurring for 200 spins in a row would not even register a "hiccup" in the overall statistical scheme of things. You will not have the bankroll necessary (infinite) to execute your betting scheme and, even if you did, the house imposes minimum and maximum bet sizes, so you are limited regardless.
These "systems" have been around for hundreds of years. They did not work then, and they still do not work today. Anyone who claims that they're selling a "newly developed" or "recently rediscovered" betting system is feeding you a line. It is probably some variation of a Martingale, Labouchere, Fibonacci, or d'Alembert system. What tends to happen with any of these systems is that you may experience a series of small wins, deceptively building a false sense of infallibility. However, as you continue to play, you inevitably will be wiped out by one catastrophic loss… and if you're really unfortunate, you could experience that disaster right away!! Let's go on to discuss the most widely used of these systems in detail.
Probably the most popular of them 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).
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.
"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.
The Grand Martingale
One notable variation to the Martingale is the Grand Martingale. The only thing "grand" about this system is the fashion in which you will lose your money. This more aggressive cousin has you double your last bet and add one more unit. If our betting unit is $5, then the progression would look like this:
1) $5 on red. You lose [20/38]^1, (52.63% of the time) and win 47.37% of the time.
The "grand" total that you are prepared to lose on just one progression with this system is [$5 + $15 + $35 + $75 + $155] = $285. Let's calculate our average net loss for the same 32 cycles of the progression:
So, this 5-bet progression loses $76.64 in "Grand" style versus the more modest $46.84 for the plain Martingale. The Grand loses 63.6% more money on a 5-bet progression!! This system is deadly, folks. Stay clear!
The Probability of a Run
Let's say that the casino that you frequent most has 6 roulette tables. On each of these tables you'll find 3 sets of even-money wagers: red vs. black, odd vs. even and high vs. low. That presents 18 possible runs at any given time. The chances of any one of these groups having a losing run of, say, 6 in a row is [20/38]^6, or 0.02126. That's about 1 chance in 47. Now factor in 3 such sets at each table times 6 tables and your chances of encountering such a run are 18/47. That works out to be 38+% of the time! A run of 6 blacks, for example, isn't as rare as you might think! How about a run of 8, in a casino with 12 roulette tables? [20/38]^8, or 0.00589 which is multiplied by 36. That's 0.21197 or better than 21% chance of finding a run of 8. These events will happen, and when they do, think back and remember what you read here. Imagine if you had been doubling up against one of these "monsters" from the get-go!!
Obviously, the less we play, the less we lose. Some methods are worse than others. My recommendations, if you must play a Martingale-type system, are as follows:
1) Wait for a particular group to be absent for 6 to 8 spins before running a progression. If there are 4 or less wheels open, then wait for a string of 6 losses. If there are 5 to 10 open wheels, then 7 will suffice. Anything over 10 will require a run of 8. Technically, this won't increase your chances of winning, but it will keep you off of the tables more (minimizing losses).
Part 2 Part 3