Stat/probability question

Question

Suppose there is an intersection in a street where crossing diagonally is allowed. The four corners form a square and there is a person at each of the four corners. Each person crosses randomly in one of the three possible directions available, at the same time. Assuming they all walk at the same speed, what is the probability that no one crosses each other (arriving at the same location as someone doesn’t count but crossing in the middle counts)

Answer 1

Let the four corners are ABCD in clock direction.

There can be at most one diagonal crossing.

Case 1, no diagonal: All people will go in the same clockwise/counter clockwise direction. (2 ways)

Case 2, one diagonal, WLOG say A crosses diagonally. C has 2 options. The person at the corner C goes to can only go to A, the other one has 2 options, so 2x2 = 4 ways. There are 4C1 (=4) ways to choose who crosses diagonally, 4x4 = 16.

Favorable outcome = 2+16=18

Total possible outcomes = 3⁴ = 81

P(no one crosses each other) = 18/81 ≈ 0.22

Answer 2

Each of the 4 walkers has 3 possible destinations, giving 3⁴ = 81 different outcomes in the sample space.

To determine the number of distinct outcomes in which nobody crosses paths, let's label the 4 corners 1-4 and focus our attention first on the walker who starts at corner 1. Suppose walker 1 heads toward corner 2: walker 2 can head to either 3 or 4. Suppose walker 2 heads to 3, then 3 likewise has 2 options and walker 4 will have 2 options that don't cross. Thus, we have 8 distinct possibilities with no crossing paths, with walker 1 heading to corner 2. By symmetry, we have 8 outcomes for all 3 of walker 1's possible destinations, and 24 in total.

P(nobody crosses paths) = 24 / 81 = 8 / 27 ~ 29.6 %