question about martingale system

In theory does the martingale system work if there is a game that has exactly 50/50 odds and no house edge and table limits?
I mean say i have $1000 to wager the chances of losing 10 times in a row are about ~0.0004% (math i used was 0.5x0.5x0.5') this is assuming im starting at a bet of $1 correct??

EDIT: this sounds like a sure way of making bank assuming i dont go on a massive losing streak correct?

Quote: limitedink

In theory does the martingale system work if there is a game that has exactly 50/50 odds and no house edge and table limits?
I mean say i have $1000 to wager the chances of losing 10 times in a row are about ~0.0004% (math i used was 0.5x0.5x0.5') this is assuming im starting at a bet of $1 correct??

EDIT: this sounds like a sure way of making bank assuming i dont go on a massive losing streak correct?

Hopefully you have more money than the casino. If you do there are better ways to make money.

Quote: limitedink

In theory does the martingale system work if there is a game that has exactly 50/50 odds and no house edge and table limits?
I mean say i have $1000 to wager the chances of losing 10 times in a row are about ~0.0004% (math i used was 0.5x0.5x0.5') this is assuming im starting at a bet of $1 correct??

EDIT: this sounds like a sure way of making bank assuming i dont go on a massive losing streak correct?

It's still 50/50.

Quote: limitedink

In theory does the martingale system work if there is a game that has exactly 50/50 odds and no house edge and table limits?

EDIT: this sounds like a sure way of making bank assuming i dont go on a massive losing streak correct?

OK. With no house edge, the Marty system neither gives you an advantage nor a disadvantage. In that situation, you are likely to make small amounts of money before losing everything. However, the chances of doubling your bankroll are the same as your chances of losing everything as per the following formula.

Probability of hitting a certain win goal P=(Initial Bankroll)/(Initial Bankroll + Target Profit) . . .OnceDear's rule of thumb.

and of course the Probability of losing all your money and all your winnings before hitting the win goal =1-P

So, for example if you are due to be put in front of a firing squad unless you can buy your freedom for $1010 and you have $1000 in your pocket, you would be wise to try to win that $10 needed by martingaling a coin toss.

Probability of turning 1000 into 1010 is 1000/(1000+10)=1000/1010=99.01%

But now lets say you had $1000 and needed $2000 to bribe your way to freedom.

Probability of turning 1000 into 2000 is 1000/(1000+1000)=1000/2000=50%
Which is exactly the same as your chance of tossing a coin once. So why bother.

Now let's say you needed 10,000 to survive (Maybe to feed and house yourself)
Probability of turning 1000 into 10000 is 1000/(1000+9000)=1000/10000=10%
and probability of losing all you had before achieving that is 100%-10%=90%

It's a bit like betting on 35 of 36 numbers of a roulette wheel: Most times you will win a tiny amount.

You cannot, of course assume you won't go on a massive losing streak. Each time you wager you add to the lifetime probability of doing so.


ps. your calculation is wrong. P of losing 10 times in a row is 0.5^10=0.000977 or 1 in 1024, which is pretty frequent if you make a few thousand wagers

Quote: OnceDear

OK. With no house edge, the Marty system neither gives you an advantage nor a disadvantage. In that situation, you are likely to make small amounts of money before losing everything. However, the chances of doubling your bankroll are the same as your chances of losing everything as per the following formula.

Probability of hitting a certain win goal P=(Initial Bankroll)/(Initial Bankroll + Target Profit) . . .OnceDear's rule of thumb.

and of course the Probability of losing all your money and all your winnings before hitting the win goal =1-P

So, for example if you are due to be put in front of a firing squad unless you can buy your freedom for $1010 and you have $1000 in your pocket, you would be wise to try to win that $10 needed by martingaling a coin toss.

Probability of turning 1000 into 1010 is 1000/(1000+10)=1000/1010=99.01%

But now lets say you had $1000 and needed $2000 to bribe your way to freedom.

Probability of turning 1000 into 2000 is 1000/(1000+1000)=1000/2000=50%
Which is exactly the same as your chance of tossing a coin once. So why bother.

Now let's say you needed 10,000 to survive (Maybe to feed and house yourself)
Probability of turning 1000 into 10000 is 1000/(1000+9000)=1000/10000=10%
and probability of losing all you had before achieving that is 100%-10%=90%

It's a bit like betting on 35 of 36 numbers of a roulette wheel: Most times you will win a tiny amount.

You cannot, of course assume you won't go on a massive losing streak. Each time you wager you add to the lifetime probability of doing so.


ps. your calculation is wrong. P of losing 10 times in a row is 0.5^10=0.000977 or 1 in 1024, which is pretty frequent if you make a few thousand wagers

Thank you this cleared everything up for me