Do a modified martingale.
Instead of betting 100, 200, 400, 800, .... until you win @ 2x.
bet 10, 10, 10, 10, 10, 20 ,20, 20, 20, 20, 40, 40, 40, 40, 40, .... until you win @ 10x.
You could guess a max lose streak of 100 before you win. So that makes 100 / 5 = 20, then (2^20)*5 = 5242880 which equals the minimum amount of Satoshi you need to bet using this technique with a starting bet of 1.
Any martingale is the same as the original x2 martingale. This method isn't 100% safe as the varience is bound to get you after playing for sometime I tried various kinds of martingale but I still ended up losing BTC. No method is safe to be honest, its a matter of time before you lose if you are lucky due to the house edge.
Its not the same as a standard martingale. In my example;
1st 40 satoshi bet wins the (40*10) - ((10*5)+(20*5)+40) = 210 win
2nd 40 satoshi bet wins the (40*10) - ((10*5)+(20*5)+(40*2)) = 170 win
3rd 40 satoshi bet wins the (40*10) - ((10*5)+(20*5)+(40*3)) = 130 win
4th 40 satoshi bet wins the (40*10) - ((10*5)+(20*5)+(40*4)) = 90 win
5th 40 satoshi bet wins the (40*10) - ((10*5)+(20*5)+(40*5)) = 50 win
I'm basically dividing the original martingale bet by 5 giving me 5 times a chance to win instead of 1. Now the chance of winning a 10x bet is much harder then a 2x times, but you only have 1 chance to win on 2x. By splitting the bet you have a range of potential gains as compared to fixed gain.
1st 80 satoshi bet wins the (80*10) - ((10*5)+(20*5)+(40*5)+80) = 370 win
2nd 80 satoshi bet wins the (80*10) - ((10*5)+(20*5)+(40*5)+(80*2)) = 290 win
3rd 80 satoshi bet wins the (80*10) - ((10*5)+(20*5)+(40*5)+(80*3)) = 210 win
4th 80 satoshi bet wins the (80*10) - ((10*5)+(20*5)+(40*5)+(80*4)) = 130 win
5th 80 satoshi bet wins the (80*10) - ((10*5)+(20*5)+(40*5)+(80*5)) = 50 win
As you can see as the lose streak increases the win range increases also.