[VA] SRC #012a - Then and Now: Probability

[Albert Chan]

(10-12-2022 12:10 PM)PeterP Wrote:  My code does deliver the correct result for R = 5, but I dont have a good way (especially right now on a plane and my work computer has no simulators installed…) to check if it is correct for R = 30, S=29. (It comes out to 1.311095094 e-13).

For S = R-1, we can treat triangle as without bottom edge (and the 2 corners).

First step from top corner, it gives equal probability to left or right side.
We can thus skip first iteration, simplified the problem without top corner.

Problem now is relatively simple, with only inside (6 ways) and edge (4 ways)
Only edge probability can "leak" to the inside; inside probabilities never "gets out".

Work out the geometric progression (not shown), with p=1/6, q=1/4, we have:

P(R, S=R-1) = ((2p)^(R-3) - q^(R-3)) / (2*p-q) * (2*p*q) + 2*q^(R-2)

(2*p*q) / (2*p-q) = 1 / (1/q-1/(2*p)) = 1 / (4-3) = 1. It simplified to:

P(R, S=R-1) = 3^(3-R) - 2^(5-2*R)

Example:

P(1,0) = 9 - 8 = 1
P(2,1) = 3 - 2 = 1
P(3,2) = 1 - 1/2 = 1/2
P(4,3) = 1/3 - 1/8 = 5/24
P(5,4) = 1/9 - 1/32 = 23/288
P(6,5) = 1/27 - 1/128 = 101/3456
...
P(30,29) = 1/3^27 - 1/2^55 ≈ 1.31109509664e-13