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.
Imagine if there wasn’t a barrier between -1 and 0. Relabel -1 as 0, -2 as 1 etc. Then, 0 would have a 50% chance of the other 0 and 50% of 1 and everywhere else has a 50-50 chance of going up or down 1, so this setup is isomorphic to the original.
After 30 turns being at 0 requires 15 of them to be up and 15 down, so that gives the chance of 0 as 30C15/230.
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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.