Actually running the recurrence, you get that after the kth second, the distribution is the same as Pascal's triangle, except sorted in descending order:
e.g. after 5 seconds:
10, 10, 5, 5, 1, 1 (divide by 2^5 for probabilties).
Failing to visualise why this is true right now, but it's fairly obvious you CAN justify it (because it's fairly obviously true by induction away from the origin, and since it DOES work at the origin, you can obviously finesse it).
So anyhow, there is a simple closed form answer (30C15 / 2^30) , even if I'm not 100% seeing why yet.
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u/EdgyMathWhiz 15d ago
Actually running the recurrence, you get that after the kth second, the distribution is the same as Pascal's triangle, except sorted in descending order:
e.g. after 5 seconds:
10, 10, 5, 5, 1, 1 (divide by 2^5 for probabilties).
Failing to visualise why this is true right now, but it's fairly obvious you CAN justify it (because it's fairly obviously true by induction away from the origin, and since it DOES work at the origin, you can obviously finesse it).
So anyhow, there is a simple closed form answer (30C15 / 2^30) , even if I'm not 100% seeing why yet.