What martingale does is to create "very high win chance" bets.
With a strategy of base amount 0.001, 2x on loss, total budget 1.024 (busted in 10 consecutive losses), what you are really betting is 99.89% (=1-0.505^10) win chance bets.
So, you are just hoping you will never see a rare event happens (10 consecutive losses in this case).
The other thing to consider is how long it takes to get there. At 0.001 BTC, to get 1 BTC you'd need to win 1000 times before that happens. The chances of this are 1/1024, but you have to consider that it's inclusive of all rolls. So you aren't rolling 1000 times. Every loss adds another to it. If you win and lose 50%, it would take 2000 rolls to get there, but there is a 1/1024 chance you're going to cap out and lose it all.
Another thing to note is that, you don't get the option of "very high win chance" bets for free.
Indeed, you are paying a high price (a much higher equivalent house edge).
Take my above case as an example, and assume the house edge is 1% (as in PD and JD).
The chance of 10 consecutive losses is indeed about 1/927 (=0.505^10, for 1% edge), much higher than 1/1024.
So, the equivalent house edge is much higher than 1% now.
If you are doing martingale with lower base amount (or higher budget, ie more rounds to get busted), the equivalent house edge would be even much higher.
Say for base amount 0.000001, total budget 1.048576 (busted in 20 consecutive losses), the chance of losing would be about 1/859354 (=0.505^20) instead of 1/1048576.
Fixed a typo.