Roulette theory (please help I would like to know the downsides to the system)

Roulette theory
First off love your site fabulous work.
Okay here we go,
You will be playing all 6 even money bets (red black even odd 1-18 19-36) at the same time but you will be using the martingale system on each of them Separately. (Stay with me here I do not agree with the martingale as much as you do cause I used to believe in it until I lost alot of money)
For instance let's make our base amount $10. Our very first bet we would put $10 on each even money spot totalling $60. Let's say the ball landed on red odd 1-18.
So the martingale strategy is to keep your winners at the base amount of $10 and double your losers. So on your next bet you would have $10 on red odd and 1-18.
The other 3 we would double to $20 (black even and 19-36). In a short summary you keep doubling the losers and keep taking the winners down to your base amount. So in theory you are winning $30 dollars per spin (unless it lands on 0 or 00 in which you would double all six even money bets) because 3 of the sections are making the $10 profit every spin. 3x $10=$30.
Now everyone knows the problem with the martingale system is if we hit a long streak it will raise your bets so high so quickly to bust your bankroll or hit the maximum on the table. So we are going to set a stop/loss on our betting at 5 spins. If on any of the six even money bets you lose 5 in at row you start it back at your base bet. My theory behind this is to be able to win the amount that you have lost when you start back over by not losing too much. So let's say again the base amount is $10. Let's say you do go the distance and you lose 5 in a row on a section. $10+20+40+80+160=$310. So you lost the $310 on the fifth spin. Let's see how many spins would it take to recover that amount. At $30 profit a spin it would take you 10.333 spins to win back what you lost but let's say 11 spins to win $330. But let's not forget that it took you 5 spins to get to that losing $310. So after you have lost you only have 6 more spins to recover your losses. Then you can keep spinning to start back on your actual profits. Losing 5 in a row is common yes but I wouldn't say that it happens every 6 to 11 spins. So overall you would have low risk of losing thousands of dollars with the martingale and you would be able to recover your losses with just a few more spins. What are the variables in this that I'm not thinking of? Or what mathematical mistakes have I made.
I have put it to the test multiple times and have came out on top everytime but I know it's too good to be true. I just wonder what would happen if it was put to the test of a million. Am I just getting lucky or is this actually working. For the credit I did come up with this all on my own I haven't told anyone about this strategy and I have never seen it on any site but I'm sure someone has tried this right?
Thanks for any comments I appreciate it.

Quote: codywhite80

Am I just getting lucky or is this actually working. Thanks for any comments I appreciate it.

Too long did not read... but you're just getting lucky.

It doesn't matter what your system is. I didn't even read it.

No systems work at roulette because every bet on the table carries a negative expectation. You can combine the bets all you like in any amount, and the end result will still be negative. All betting systems do is rearrange the amount and frequency of wins and losses. In the end every system loses at a rate equal to the house edge for the bets it makes.

Quote: codywhite80

but I know it's too good to be true...Am I just getting lucky or is this actually working. it.

If this worked the casinos would have
shut down roulette 150 years ago, and
they have not.

The short version: six simultaneous Martingales, with a stop point of 5 losses in a row.

Quote: codywhite80

So in theory you are winning $30 dollars per spin (unless it lands on 0 or 00 in which you would double all six even money bets) because 3 of the sections are making the $10 profit every spin. 3x $10=$30.

The problem is, when you do lose the same bet five times in a row, you lose $280 - and you run the risk of doing that on three different bets. You are also not taking into account the possibility that two bets can reach the $310 loss point simultaneously. (For example, if the numbers 1, 15, 7, 9, and 3 come up, that's five straight losses on both the even and high bets.) It's not common, but it's not impossible either. Replace that 15 with 1, 3, 5, 7, or 9, and you lose $310 on the black bet as well.

You are just getting lucky.

Quote: codywhite80

...playing all 6 even money bets...at the same time...

That's all you need to read about this fairly stupid system. The rest is a stop after five losses Martingale.

The basic concept is you will occasionally lose a lot (in this case stop-loss of $310) compared with winning $10 most the time. Sadly, as with all similar flawed systems, while the chances of losing is low and you keep reaping in $10's (i) in the long term you lose because you cannot get back your rare losses (in this case $310) with enough $10's before another loss comes along (ii) in the long run a loss occurs more often than 1 in 32.

Quote: codywhite80

Roulette theory
First off love your site fabulous work.
Okay here we go,

Welcome. I'll be gentle with you. Others might not.

Quote:

You will be playing all 6 even money bets (red black even odd 1-18 19-36) at the same time

That's first error. Putting money on red and black at the same time, even if different amounts is saying to the casino 'here: take 1/37th of the lower of the two amounts for no benefit to me: Do the maths on that in isolation.

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..you will be using the martingale system on each of them Separately. (Stay with me here I do not agree with the martingale as much as you do cause I used to believe in it until I lost alot of money)

You will indeed continue to do so.

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snip %<
What are the variables in this that I'm not thinking of? Or what mathematical mistakes have I made.

OK. consider yourself to be heading up a team of 3 players. One Martingales odd/even. One Martingales red/black and one Martingales High/Low with exactly the stop losses and quitting points you desxribe. Now, each of those team members will, on average bust out before doubling his money. We can calculate the probability, indeed you'll find such calculations charted on this site. Is any one of your team beating the house edge? Is any one of your teame getting a better result than the other two? No. Because if he was, you would cut your team down to one man and one strategy. Effectively you are heading up a team where each team member has less than a 50% chance of doubling his money. The expected team ( dollar) outcome is exactly the sum of the outcomes of the team members.

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I have put it to the test multiple times and have came out on top everytime but I know it's too good to be true.

You have tested a hedged Martingale. Martingales do win more often than they lose. You need to test how often your 'team' doubles its combined bankroll before going bust.

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I just wonder what would happen if it was put to the test of a million.

You would die of old age. You would stand approximately 45% chance of doubling your huge bankroll before losing it. To win a significant amount, you would need to bankroll an equally significant amount, with total loss being probable.

why waste a spin betting 60 on all when you walk up to the table go off the last spin and bet 30 on what didn't hit. then go from there. gives you one less green fade.