Bullet-proof martingale!?!

Here's a third way of looking at it - through mathematical expectation, which is (win% * avg. win) - (loss% * avg. loss).

Assume 50% win ratio and 1:1 reward:risk.

If your strategy is to stop after 3 losses in a row, then your loss % is: 0.5^3 = 12,5%
Your average loss is: 1+2+4=7

Win % is 100-12.5=87.5%
Average win: 1

Math expectation: 0.875*1 - 0.125*7 = 0

Math expectation WITHOUT martingale: 0.5*1 - 0.5*1 = 0

Now you see that martingale does not change expectation in any way. It only makes you have the smallest positions when you win in the first trades, thus reducing your actual reward:risk ratio, as I tried to show in the previous post.
What it does to YOU: it fools you, makes you believe that you cannot lose and it is psychologically very pleasing.
 
Quote from Indrionas:

Now you see that martingale does not change expectation in any way. It only makes you have the smallest positions when you win in the first trades, thus reducing your actual reward:risk ratio, as I tried to show in the previous post.
What it does to YOU: it fools you, makes you believe that you cannot lose and it is psychologically very pleasing.
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Your math apllies perfectly to coin tosses and probably to constant step grid blind entries. No matter the grid settings. All you said is correct. But market opens a wide range of possibilities. When you keep a close eye in the main trend (don't care about short term volatility, stick to the persistent moves) you will rarely reach the third layer. Having mistaken the perfect entry (the market goes against you) you simply wait. If you are in the right side, the market will soon revert. Your drawdown will be very small. You have not to pick the exact reversal point. Is the trend still valid? That is the question you must answer before making the second entry. Ready to revert, the market still goes against you, then you double up again? No. There is a warning sign. Maybe the trend is reversing. Trends use to revert in a high volatility fashion and you may profit nicely with the ups and downs. Market doesn't have to come back to the first entry price for you to win. There is much more than 50% probability of you closing the positions with a profit. I don't advocate a fixed rules use but a highly discretionary one. It must be highly dependent on the trader's experience. I think it makes all the difference.
 
Quote from Xtrader59:

Your math allies perfectly to coin tosses and probably to constant step grid blind entries. No matter the grid settings. All you said is correct. But market opens a wide range of possibilities. When you keep a close eye in the main trend (don't care about short term volatility, stick to the persistent moves) you will rarely reach the third layer. Having mistaken the perfect entry (the market goes against you) you simply wait. If you are in the right side, the market will soon revert. Your drawdown will be very small. You have not to pick the exact reversal point. Is the trend still valid? That is the question you must answer before making the second entry. Ready to revert, the market still goes against you, then you double up again? No. There is a warning sign. Maybe the trend is reversing. Trends use to revert in a high volatility fashion and you may profit nicely with the ups and downs. Market doesn't have to come back to the first entry price for you to win. There is much more than 50% probability of you closing the positions with a profit. I don't advocate a fixed rules use but a highly discretionary one. It must be highly dependent on the trader's experience. I think it makes all the difference.
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The parameters (50% win rate and 1:1 reward:risk) don't matter much, because you can change them any way, and you will reach the same conclusion: martingale doesn't change math expectation. I might as well used x and y for proving my point instead of randomly choosing 50% and 1.0


On the other hand, martingale tactics are valid if it is somehow part of your trading plan. Example: you treat 3 consecutive trades as one, you may split the risk among these 3 attempts, e.g. risk 0.1% on the first attempt, 0.2% on the second and 0.4% on the third, this way you will know you won't lose more than 0.7% per trade (triad).

Also, always remember that markets do not follow normal distribution. There are plenty of 5-sigma, 6-sigma days, even 8-sigma and more. And some day, sooner or later you will be caught on the wrong side of such price move and your position will probably be loaded because of martingale sizing. Don't believe the hype, I know there are plenty of threads about so called "grid trading" in forex related forums. In my honest opinion, it's a bullcrap.
 
Quote from Indrionas:

Also, always remember that markets do not follow normal distribution. There are plenty of 5-sigma, 6-sigma days, even 8-sigma and more.
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Completely agree. I do my best to not bother much about intraday moves (time frames from M1 to H4), trying to stick with D1 to MN trends. Don't go against a strong trend and I am always ready to change side if the trend clearly changes. 300 pips against me take me down USD 30,00. Perfectly comfortable.
 
Quote from Indrionas:

It doesn't matter where you take your loss. You're just exponentially decreasing your reward:risk ratio with every next trade after loss, while keeping win rate the same (strategy remains the same).

Here's example:

Assume you have 50% win ratio and 1:1 reward to risk ratio (i.e. risk one unit to gain one).

If you apply martingale strategy, you have these odds:

50% (1 win) that you risk 1 units and gain 1 unit

25% (1 loss, 1 win) that you risk 2 units and gain 1 unit

12,5% (2 losses, 1 win) that you risk 4 units and gain 1 unit

6,25% (3 losses, 1 win) that you risk 8 units and gain 1 unit

3,125% (4 losses, 1 win) that you risk 16 units and gain 1 unit

etc.

Now let's look at the big picture:

Your average reward to risk ratio: 0.5*1 + 0.25*(1/2) + 0.125*(1/4) + 0.0625*(1/8) + 0.03125*(1/16) + ... = sum of 0.5^n * 1/(2^(n - 1)) where n goes from 1 to infinity, this function converges to 0.6666, that is, your average expected reward to risk ratio is 2:3, which is less than 1:1 which you would be getting without martingale.

No matter how you look at it, martingale with stop after x losses is the same as strategy with constant risk, but with martingale you're only reducing your gains.
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Can you do the same analysis for an anti-martingale process?
Is there a way to make anti-martingale process overtake martingale process (by having both long & short positions of the same instrument) mathematically such that the net is profitable?
 
Quote from scalpmaster:

Can you do the same analysis for an anti-martingale process?
Is there a way to make anti-martingale process overtake martingale process (by having both long & short positions of the same instrument) mathematically such that the net is profitable?
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First, you need to define what is anti-martingale. Martingale is known as doubling (tripling, quadrupling, etc.) risk to recoup losses and win some trivial gain. While anti-martingale is not understood as lowering risk twice, thrice, etc. after a loss. It usually means keeping risk constant - simple as that - some fraction of the equity.

You may analyze anti-martingale vs. martingale, but the outcome is clear to me. The first thing that comes to my mind is to check risks of ruin. A risk of ruin for martingale strategy is very high, and anti-martingale is impossible to ruin (theoretically).
The second point, is that reward to risk ratio with martingale tactics is reducing exponentially, while anti-martingale keeps reward-to-risk ratio constant. That's the main differences between these two.
 
Quote from Big Game Hunter:

Sorry i mean what percentage it would lose more than 5 times assuming 750 separate event cycles...
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why would the prob be dependent on the number of separate events?
 
Quote from rosy2:

why would the prob be dependent on the number of separate events?
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Suppose it were 75/25% with 2 events. The probability of 2 losses in a row would be 0.25 x 0.25 = 6.25%.

Now suppose it were 1 million events. I don't know exactly how to calculate, but the probability of 2 losses in a row would be very close to 100%.
 
Quote from scalpmaster:

Is there a way to make anti-martingale process overtake martingale process (by having both long & short positions of the same instrument) mathematically such that the net is profitable?
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To pyramide in both directions at the same time would deliver big profit in sideways markets. But when the market breaks out and begins to trend your negative positions will turn into a nightmare if the pullbacks are not enough to allow you to close them. You may take the losses before they grow too much but I am not sure if the risk/reward is attractive. I believe not.
 
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