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
