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Started By
Message
Roulette - Martingale System
Posted on 2/9/18 at 2:09 pm
Posted on 2/9/18 at 2:09 pm
Let me start by saying I know I can't beat the casino and I know this will never work in the long run.
But can anyone give me the odds on the best time to quit on the double down method considering I have 7 lost spins of the roulette wheel to play?
So if I play $10 hands using the Martingale and have a bankroll of $640 (7 spins) when should I cash out? With this bankroll, I was thinking around $250-400. Can't find it anywhere and I know TD will come thru in the clutch.
But can anyone give me the odds on the best time to quit on the double down method considering I have 7 lost spins of the roulette wheel to play?
So if I play $10 hands using the Martingale and have a bankroll of $640 (7 spins) when should I cash out? With this bankroll, I was thinking around $250-400. Can't find it anywhere and I know TD will come thru in the clutch.
7
Posted on 2/9/18 at 2:26 pm to AmericaOverParties
quote:
But can anyone give me the odds on the best time to quit on the double down method considering I have 7 lost spins of the roulette wheel to play?
The moment you win.
ETA - there is no formula that will give you an optimal time to cash out. The expected value of the martingale system with roulette is negative. If you're ever positive, you've beaten the system.
This post was edited on 2/9/18 at 2:27 pm
Posted on 2/9/18 at 2:29 pm to slackster
quote:
The moment you win.
I'm meaning how many spins? I figured out theres a 1.115% chance that I would lose 7 spins in a row but this is out of 7 spins.
I need to know how many spins would it take to get reach the 30%...40%....50%? I would never go past the 50% mark.
Posted on 2/9/18 at 2:33 pm to AmericaOverParties
Assuming no double zero, the chances of you doubling your money would be 30%. As slackster said, -EV from the start.
Posted on 2/9/18 at 2:36 pm to AmericaOverParties
98%
97%
94%
87%
75%
50%
Pct of initial bankroll (640) left after consecutive losing spins
97%
94%
87%
75%
50%
Pct of initial bankroll (640) left after consecutive losing spins
This post was edited on 2/9/18 at 2:38 pm
Posted on 2/9/18 at 2:42 pm to slackster
quote:
If you're ever positive, you've beaten the system.
Well, you're def positive from the beginning. I have greater than a 98% chance that I could win 1 of 7 spins so as long as I have the funds to double down and play 7 spins then I would have greater than a 98% chance to do so.
Posted on 2/9/18 at 2:47 pm to AmericaOverParties
His point is that your win expectancy converges to 1, but never gets there.
Meaning the limit of the win expectancy as your number of tries goes to infinity is 1.
Meaning the limit of the win expectancy as your number of tries goes to infinity is 1.
Posted on 2/9/18 at 2:49 pm to castorinho
quote:
Pct of initial bankroll (640) left after consecutive losing spins
Thats just doubling the percentage and not what I'm asking.
I have a 1% chance to lose 7 times in 7 spins. I know that my chances to lose will increase the number of spins I play. Therefore, I would be more likely to lose 7 times in 8 spins, 9 spins, etc. Can anyone give me the percentage chance for each?
An example would be, I might have a 30% chance that I'd lose 7 spins consecutively out of 50 spins total. <--------That % is not correct but thats the number I'm looking for.
Posted on 2/9/18 at 2:54 pm to AmericaOverParties
The percentage reflect the bankroll exactly because you chose 640 based on the 7 spins. If you chose a random number, say 700, it would not match.
Are you saying that in this hypothetical, you'll keep betting even if you win? Do you reset to 10 after a win?
Your question isn't clear
Are you saying that in this hypothetical, you'll keep betting even if you win? Do you reset to 10 after a win?
Your question isn't clear
Posted on 2/9/18 at 2:57 pm to castorinho
quote:
Your question isn't clear
My mistake, yea martingale goes like this:
Bet $10, if win bet $10, if lose double and bet $20
Continue to double down until you win the hand then return to $10 and start again.
Posted on 2/9/18 at 3:14 pm to AmericaOverParties
quote:
An example would be, I might have a 30% chance that I'd lose 7 spins consecutively out of 50 spins total. <--------That % is not correct but thats the number I'm looking for.
That's actually not too far off in my estimation. The chance of losing 7 straight times with double zeros is 1.12%, based on a binomial distribution calculator. Next, take that 1.12% as the likelihood of success (failure in practice), and find the probability of it happening >0 times over 44 trials, and you'll get 39%.
44 trials would be 44 rolling 7 trial periods, which would be the first 7 spins, then 43 more spins after that, so 50 spins total.
The probability of losing your bankroll in 50 spins is 39%.
The math is solid, but my logic could be off.
Posted on 2/9/18 at 3:14 pm to AmericaOverParties
So your max win is $10 in that scenario, and you hope that your string of losses runs ends before you hit the table max?
Posted on 2/9/18 at 3:24 pm to slackster
quote:
The probability of losing your bankroll in 50 spins is 39%.
The math is solid, but my logic could be off.
Appreciate it bro, thats what I'm looking for. So I probably could spin about 60 times before reaching that 50% mark. Relatively speaking, I'd have about 50-55 spins to do my damage before cashing out my earnings while keeping the "odds" in my favor of not losing all my bankroll.
Posted on 2/9/18 at 3:27 pm to ATLdawg25
quote:
So your max win is $10 in that scenario, and you hope that your string of losses runs ends before you hit the table max?
The max win would be $640, if I lost 6 consecutive hands I would be betting my 7th hand at $640. At the tables I've seen, $1,000 is max bet.
Posted on 2/9/18 at 3:31 pm to AmericaOverParties
quote:
The max win would be $640, if I lost 6 consecutive hands I would be betting my 7th hand at $640.
But you're not actually winning $640, are you? That will only get you back to the original $10 bet that you lost.
Posted on 2/9/18 at 3:35 pm to ATLdawg25
quote:
But you're not actually winning $640, are you? That will only get you back to the original $10 bet that you lost.
Right. Chances are I could play 50-60 spins and not lose 7 consecutive hands, however this does not tell me the amount of money I would win on 50-60 spins.
This post was edited on 2/9/18 at 3:36 pm
Posted on 2/9/18 at 3:38 pm to AmericaOverParties
It just seems like a ton of risk. If the table max is $1,000, you're going to lose your bankroll. It's just a matter of time.
If you have one bad streak and lose up to the $1000 table max, it would take 100 winning rolls to offset that.
If you have one bad streak and lose up to the $1000 table max, it would take 100 winning rolls to offset that.
Posted on 2/9/18 at 3:41 pm to AmericaOverParties
quote:
Appreciate it bro, thats what I'm looking for. So I probably could spin about 60 times before reaching that 50% mark. Relatively speaking, I'd have about 50-55 spins to do my damage before cashing out my earnings while keeping the "odds" in my favor of not losing all my bankroll.
Around the 65 spin mark it's more likely than not you'll have hit 7 straight losers. That doesn't mean you're likely to be up any money during that time though.
Posted on 2/9/18 at 3:44 pm to AmericaOverParties
quote:
Let me start by saying I know I can't beat the casino
I don't think you do based on this thread and your responses....
Posted on 2/9/18 at 3:55 pm to ATLdawg25
I would like to give myself a 60% or greater chance to win so that would mean 50 spins. With a 47% winning chance for each spin that would put me at 23.5 wins with a $235 profit on $10 hands.
So I would be risking $1270 to win $235 on avg.
Def not worth it being a 40% chance to lose $1270 vs a 60% chance to win $235. Pretty bad odds. Might as well sports bet.
So I would be risking $1270 to win $235 on avg.
Def not worth it being a 40% chance to lose $1270 vs a 60% chance to win $235. Pretty bad odds. Might as well sports bet.
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