In this case for doubling the balance you must win 1000 times and you can lose up to 10 times in a row. As o_e_l_e_o said, the chance of losing 10 times in a row is only 0.0010787. But the chance of not happening this 1000 times is (1-0.0010787)1000 = 0.34 = 34%
Your calculation gives the odds of not getting a run of 10 losses when considering separate sets of 10 rolls each. Provided there is a win at some point in those 10 rolls, then that set doesn't count as an overall loss and you look at the next set of 10 rolls. That's not how Martingale works though. Martingale is a continuous set of rolls, and isn't subdivided in to sets of 10. In your calculation, if the first set of 10 had a win on the third roll, and the second set of 10 had a win on the seventh roll, that isn't accounted for despite there being 13 losses in between the two wins.
I completely understand you. I thought about this when I was calculating the probability.
Assume that I am using martingale strategy. Whenever I lose a bet and, I double the bet amount and whenever I win a bet, I return to the base a bet.
Given my balance, after I return to the base bet, I can lose 10 times in a row.
After I win a bet and return to the base bet, I must win at lease one of 10 following bets.
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. Now I need to win one of the 10 bets of the second set.
Once I win a bet, the set is finished. I can skip the remaining bets of the set and go to the next set.
As all bets are independents from each other, I think my assumption is true. Maybe I am wrong.