Quote from: freeyourmind on March 20, 2015, 10:43:32 PM

Yeah, good point, so the question is what is Y?

If I look at a high/low game with 47.5% odds of doubling your bet, after the house takes its cut, then let's say I'm willing to go 11 consecutive bets before I lose it all.

Bet Total Loss
1 1
2 3
4 7
8 15
16 31
32 63
64 127
128 255
256 511
512 1023
1024 2047

So assuming an initial bet of 1 satoshi, using martingale, my 11th bet would be 1024 and at that point I lose 2047 satoshi. The odds of losing 11 bets in a row is (0.525^11) = 0.0835% or 1/1197 bets. So for every 1197 bets, there will be one losing streak, on average, which will lose 2047+ on making 1196...not the best strategy in the long run.

Hopefully the math makes sense - feel free to correct if I made a mistake.

The math looks good, though I didn't check your calculations. There's a problem with it however: it doesn't answer your question: "what is Y?"

We want to know your chance of doubling up using this system. You seem to have 2047 satoshi, since you're able to lose 11 times before busting. So in order to double up, you need to win 2047 times. The probability of winning 2047 sequences in a row is (the probability of winning one sequence) to the power of 2047. The probability of winning one sequence is (1 minus the probability of losing 11 rolls in a row), and the probability of losing 11 rolls in a row is, as you said, 0.525^11.

So Y as a probability (not a percentage) is:

(1 - 0.525^11)^2047 = 0.1808

So Y is 18.08%. Considerably less than 50%, and so it's a losing strategy.

Most Bitcoin games (at least the ones I'm familiar with have a house edge of 1%, and so give you a 49.5% chance of doubling your stake. In that case, Y is:

(1 - 0.505^11)^2047 = 0.3277, or 32.77%

Much better, but still less than 50%.

Interestingly, even with a 0% house edge the chance of doubling up before busting is:

(1 - 0.5**11)**2047 = 0.3679 or 36.79%

This came up recently and surprised me (*). How can it be less than 50% with a 0% house edge?

It turns out the reason is that by the time you lose your 2047 satoshis, you've probably already won a bunch of times, and you get to keep those winnings. So it turns out you're not choosing between doubling up or busting completely, you're choosing between doubling up, and getting left with only however much you were able to win before you hit your big losing streak.

So this makes me think I was wrong here:

Quote from: dooglus on March 20, 2015, 03:55:00 PM

No, it's not unusual. It just has to happen less than half the time on average, due to the house edge. Otherwise this simple strategy would be profitable:

* start with X; martingale until doubling; quit

If you successfully double Y% of the time, your expected profit would be:

(Y*X - (100-Y)*X) / 100
= X(2Y - 100) / 100

which is positive exactly when 2Y > 100, ie. when Y > 50%

because I was thinking that there were only two possible outcomes:

a) you double up
b) you lose everything

but by far the most common outcome is:

c) you hit a big losing streak and can't afford to make that last bet, even though you're not bust yet

What this does tell me is that in theory you should find that you are able to double up less than 37% of the time if you stop and call it a loss when you can't afford to double to make your next bet.

(*) Here's where it came up before. I'm talking about a bankroll of 15 instead of 2047, and a house edge of 2.702%, but other than that it's the same issue. I found I doubled up just 38% of the time, busted only 6% of the time, and the rest of the time ended up with from 1 to 14 satoshis:

Quote from: dooglus on March 09, 2015, 10:55:43 PM

Code:

3041169 wins out of 8000000 rounds (38.015%)
0:6.24%, 1:5.86%, 2:5.49%, 3:5.13%, 4:4.82%, 5:4.54%, 6:4.24%, 7:3.98%, 8:3.73%, 9:3.50%, 10:3.27%, 11:3.06%, 12:2.87%, 13:2.71%, 14:2.54%, 30:38.01%
expectation: 15.008940625

So only 6.24% of the time do you actually bust.
5.86% of the time you end up with 1 unit,
5.49% of the time, 2 units, etc.