Is it possible to beat 50/50 with equal risk/reward?

Okay, so I appreciate fixed TP and SL have been discredited in other posts, but if we imagine a system that targeted some predefined constant == takeprofit == stoploss, how possible would it be to achieve a win% of say 55%+ over a period of time?

I just figured that if risk == reward (after commission, slippage, spread, rollover etc) and win% > 50, then surely the crazy old martingale system actually makes sense..?

I've tried some very basic decision making algorithms but with my limited experience have only been using the standard indicators in my bots and have never yielded anything significantly consistent in WFO; something I painstakingly did in MT4 before I discovered the magic of Zorro.

I note in the Zorro Manual it mentions something about a machine learning approach that was correct 57% of the time on EUR/USD H1 but can't really figure out under what conditions (https://manual.zorro-project.com/ under Main Topics > Trading Strategies, the blue graph bit).

But the question remains. Is beating equal risk/reward trading with fixed TP/SL possible?

I don’t think stop losses were discredited.

The real issue is TP. You need to determine on a case by case basis whether it is appropriate for your trading system.

One way to check this is to Test without TP and produce and analyze MAE and MFE plots.

Can you do well with TP? Maybe, but you might have better options.

Okay, but I was fishing around with the idea of seeing whether the Martingale strategy might actually be viable (albeit prone to occasional significant drawdown) if you had equal risk reward, but greater than 50% win rate.

In the above scenario we could apply expected value theory and show a projected long term return.

Example

a system with equal net risk/reward = 1
winrate 55%
maximum affordable consecutive losses : 4 (on the 5th loss you cut your losses and start again)

Without Martingale we have 55 win - 45 loss = net +10 after 100 positions, therefore each position has an expected value of +10/100 = 0.1

With Martingale an absolute loss (5 consecutive losses) is worth -31 and has a likelihood of 0.45^5 =~0.0185
a non-absolute loss therefore has a likelihood of 1 - ~0.0185 = ~0.9815 and is worth 1
The net win - loss is then (0.9815 x 1) - (0.0185 x 31) = 0.41 for each position based on expected value theory.

0.41 > 0.1

This number is improved with a higher max loss count as you might expect.

Well, Raisinbran, I think you have some wrong hypothesis here:

First you are assuming that your signal win rate will keep constant in the future and I think this is a big mistake. Try to backtest your signal in different periods, i.e. from 2013-2015 and 2014-2016. Does it give you the same win ratio in both scenarios?

Second, you are not thinking on real execution issues. Even if your TP/SL orders are placed in a "equally" distance, at the end will be executed as market stop orders, so you should assume some slippage. Also, you are not counting trading costs: spread and brokerage commissions, and depending on the instruments you may have other additional charges. So in practise your theorical 50-50 R/R won't be symmetric.

There are many kind of "Martingales", some of them interesting depending on the system. I recommend you to study them and to understand the exponetial risk increase of each one.

Cheers,

This is interesting. i dont think that higher stake can increase the expectancy of a trade, which is still 0.1 with martingale and not 0.41. Something with the math is wrong, but where is the mistake and how is it calculated correctly?

The math is not wrong, the mistake is that the 0.41 are not from higher expected value but from higher investment. The martingale need $32 capital for surviving 5 losses. If you invest the $32 without martingale, you had $6 per trade for surviving 5 losses, so you have 0.6 profit per trade, not 0.41. So, martingale loses even with 55% win rate.

Interesting, I hadn't considered this. Essentially for a given starting position size the martingale wins, but actually you can afford a much higher starting size for non-martingale if we assume the same max loss for each.

So after 100 positions at $1 each, max loss ~$32, martingale is up $41. Without martingale you can afford $6 each for the same maximum drawdown and that would yield 10 net wins at 55% win rate, which is $60.

I guess the only difference there then is that with martingale the losses have to be consecutive, whereas without it the losses are cumulative and hence the max DD is much more likely since any given win only takes you one step back towards breakeven, whereas martingale would bring you back to + 1 initial position size (barring incidents after max loss). Although I'm not sure that last statement is entirely true

In any case, thanks for your replies. I guess the only way to test this without nailing the theory is to run simulations over a large number of controlled trades with different position sizes for each strategy to balance out max DD as above. I guess R would be good for this but I'm not confident with it for now so it might have to be a basic excel bodge XD

I am wondering does SL/TP really matter at all for this question. If you are looking for better than 50/50 that has to presume that you have some method of entry and/or exit that generates a higher probability of a win than loss?

On a 50 point bet (tp=50, sl=50), with a 1 point spread (2%) you would need a 51% win rate to break even. If your 'method' gives you > 51% you would have something that can make money.

If you/system cannot see through the 'noise' then the market will be random to you and you will not do better than 50/50. If you can only do 50/50 you are losing money because of costs.

I guess the 'noise' is why we have algorithms that try and filter. In reality though, are those filtered features anything more than random over the long term?

If you are doing less than 50/50 then you have either an emotional problem, or a brilliant system that is running upside-down!

The magnitude of the bet is irrelevant yes. I was simply exploring the notion that if you can beat that breakeven point (>51% win in your example WickedBoy), would Martingale make sense from a mathematical point of view in expected return per bet since all non-max losses are wiped out with scaled wins, and the occasional max loss is covered by the intermittent wins.

It just feels logical that if you've been able to improve your odds of winning without affecting the payout, then it makes sense to compound the "unlikliness" of losing anything by using Martingale.

I think the significant thing that I overlooked here though is that in order to be able to survive a maxloss, your initial bet has to be much lower since it ramps up exponentially. Whereas without martingale you can stay somewhere in the middle of that bet size range and net a better result (see 0.41 vs 0.6 comment from Petra).

Also the point you make about finding a great system upside down got me excited back in the day too! I was still working on the idea of this Martingale thing working and so just needed any old system that was consistent, win or lose, and I'd reverse the signals if it was consistently losing XD. Couldn't even manage that. Everything I tried seemed to be nearly equally random, with any improvements between systems almost certainly put down to over fitting to historical data. I managed to get a couple of ideas to give good results in a walk forward optimization but as they say; throw enough sh*t at the wall and something will stick, so I wasn't convinced...