Can you evaluate this roulette betting strategy?

So I haven't really adjusted the Martingale strategy but instead made a distinction on what would be an acceptable and favorable outcome in terms of winning a specific amount of money.

Here is where I need the help

If I decided to have a bank roll of $5110 and a progessive doubling bet starting at $10 maxed at $2560 enough for 9 losses would it be profitable on a single 0 table if I always walked away with a $100 profit before returning to the table the next day?

I found a website that allowed me to try this system but only gave out results for a single days attempt and in 100 attempts I never didn't make a $100 profit. I didn't keep track of the minimum/max or average number of spins needed or the average bet just if I was successful. This gives the potential for a profit $35,600 if succesful in 356 consecutive days.

For some people I guess that kind of profit isn't worth the trip to the casino but I'm ok with that return.

So here is the test,

Gambler A has decided to play roulette for 50 years (this particular gamblers lifetime) and only plans to stop once he's lost all his money. He begins his gambling career with $5110. Each day Gambler A goes to the Casino and plays roulette with a starting bet of $10, doubling on all loses to a max of $2560 since the table maxes out their bets at $5000. Gambler A only plans to play until he wins $100, no more and no less.

What is Gambler A's first years profits? What is his 10 year profit (assuming he is successful in avoiding a total loss of all his funds that he won and began with)? What is his lifetime profit (assuming he is again successful in avoiding a total loss of all his funds that he won throught his life)?

Statistics desired:
First year's profits:
10 year profits:
50 year profits:
Total number of lifetime positive profits out of 1000 lifetimes:
Total number of years this strategy yields a profit before their is a total loss:

Betting progression is as follows (assuming there is no total loss in the first few attempts) 10, 20, 40, 80, 160, 320, 640, 1280, 2560, back to 10 because of table max. Upon a win the bet returns to $10.

Thanks

ps Please also test a system where the betting progression is 25, 50, 100, 200, 400, 800, 1600, 3200, back to 25 (Bank Roll 6375 and same profit outcomes). And lastly a system that bets a progression of 100, 200, 400, 800, 1600, 3200 ( Bank Roll of 6375 and same profit outcomes).

Quote: natecbc

What is Gambler A's first years profits? What is his 10 year profit (assuming he is successful in avoiding a total loss of all his funds that he won and began with)? What is his lifetime profit (assuming he is again successful in avoiding a total loss of all his funds that he won throught his life)?

Not a mathematician here, like many others on this site... I read a few paragraphs and the thread title...

Eventually, he will lose. There is no way to beat a game with a built-in house edge.

If there's a statistician out there who can disagree, be my guest.

Each spin is independent irrespective of same wheel or same day or next day or next wheel. You count up the number of slots on the wheel and that's your possibilities today and tomorrow, after winning 100.00 and after winning zilch... The wheel never knows you left and came back, the little white ball doesn't know if its day or night.

Please also test a system where the betting progression is 25, 50, 100, 200, 400, 800, 1600, 3200, back to 25 (Bank Roll 6375 and same profit outcomes). And lastly a system that bets a progression of 100, 200, 400, 800, 1600, 3200 ( Bank Roll of 6375 and same profit outcomes).

Quote: FleaStiff

Each spin is independent irrespective of same wheel or same day or next day or next wheel. You count up the number of slots on the wheel and that's your possibilities today and tomorrow, after winning 100.00 and after winning zilch... The wheel never knows you left and came back, the little white ball doesn't know if its day or night.

true but I suspect that if we generated a billion spins and you left after a $100 win and re entered at a varying number of spins (lets say in a day a live table spins 500-750 times, and you leave and enter within a range of 350-600) that does affect when you might face recurring lossess and wins. I think based on what you are saying you sit for 1 billion spins to see if it generates a profit in which case you don't find yourself avoiding the span of 10 consecutive losses. The variable can be when you re-enter the sequence of 1 billion numbers. I would imagine this is true but I'm not certain...also I would imagine then it would be impossible to calculate your winnings and loses with such a variance...but with a randomized entry int to the 1 billion numbers within that span of spins (350-600) you'd at least get a figure.

Quote: natecbc

So I haven't really adjusted the Martingale strategy but instead made a distinction on what would be an acceptable and favorable outcome in terms of winning a specific amount of money.

The martingale is a net loser. It's been covered in many places. Losing 9 coin flips in a row is a 1 in 512 chance. Losing 9 spins in a row is a 1 in 322 chance. That means in 1 year, you'll have about 1 or 2 sequences where you wipe out.

Here's the secret about the Martingale (and other system were you increase your bet on a loss) - you look like you've found a great system at first, as you keep having a winning sequence, time after time. You'll many times more often than you lose.

Problem occurs is when you lose, you lose big, and potentially wipe out all your winnings so far.

Quote:

Here is where I need the help

If I decided to have a bank roll of $5110 and a progessive doubling bet starting at $10 maxed at $2560 enough for 9 losses would it be profitable on a single 0 table if I always walked away with a $100 profit before returning to the table the next day?

No. If by profit you mean "can I expect to make a profit on this"? If you play it 30 times, you probably will make a profit of $300. There's about a 1 in 10 chance you'll lose all $5110. 9 times you win $300, 1 time you lose $5110.

Quote:

I found a website that allowed me to try this system but only gave out results for a single days attempt and in 100 attempts I never didn't make a $100 profit. I didn't keep track of the minimum/max or average number of spins needed or the average bet just if I was successful. This gives the potential for a profit $35,600 if succesful in 356 consecutive days.

For some people I guess that kind of profit isn't worth the trip to the casino but I'm ok with that return.

I'm not surprised you played for 100 days and won them all. That will happen about 1 time 4.

Quote:

So here is the test,

Gambler A has decided to play roulette for 50 years (this particular gamblers lifetime) and only plans to stop once he's lost all his money. He begins his gambling career with $5110. Each day Gambler A goes to the Casino and plays roulette with a starting bet of $10, doubling on all loses to a max of $2560 since the table maxes out their bets at $5000. Gambler A only plans to play until he wins $100, no more and no less.

What is Gambler A's first years profits? What is his 10 year profit (assuming he is successful in avoiding a total loss of all his funds that he won and began with)? What is his lifetime profit (assuming he is again successful in avoiding a total loss of all his funds that he won throught his life)?

Statistics desired:
First year's profits:
10 year profits:
50 year profits:
Total number of lifetime positive profits out of 1000 lifetimes:
Total number of years this strategy yields a profit before their is a total loss:

Betting progression is as follows (assuming there is no total loss in the first few attempts) 10, 20, 40, 80, 160, 320, 640, 1280, 2560, back to 10 because of table max. Upon a win the bet returns to $10.

I could sit down and code all this up and work it out, but there's not much reason why. I know a 9 step martingale loses about 1 time in 322 on the Roulette wheel. A finger in the air estimate suggests Gambler A will have lost $2,150 each year (winning 364.88 days, losing 1.12 days). He might be wiped out if he gets two 9 step streaks, or have lost his entire bank roll early on after a few small wins.

After 10 years, he will probably have no profit at all.
After 50 years, doing this daily, I'd be surprised to see a single lifetime profit.

Quote:

Thanks

ps Please also test a system where the betting progression is 25, 50, 100, 200, 400, 800, 1600, 3200, back to 25 (Bank Roll 6375 and same profit outcomes). And lastly a system that bets a progression of 100, 200, 400, 800, 1600, 3200 ( Bank Roll of 6375 and same profit outcomes).

8 step Martingale loses 1 time in 170, 7 step Martingale, 1 time in 89.

Give or take. Assuming a American Roulette wheel.

In short ... no series of negative expectation bets can add up to a positive expectation.

Quote: natecbc

true but I suspect that if we generated a billion spins and you left after a $100 win and re entered at a varying number of spins (lets say in a day a live table spins 500-750 times, and you leave and enter within a range of 350-600) that does affect when you might face recurring lossess and wins. I think based on what you are saying you sit for 1 billion spins to see if it generates a profit in which case you don't find yourself avoiding the span of 10 consecutive losses. The variable can be when you re-enter the sequence of 1 billion numbers. I would imagine this is true but I'm not certain...also I would imagine then it would be impossible to calculate your winnings and loses with such a variance...but with a randomized entry int to the 1 billion numbers within that span of spins (350-600) you'd at least get a figure.

Nope.

Your entry point into a list of random numbers does not matter. If they are truly random, there is no difference if you look skip 100 numbers, take every third number or only start after 5 consecutive numbers less than 13.

Quote: thecesspit

ps Please also test a system where the betting progression is 25, 50, 100, 200, 400, 800, 1600, 3200, back to 25 (Bank Roll 6375 and same profit outcomes). And lastly a system that bets a progression of 100, 200, 400, 800, 1600, 3200 ( Bank Roll of 6375 and same profit outcomes).

8 step Martingale loses 1 time in 170, 7 step Martingale, 1 time in 89.

Give or take. Assuming a American Roulette wheel.

In short ... no series of negative expectation bets can add up to a positive expectation.

I'm assuming European rules and single 0. The $100 min progression to the max bet of $3200 I guess you could extend it to $6400 but you'd have to win 127 times with no losses just to be ahead of one potential loss. I haven't really factored in imprisonment into the calculation as the sites I'm using fail to lend the option.

Okay for a single-0 wheel (no half en prison rule) :

9 step Martingale loses 1 time in 402
8 step loses 1 time in 206
7 step loses 1 time in 106

En prison would adjust these slightly, and a half push on a 0 would mean you'd need to make adjustments to the strategy slightly. These would raise the number of attempts before a failure (but still mean it'd happen too often to make a profit).

Well I tried this test. With a bankroll for $100000, a min $100, a max bet of $25600, I was successfully able to acheive a win of $35600 in 700-1000 spins 41 times out of 50. So this changes the original premise slightly in that you could lose the max bet and continue by trying to recoop all the money for a max bet loss in a series Although in a max bet loss the money was unrecoverable before subsequent max losses resulted in a complete loss of bankroll and winnings. Again I've not included En prison rules. This essentially accomplished nothing in that who has $100000 isn't staking it for $100 wins.

Here are my numbers

Quote: natecbc

So I haven't really adjusted the Martingale strategy but instead made a distinction on what would be an acceptable and favorable outcome in terms of winning a specific amount of money.

Here is where I need the help

If I decided to have a bank roll of $5110 and a progessive doubling bet starting at $10 maxed at $2560 enough for 9 losses would it be profitable on a single 0 table if I always walked away with a $100 profit before returning to the table the next day?

No.
you only have a 97.544616% chance of winning $100 before total starting bankroll ruin.
(1-(19/37^9)) = 99.751706% to win one time ($10) and 99.751706%^10 = 97.544616%
to win 10 times in a row for $100 profit.

2.455384% chance of a progression ruin or 1 in 40.7 attempts.
To me, way too high of a risk of ruin.

Quote: natecbc

I found a website that allowed me to try this system but only gave out results for a single days attempt and in 100 attempts I never didn't make a $100 profit.

97.544616%^100 = 8.323965% or about 1 in 12. You were on the good side of luck.
($10,000 profit)

Quote: natecbc

I didn't keep track of the minimum/max or average number of spins needed or the average bet just if I was successful.

To win $100
9 step Marty
average bet = $100.37
average number of trials = 20.5

Some data:
Probability of doubling a $5110 bankroll ($10 starting bet - 9step Marty) before losing $5110 = 99.751706%^511 or 28.072996% or 1 in 3.56 (280,730 out of 1 million playing this way will double, Yahoo!)

Out of 1 million players playing this way about 719,270
would bust out at different levels, never ending with a profit after losing a $5110 progression. (Most had some money left over, not $5110)
Not the 24,554 that busted their starting bankroll before EVER winning!
We call them losers.
They are NOT impressed with natecbcSRS (super roulette system)

Quote: natecbc

This gives the potential for a profit $35,600 if succesful in 356 consecutive days.

97.544616%^365 = 0.000114606 or 1 in 8,725.58 or 115 out of 1 million players playing your method would be driving a new car to the bank.
Now will they do it again for the next year? Sure why not! They now have a bankroll.

Quote: natecbc

For some people I guess that kind of profit isn't worth the trip to the casino but I'm ok with that return.

So here is the test,

Gambler A has decided to play roulette for 50 years (this particular gamblers lifetime) and only plans to stop once he's lost all his money. He begins his gambling career with $5110. Each day Gambler A goes to the Casino and plays roulette with a starting bet of $10, doubling on all loses to a max of $2560 since the table maxes out their bets at $5000. Gambler A only plans to play until he wins $100, no more and no less.

What is Gambler A's first years profits? What is his 10 year profit (assuming he is successful in avoiding a total loss of all his funds that he won and began with)? What is his lifetime profit (assuming he is again successful in avoiding a total loss of all his funds that he won throught his life)?

Statistics desired:
First year's profits:
10 year profits:
50 year profits:
Total number of lifetime positive profits out of 1000 lifetimes:
Total number of years this strategy yields a profit before their is a total loss:

Betting progression is as follows (assuming there is no total loss in the first few attempts) 10, 20, 40, 80, 160, 320, 640, 1280, 2560, back to 10 because of table max. Upon a win the bet returns to $10.

Thanks

ps Please also test a system where the betting progression is 25, 50, 100, 200, 400, 800, 1600, 3200, back to 25 (Bank Roll 6375 and same profit outcomes). And lastly a system that bets a progression of 100, 200, 400, 800, 1600, 3200 ( Bank Roll of 6375 and same profit outcomes).

Now, given my math formulas, you can answer the rest of your questions.
Easy stuff

Sally

if you really want to martingale, at least play a game with a lower house edge and better chance of winning like the player bet in baccarat or the banker bet in EZ baccarat.

Quote: natecbc

Well I tried this test. With a bankroll for $100000, a min $100, a max bet of $25600, I was successfully able to acheive a win of $35600 in 700-1000 spins 41 times out of 50.

Seems too high. I say more like 41 out of 100.
The max bankroll needed for the same 9 step Marty is only $51,100
Each win is $100 net.
So you only need 356 wins in a row.

x= (19/37)^9 = losing 9 in a row
1-x = 99.751706% prob of not losing 9 times in a row and winning $100 one time.
99.751706%^356 (you need 356 wins in a row)= 41.270295%
Hey not bad.

$1,003.69 average bet
730 avg # of spins

Player in Baccarat would give a
45.547469% chance to achieve a win of $35600

But you get the idea on how to do this now.
One can easily make a table in a spreadsheet, like Excel.

You can even calculate the chances of losing, say 50% of your starting bankroll or ending up at with any value of a bankroll.
This assumes you follow a proper x-step Marty sequence where you win the same amount at each step.