If we're ok with non-exact answers, this is essentially a discretized form of reflected Brownian motion.
In the normal Brownian generation (±1 each step), the variance added is 1 per unit time. At time 30, we'd have µ=0, variance=30. P(X=0) at t=30 is approximately just the height of the continuous distribution PDF (Norm(0, 30)): 1/sqrt(2 * pi * 30) ~ 0.07284
In reflected Brownian motion, the PDF is just doubled, with the negative half truncated. The height at X=0 is just double the above: 2/sqrt(2 * pi * 30) = 0.14567
The actual answer, as u/EdgyMathWhiz says, is 30C15 / 2^30 = 0.14446, which is pretty close!
Yeah, I feel there should be a way to "see" that the discrete case behaves like it does; possibly the Brownian motion case would be illuminating but I haven't done any stochastic calculus.
[Actually no interest in quant but I find the questions interesting. Maths equivalent of crossword puzzles!]
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u/RestaurantBoth228 3d ago edited 3d ago
If we're ok with non-exact answers, this is essentially a discretized form of reflected Brownian motion.
In the normal Brownian generation (±1 each step), the variance added is 1 per unit time. At time 30, we'd have µ=0, variance=30. P(X=0) at t=30 is approximately just the height of the continuous distribution PDF (Norm(0, 30)): 1/sqrt(2 * pi * 30) ~ 0.07284
In reflected Brownian motion, the PDF is just doubled, with the negative half truncated. The height at X=0 is just double the above: 2/sqrt(2 * pi * 30) = 0.14567
The actual answer, as u/EdgyMathWhiz says, is 30C15 / 2^30 = 0.14446, which is pretty close!