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Martingale's betting method can be seen here:https://www.investopedia.com/articles/forex/06/martingale.asp My question is if there is a way to put a non-exploding martingale, [There is one attempt to do this but without much success you can look here:https://www.mql5.com/en/articles/1800 ]

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No.

Suppose you have a capital base, $C$, and zero sum game, $G$, where for a unit play you can either win $x$ or lose $y$, such that $E[G] = p_x x + p_y y = 0$.

You devise any kind of strategy to play any number of units, $\alpha_i > 0$ for each successive game after a win. Now there always exists a chance that you will lose your entire capital base, because:

$$ C \leq \sum_i^N \alpha_i y $$

for some $N$, where that probability of that happening is $ \prod_i^N p_y $.

Example

The game is 50-50 win 1 or lose 1, and your capital base is 30. Your strategy is double bet on a loss, i.e. $\alpha=[1,2,4,8,16,..]$ Then: $$ 30 \leq (1+2+4+8+16) $$ and the likelihood of this happening is $2^{-5}\approx 3\%$, and the other 97% of the time you win 1 unit.

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  • $\begingroup$ If it always explodes why can't it be reversed [on a very small part]? If it always explodes the other will always earn? $\endgroup$ Jun 29 '20 at 21:25
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Position sizing, in and of itself, is insufficient to manage risk trading any financial market. You also need to be well informed about the expected ranges of prices on the particular instrument you are using.

You can use aggressive position sizing, but you'd have to modify the martingale so that you are not strictly using 2x, 3x, etc on each level. Otherwise you will run out of capital too quickly.

Would also recommend backtesting (with tick data) different market conditions to find out about different ranges of prices that occur.

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