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

[Albert Chan]

(10-14-2022 10:24 AM)J-F Garnier Wrote:   30 ! the nodes are coded from P=1 to 15:
 40 !        1
 50 !       2 3
 60 !      4 5 6
 70 !     7 8 9 10
 80 !   11 12 13 14 15

there are indeed 16 paths over 144 that end in the last row, but all paths are not equally probable,
we can't just do 16/144.

Yes. all paths are not equally probable.
Above triangle, from 1 to bottom row in 4 steps, all steps must go down a row.

In 2 steps:
P(1 to 4) = 1/2*1/4 = 1/8
P(1 to 5) = 1/4

In 2 steps:
P(4 to bottom) = 1/4*1/2 + 1/4*1/3 = 5/24
P(5 to bottom) = 1/3*1/3 = 1/9

In 4 steps:
P(1 to 4 to bottom) = 1/8*5/24 = 5/192 ≈ 2.604% = P(1 to 6 to bottom)
P(1 to 5 to bottom) = 1/4*1/9   = 1/36   ≈ 2.778% , slightly more probable.

  →  P(1 to bottom) = 2*5/192 + 1/36 = 23/288 ≈ 7.986%

Or, we can simply list all paths, then sum the probabilities.

Paths            1/Probability
1 2 4  7 11      2*4*4*4 = 128
1 2 4  7 12      2*4*4*4 = 128
1 2 4  8 12      2*4*4*6 = 192
1 2 4  8 13      2*4*4*6 = 192
1 2 5  8 12      2*4*6*6 = 288
1 2 5  8 13      2*4*6*6 = 288
1 2 5  9 13      2*4*6*6 = 288
1 2 5  9 14      2*4*6*6 = 288
1 3 5  8 12      2*4*6*6 = 288
1 3 5  8 13      2*4*6*6 = 288
1 3 5  9 13      2*4*6*6 = 288
1 3 5  9 14      2*4*6*6 = 288
1 3 6  9 13      2*4*4*6 = 192
1 3 6  9 14      2*4*4*6 = 192
1 3 6 10 14      2*4*4*4 = 128
1 3 6 10 15      2*4*4*4 = 128

P(1 to bottom) = 4/128 + 4/192 + 8/288 = 23/288