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

[PeterP]

(10-21-2022 04:10 PM)Albert Chan Wrote:  

(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

Very neat Albert! It converts the summation into a formula for the sum as its a geometric progression (which I did not recognize). You clearly did not need a computer and could have proven the result to be correct on an airplane with just a simple calculator, your pen and pencil :-) Thank you for sharing.