Thanks for the explanation! Isn't it basically "dollar cost averaging"?
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How much money would you need to Martingale your way to profits?
- Thread starter 1a2b3cppp
- Start date
Do you realize that martingale isn't strictly the name for the betting strategy? A system is said to be martingale if E[X(t+1)|E(0),... X(t)] = X[t]? That under any rational economic system, the double-down strategy is percluded as axiomatically possible just as true riskless arbitrage is?
Anyway, to save you some time, your original question belongs to a class of problems known as Stopping Time Problem (http://en.wikipedia.org/wiki/Optional_stopping_theorem). Yes - Stopping Time - as in, if you keep double down, even in a perfectly random martingale setting, what is the time at which stop out because you ran out of money.
In any case, you are theoretically guaranteed to lose. Not practically, but theoretically.
Anyway, to save you some time, your original question belongs to a class of problems known as Stopping Time Problem (http://en.wikipedia.org/wiki/Optional_stopping_theorem). Yes - Stopping Time - as in, if you keep double down, even in a perfectly random martingale setting, what is the time at which stop out because you ran out of money.
In any case, you are theoretically guaranteed to lose. Not practically, but theoretically.
Quote from 1a2b3cppp:
Can you share how you arrived at those numbers?
How are you coming up with these %s and profit targets?
Quote from 1a2b3cppp:
Can you share how you arrived at those numbers?
How are you coming up with these %s and profit targets?
These are just quick generic numbers if I make some assumptions, and are only for purposes of quick explanation to demonstrate the point in real number terms.
I ran probability stats for a few different instruments under several scenarios. A short list would be;
Trading S&P mini futs with a 10-point target, which would also require a 10-point doubling factor on the losing side.
I ran them again with a 2-point target, which changed the numbers some.
I tried it with SPY options. Selling calls, buying calls, and also with puts.
The numbers all change for sure. But the basic fact is that it is all based on how often you want to allow a blowup event to occur, statistically. A probability of 0.1% would be once in a thousand trades. If you are trading 10 times a day, then a blowup event will happen at a frequency of about once every 5 months. OTOH, if you were trading only once every week, then blowup would be expected once during a 20 year period.
The numbers that suggested a 2.5% annual return carried a 0.05% blowup probability. Which coincided with a 40 year period. Problem is that you still don't know when in that period it might be.
Quote from sjfan:
Do you realize that martingale isn't strictly the name for the betting strategy? A system is said to be martingale if E[X(t+1)|E(0),... X(t)] = X[t]? That under any rational economic system, the double-down strategy is percluded as axiomatically possible just as true riskless arbitrage is?
Anyway, to save you some time, your original question belongs to a class of problems known as Stopping Time Problem (http://en.wikipedia.org/wiki/Optional_stopping_theorem). Yes - Stopping Time - as in, if you keep double down, even in a perfectly random martingale setting, what is the time at which stop out because you ran out of money.
In any case, you are theoretically guaranteed to lose. Not practically, but theoretically.
You won't blow up if you make the scale wide enough. It's mathmatically impossible. But the down side is you may sit on an open position for years waiting for it to go back up.
Christ you are dense. It's mathematically impossible only if you either have inifinite wealth AND infinite time. In fact, it's mathematically impossible for you to be successful using this strategy so long as you not god. Math is not your friend here.
To wit:
"The optional stopping theorem can be used to prove the impossibility of successful betting strategies for a gambler with a finite lifetime (which gives condition (a)) and a house limit on bets (condition (b)). Suppose that the gambler can wager up to c dollars on a fair coin flip at times 1, 2, 3, etc., winning his wager if the coin comes up heads and losing it if the coin comes up tails. Suppose further that he can quit whenever he likes, but cannot predict the outcome of gambles that haven't happened yet. Then the gambler's fortune over time is a martingale, and the time ô at which he decides to quit (or goes broke and is forced to quit) is a stopping time. So the theorem says that E[Xô] = E[X1]. In other words, the gambler leaves with the same amount of money on average as when he started. (The same result holds if the gambler, instead of having a house limit on individual bets, has a finite limit on his line of credit or how far in debt he may go, though this is easier to show with another version of the theorem.)"
To wit:
"The optional stopping theorem can be used to prove the impossibility of successful betting strategies for a gambler with a finite lifetime (which gives condition (a)) and a house limit on bets (condition (b)). Suppose that the gambler can wager up to c dollars on a fair coin flip at times 1, 2, 3, etc., winning his wager if the coin comes up heads and losing it if the coin comes up tails. Suppose further that he can quit whenever he likes, but cannot predict the outcome of gambles that haven't happened yet. Then the gambler's fortune over time is a martingale, and the time ô at which he decides to quit (or goes broke and is forced to quit) is a stopping time. So the theorem says that E[Xô] = E[X1]. In other words, the gambler leaves with the same amount of money on average as when he started. (The same result holds if the gambler, instead of having a house limit on individual bets, has a finite limit on his line of credit or how far in debt he may go, though this is easier to show with another version of the theorem.)"
Quote from 1a2b3cppp:
You won't blow up if you make the scale wide enough. It's mathmatically impossible. But the down side is you may sit on an open position for years waiting for it to go back up.
Quote from sjfan:
Do you realize that martingale isn't strictly the name for the betting strategy? A system is said to be martingale if E[X(t+1)|E(0),... X(t)] = X[t]? That under any rational economic system, the double-down strategy is percluded as axiomatically possible just as true riskless arbitrage is?
Anyway, to save you some time, your original question belongs to a class of problems known as Stopping Time Problem (http://en.wikipedia.org/wiki/Optional_stopping_theorem). Yes - Stopping Time - as in, if you keep double down, even in a perfectly random martingale setting, what is the time at which stop out because you ran out of money.
In any case, you are theoretically guaranteed to lose. Not practically, but theoretically.
Yes... it all comes down to basic game theory. You ask a person if it is possible to call a coin flip correct 1000 times in a row. They say no, but it actually is possible.
Here's the backtest on an FX pair over a certain period of time (hope the attachment works).
Basically, on an initial bet of $1000 (looking to target XX% of that as P/L) you'd need $64,000 to withstand any drawdowns. Over the time period it works out to slightly better than .55% return on equity (actually over ~10months), so slightly higher than the RFR.
There are many parameters you can use, this was one quick and dirty version - I'm sure you can probably get slightly better results. Point being, it doesn't seem worth the capital imo.
Basically, on an initial bet of $1000 (looking to target XX% of that as P/L) you'd need $64,000 to withstand any drawdowns. Over the time period it works out to slightly better than .55% return on equity (actually over ~10months), so slightly higher than the RFR.
There are many parameters you can use, this was one quick and dirty version - I'm sure you can probably get slightly better results. Point being, it doesn't seem worth the capital imo.