The probability of "11111" at any given position in the sequence can be calculated as (2/3)^5 ≈ 0.13169. The number of occurrences is close to 10000 times this: 1316.9. To be more precise, the expected number of occurrences is about 0.13169 x 9996 ≈ 1316.3, because there are only 9996 places for a subsequence of length five in a sequence of 10000. The actual number will usually (in fact, with over 99% probability) be somewhere between 1230 and 1404. We check the solution allowing for an even wider margin that covers 99.99% of the cases.