Welcome guest, is this your first visit? Create Account now to join.

Welcome to the Australia Sports Betting Forum.

If this is your first visit, be sure to check out the FAQ by clicking the link above. You may have to register before you can post: click the register link above to proceed.

Thread: Martingale System plus online betting?

1. Martingale System plus online betting?

Someone please explain to me, why this, in theory, shouldn't result in an overall "Win".

Ok, lets say you only bet on football games. You always bet on the favorite with the lowest odds always being somewhere around 2.0-2,35. You do not bet on X, nor the underdog team. You ALWAYS bet on the favorite team with odds being around 2.0-2,35 all the time.

Now, you start betting on one match for lets say 100bucks.

IF you lose, you bet on a new match with exactly the same conditions as mentioned above, and bet the 100bucks you lost, PLUS another hundred bucks. If you win now, (for people not familiar with the martingale sys.) You win your loss, PLUS additional 100 buck. And then you just rinse and repeat, always making sure to double you loss for the next bet PLUS a little extra for pure winnings. another example, 1st loss: bet 100bucks 2nd loss: bet 200bucks 3rd loss: bet 400bucks 4th loss: bet 800bucks. As the martingale is made for casino only, it wont work due to the casino max bet. In online betting you are not restricted to any max bet. You can go bet 50.000 at bet365, and 50.000 on unibet, so lets say you play on 4 different sites, totaling for up to 100.000 usd.

THis means that you would be able to lose a game 10 times in a row, and pardon my mathematics, I would anticipate the odds of being one third every time, and divide by two for every new bet.

My calculations would look like this.

33% chance for your team to win.

33% / 2
=15%

now, i just need to do this ten times in a row to find the percentage chance for me losing 10 times in a row.

33%/2/2/2/2/2/2/2/2/2/2
=0,03%

this means that this would happen to me once every 3333 time.

in the meantime, i would have been able to make money many times, but since the martingale system takes a lot of "bets" to make a revenue, i would think that every day, it would take me approx 4 spins to earn me 100 bucks. considered the 365 days per year, this means that i would spent 1460 bets on winning 100bucks every single day for a year which amounts to about
36500USD!!

I don't know how to end this properly since I myself find it a bit hard to grasp, so if anyone could point out the positive and negative factors around this I would appreciate it

______________
valuepointdistribution.com

2. Hi chrisrieom, there are a few things I can point out:

1. If you lost nine bets in a row, you would need to wager \$51,200 for your 10th wager - the equation is \$100 * 2^(10-1). When you include the 1st through 9th wagers, your total spend is \$102,300, so a lot of money is required if you hit a bad streak.

2. The chance of losing the first bet is (1-0.33) = 67%. The chance of losing two consecutive bets is (1 - 0.33)^2 = 44.9%, etc., so the chance of losing ten consecutive bets is (1- 0.33)^10 = 1.82%, rather than 0.03%.

3. If you employ the Martingale system for a short enough amount of time then providing you have \$102,300 to wager with then it's unlikely you will lose ten times in a row, however if you play the system long enough you would be surprised at the high likelihood of hitting a 10 bet losing streak. Here are the mathematics behind it - source: http://en.wikipedia.org/wiki/Marting...etting_system)

Suppose you have \$102,300 to use on the system. This enables you to sustain nine consecutive losses, but you would lose everything if you lost ten times in a row.

The odds of losing once is 67%. If you place a total of ten bets the odds of losing all of ten are 0.67^10 = 1.82%. Therefore, the chance of being solvent (i.e. having enough to keep playing) is 1 - 1.82% = 98.18%.

However if you play again and again the chance of losing ten times at each subsequent bet is (1 - 0.67)*(0.0182) = 1.22%. If you played this strategy for 65 bets the probability of remaining solvent is 0.9818 * (1-0.0122)^(65-10) = 49.9%, so the chance of becoming bankrupt is 1 − 0.499 = 50.1%. If you played the strategy for 200 bets the probability of becoming bankrupt is 90.5%.

Have a play around with some numbers in Excel. I simulated the strategy and went bankrupt regularly when simulating a few hundred bets.

3. In all honesty.. martingale might seem great and work really well for awhile.. but you are limited by your bankroll and you will reach a point where you can't make the next bet. This will occur and you will probably need to learn the hard way.. also your book will only allow you to bet up to a certain amount. Either way you will be capped… losing streaks happen.

I use a modified labrouchere money management system for my plays… works a treat. Less risk and still decent reward.. if you google you will come up with plenty of reading material.. it may seem complicated at first but is actually quite easy.

BOL

Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•