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%
The chance of getting 10 losses in a row in such a large run is actually far higher than that.
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.
For example, the chance of flipping heads 10 times in a row is 0.5
10 = 0.0009766
Using your equation (1-0.0009766)
100 looks at 100 different sets of 10 flips, and probability of getting 10 heads in any one set is 0.0930861.
If instead of looking at 100 sets of 10 flips we look at 1000 continuous flips, then the chance of getting 10 heads in a row is now 0.38545, which is obviously far higher.
Payout 4x and you got over 10 streak losses is very normal to happen.
Payout of 4x is a 24.8% win chance, and so a 75.2% loss chance. So 0.752
10 = 5.78% chance of happening. So yes, pretty common.