Quote from: dooglus on July 29, 2013, 04:05:00 PM

Quote from: infested999 on July 29, 2013, 01:03:12 PM

If I have bankroll of 0.512 btc and I martingale with min bet 0.001 with the only purpose of doubling that 0.512:

Chance to double: 45.315736365781955072%
Chance to lose all: 54.684263634218044928%

Correct?

Let's say you lose it all if you ever get 9 losses in a row (1,2,4,8,16,32,64,128,256). This isn't quite true, because even if your first 9 bets are all losses you still have 0.001 left, and if you've won some sequences before your 9 losses you'll have even more left.

And you double if you have 512 successful sequences before you hit 9 losses in a row.

I assume you're playing the 49.5% 2x game, and so your probability of hitting 9 losses in a row is 0.505**9
So your probability of any single sequence being successful is (1 - 0.505**9).
And your probability of having 512 successful sequences in a row is (1 - 0.505**9) ** 512

That's 0.3345881014449683, or about 1 in 3.

I wonder why we differ so much.

I got the same as you, dooglus - except that you've included all 512 rolls, when it's only on the 9th that the first time 9 losses in a row can occur - so you can only include rolls 9 to 512. That changes the result slightly: (1 - 0.505**9) ** 504 = 0.3403612.

A couple of extras: Since this is a series of trials with the probability of 9 losses in a row occurring being 0.505**9, the number of rolls until 9 in a row occurs is a geometrically distributed variable for X = 1, 2, 3... - which makes expected values and confidence intervals easy to calculate.

Expected number of rolls until 9 losses in a row = 1/ 0.505**9 = 468.142

95% confidence interval for number of rolls until 9 losses in a row = 12 to 1726 rolls.

If the attempt is made multiple times, 50% of the time 9 losses in a row will occur before 325 rolls.

I wouldn't use this as an investment vehicle