I disagree, but will allow you to prove me wrong...
Suppose I have 127 BTC and want 128. I propose betting 1 BTC at 49.5% (2x payout). If it wins, stop, else double and repeat until it wins. So bet:
1, 2, 4, 8, 16, 32, 64
until one of them wins. If none wins, I've lost all 127 BTC. If any wins, I've got the 128 BTC I wanted.
This strategy will work so long as any one of the maximum of 7 bets wins. ie. unless all 7 lose. The probability of all 7 losing is 0.505^7 = 0.008376, so I have a 100 * (1 - 0.008376) = 99.1624% chance of success.
It's pretty likely that I'll end up betting less than the full 127 BTC. My expected total stake is less than 127. So my expected loss is less than 1.27 BTC.
What's your equivalent single bet strategy? How much does it expect to risk? How much does it expect to lose?
Jup. You're right. The equivalence only holds for betting without an edge. If there's an edge against the player, Martingale is slightly more favourable than a single bet with equal profit.