[h=2]
Destroying The Martingale System Myth[/h] Nearly every gambler will come across the Martingale System at some point in their lives. The premise is so simple that a large proportion of players will actually 'invent' this system themselves. But like most betting systems, the Martingale System is flawed at the most basic level.
[h=3]What is the Martingale System?[/h] The Martingale System can be used in many situations, but it is most often associated with roulette. The concept involves betting one outcome of red/black, odd/even or high/low and doubling the amount wagered every time the bet loses.
For example, a player places a bet of $1 on Black. If the outcome of the spin is Black, the player receives $2 - consisting of their $1 bet and a $1 profit. If the outcome of the spin is not Black, the player would lose their $1 stake. On the next spin they would double their bet to $2. Now if the outcome were Black they would win $4 - consisting of their $2 bet, $1 previous bet and $1 profit. If they lose the bet would double to $8, and so on.
As you may be able to see, the profit is limited to the value of the initial bet (in this case $1).
[h=3]Does It Work?[/h] Martingale System advocates will use the following logic to 'prove' the systems validity: The probability of a roulette wheel
never coming up black is zero, assuming that the wheel has a fair and unbiased spin. Or to be more accurate, the probability of the result of a roulette wheel never being black approaches zero as the number of spins approaches infinity.
The problem with this logic is that an infinite number of spins requires an infinite amount of money, and a casino willing to take bets of an infinite value. Both of which do not exist.
[h=3]Lets See The Numbers[/h] To help you better understand the numbers involved, the table below shows the amount of money required to last for up to 10 spins, doubling each time. Notice that by spin number 10 your initial $1 bet has turned into a bet of $512, and you have racked up total bets of $1,023. Bear in mind that if your $512 bet wins, you still only come out $1 up overall.
| Spin | Bet | Cumulative Bet | Profit |
| 1 | $1 | $1 | $1 |
| 2 | $2 | $3 | $1 |
| 3 | $4 | $7 | $1 |
| 4 | $8 | $15 | $1 |
| 5 | $16 | $31 | $1 |
| 6 | $32 | $63 | $1 |
| 7 | $64 | $127 | $1 |
| 8 | $128 | $255 | $1 |
| 9 | $256 | $511 | $1 |
| 10 | $512 | $1,023 | $1 |
In the grand scheme of things, $1,023 isn't a huge amount of money. And if you had $1,023 to 'invest' you may feel (or be conned into thinking) that the Martingale System could work for you. After all, what are the chances of a roulette wheel not coming up black 10 times in a row? Lets find out!
[h=3]You Cannot Escape The House Edge[/h] As you probably know, there are three colour outcomes on a roulette wheel - red, black and green. As you only win when the outcome is black and lose if it is red or green, your chances of winning are not 50-50. Taking the European 'single zero' roulette wheel as an example, your chances of winning are 18 (the number of black segments on a roulette wheel) in 37 (the total number of segments: 18 black, 18 red, 1 green). We can write this as:
P(win) = 18/37
P(win) = 0.486
Which makes the probability of you losing:
P(lose) = 1 - P(win)
P(lose) = 1 - 0.486
P(lose) = 0.514 (can be written as 1 in 1.95)
The probability of you losing two spins in a row can be worked out by taking the probability of one loss and multiplying it against it self, whilst three losses would be the probability of 2 losses multiplies by the probability of one loss, and so on. If we insert this information into the table above it looks like this:
| Spin | Bet | Cumulative Bet | Profit | Chance of loss |
| 1 | $1 | $1 | $1 | 1 in 1.95 * |
| 2 | $2 | $3 | $1 | 1 in 3.79 * |
| 3 | $4 | $7 | $1 | 1 in 7.38 * |
| 4 | $8 | $15 | $1 | 1 in 14.4 * |
| 5 | $16 | $31 | $1 | 1 in 28.0 * |
| 6 | $32 | $63 | $1 | 1 in 54.5 * |
| 7 | $64 | $127 | $1 | 1 in 106 * |
| 8 | $128 | $255 | $1 | 1 in 207 * |
| 9 | $256 | $511 | $1 | 1 in 403 * |
| 10 | $512 | $1,023 | $1 | 1 in 784 * |
* To 3 significant figures. If you wish to do the calculations yourself, start from the first step and use the un-rounded values.
So as you can see, the chance of you losing 10 spins in a row is approximately 1 in 784. Therefore, for every 784 sets of spins you would expect a run of 10 loses once. Seeing as you make a profit of $1 for every win, and a 10 loss run would cost you $1,023, then in theory for every $783 you win, you would lose $1,023 - making a net loss of $240 (783 winning spin sets minus 1 losing spin set). This assumes that you run out of money after the 10th losing spin and revert back to your initial $1 bet.
But this number comes as no surprise, and is directly related to the house edge. For those of you who are unfamiliar with this term, the house edge is the way in which casino tips the tables in their favour - for roulette this is the zero on the wheel. The house edge of a roulette table with a single zero is 1/37 - approximately 2.7%.
The table below calculates the theoretical total wagers for the 784 roulette spins mentioned above, which comes to $8,868.97. As we know that the house edge for this particular game of roulette is 2.7% we can calculate the expected loss as approximately $240 (to the nearest dollar) - which is the same number that we have already come to.
| Spin | Losses | Wins | Probability (as %) | Number of occurances | Total Stake |
| 1 | 0 | 1 | 48.65% | 381.55 | 381.55 |
| 2 | 1 | 1 | 24.98% | 195.93 | 587.79 |
| 3 | 2 | 1 | 12.83% | 100.61 | 704.3 |
| 4 | 3 | 1 | 6.59% | 51.67 | 775 |
| 5 | 4 | 1 | 3.38% | 26.53 | 822.47 |
| 6 | 5 | 1 | 1.74% | 13.62 | 858.33 |
| 7 | 6 | 1 | 0.89% | 7 | 888.52 |
| 8 | 7 | 1 | 0.46% | 3.59 | 916.13 |
| 9 | 8 | 1 | 0.24% | 1.84 | 942.73 |
| 10 | 9 | 1 | 0.12% | 0.95 | 969.16 |
| 10 | 10 | 0 | 0.13% | 1 | 1023 |
| | | | Total: | 8,868.97 |
jer bas zato jer su sanse 50% 50% nemoze u nedogled ici 20 crnih... tj. moze ali sanse su za to jako male...
