First, I love how martingale charts look. A perfect martingale will be a straight line with periodic drops of varying magnitude, where deeper drops are exponentially more rare.
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?
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.
Oh, you're right. Normal martingale is only a straight line if you ignore the losses
That's an interesting variation on the martingale, too. It busts much quicker, but it's also a faster earner. Maybe something to try with tiny amounts.