Quote from: hosseinimr93 on February 05, 2020, 09:01:45 PM

If I have a win on the third bet, the first set of 10 is finished and now I am on the second set. Because in the first set, I need at least one win and I have already made that.

This statement is correct, that the run of 10 obviously resets when you win, but that's not what your (1-x)^y equation above calculates. I'll try to give examples below.

Your equation of (1-x)^y assumes each set of 10 is entirely independent of each other set of 10. In 1000 rolls, there will be exactly 100 sets of 10 consecutive rolls.

___________	__________________________
First set	01 02 03 04 05 06 07 08 09 10
Second set	11 12 13 14 15 16 17 18 19 20
Third set	21 22 23 24 25 26 27 28 29 30
___________	__________________________

And so on. This is obviously not how Martingale works, because if you win on roll 3 and 17, and lose on all the others, then you will have a >10 loss streak, but your calculation doesn't take that in to account. Your calculation looks at the chance of losing 10 rolls out of 10, over 100 completely separate sets of 10 rolls which are independent of each other.

What actually happens is that any consecutive 10 numbers can make up their own set, as such:

___________	__________________________
First set	01 02 03 04 05 06 07 08 09 10
Second set	02 03 04 05 06 07 08 09 10 11
Third set	03 04 05 06 07 08 09 10 11 12
___________	__________________________

And so on.

That's why it is actually far more likely that Martingale will result in your going bust than your calculation would suggest.
(1-0.0009766)¹⁰⁰ (100 separate sets of 10 rolls) gives a chance of one of those sets of 10 being all losses of 9.31%
The real chance of rolling 10 losses in a row in 1000 consecutive rolls is 38.55%