Second, is there any way to calculate the probability of a bust on a martingale procession while taking into account the ever-increasing bankroll it provides?
I think it wouldn't be exactly straight, even connecting the peaks, because a:
-1 -2 +4
and a
-1 -2 -4 -8 -16 +32
both add 1 to the total bankroll, but take different numbers of steps to do so.
I've seen people play like:
-1 -3 -7 -15 -31 +63
before, which *does* lead to a perfect straight line, since the net gain is equal to the number of steps. The 6 numbers in the previous line sum to 6. It's equivalent to starting a new martingale of size 1 each bet:
First bet, bet 1 (and lose)
Second bet, bet 2 for the first martingale, and 1 for the new one you're starting: so 3 in total
Third bet, bet 4 for the first, 2 for the 2nd and 1 for the new one: 7 in total
...
Sixth bet: bet 32 for the 1st, 16 for the 2nd, ..., and 1 for the new one and this time we win, finishing off all 6 concurrent martingales at once.
I think the first person I noticed playing this way was the owner of the BTC Guild poll, Eleuthria. It didn't work out too well for him in the end, and unfortunately I can't remember his account number to show the chart.
Edit: I forgot your 2nd question. I'm sure there must be, but I don't know of it. The number of rolls you can withstand before busting only goes up in whole numbers. Most wins don't help you survive longer, because they don't give you enough extra to make a whole extra bet. So I'm not sure how to work out the probability. Other than that the probability of busting is 1 if you play long enough...