felt bored and calculated it
so imagine you have 7 btc and want to make it 8 btc and there is a house edge of 2%
chance you'll be able to do this with martingale is 86.375%
chance you'll be able to do this with a bet of 7 btc is 85.75%
just a difference for exactly the same outcome
Your calculations are wrong, with a 0% house edge doing it with martingale is a 87.5% and doing it all in with 7 btc would be also 87.5% so it doesnt matter wheter you use martingale or all in, the only thing you win is time
This is also false. The distributions are not the same, even though the end effect is the same.
Consider one family with a 1.80m tall father, 1.60m tall mother, and a 1.70m tall son. The "expected height" of the family is 1.70m.
Consider another family with a 2.00m tall father, a 1.60m tall mother, and a 1.50m tall child. The Expected height is also 1.70m, but obviously the distribution is different.
Your example doesn't apply here. In OP's example you end up with either 0 or 8 BTC. There's no "son" who is somewhere between mother and father. There's just "lose 7" and "win 1" as the two possible outcomes, and they have different probabilities depending on which strategy you use. In that sense one strategy is clearly better than the other, in that it wins the same amount as the other but with higher probability.
It's the same thing with gambling strategies. The EV is the same, but distribution is different. In the case that you want to win 1BTC and EXACTLY 1BTC, no more, no less, with EXACTLY a 7BTC bankroll*, there are two optimal routes**:
1. Bet all-in at high odds
2. Martingale with a BASE BET of 1BTC.
These result in the same odds for earning exactly 1 BTC. Any other variation of the martingale would result in a lower chance.
*Any difference in ANY of the variables would change this conclusion.
**There are possibly other routes that would arrive at the same % chance, but these are not within the topic of martingale vs all-in.
The two strategies OP proposes:
A) bet 7 BTC at a multiplier of (8/7)x
B) bet 1 BTC at 2x, if you lose, bet 2 BTC at 2x, if you lose, bet 4 BTC at 2x
Hopefully you can see that they both have only two possible outcomes: you end up with 0 BTC or 8 BTC.
OP claims that the probability of ending up with 8 BTC instead of 0 BTC is higher with strategy B is higher than with strategy A. You claim that they are the same. You are wrong, and he is right (unless the house edge is 0% of course, in which case both strategies have a probability of success of 7/8). Do the math for yourself and you will see it.
Exactly. The OP can't do math, yet keeps telling others they're wrong. eh.
That seems unnecessarily rude. Especially when OP is right.
If you don't want to do the math, perhaps check some of the many threads where this has been discussed before. Light links to one such thread:
dig up an old conversation me a dooglus had about losing 'less' by betting less (ie. rather than all in betting, splitting your bets into n portions where you aim to get to x as a value then stop)
