Differences = 3n

[Pavel314]

We have a weather center gizmo in the livingroom that gives the outside temperature and humidity as well as the inside temperature and humidity. These four numbers are displayed in a 2x2 grid and change throughout the evening if the wind direction shifts or if we throw another log into the woodstove. Yesterday evening, I noticed that the differences between any two of the numbers were all divisible by three.

I consider the weather center numbers independent of each other, at least for a guy sitting in a recliner chair in front of the fire. Given two random integers, I would say that the odds of the difference between them being divisible by three is 1/3.

Since there are six ways to make a pair of numbers out of the four showing on the gizmo, would the probability of the differences all of the pairs being divisible by three be (1/3)^6?

I have a vague feeling that something like degrees of freedom should enter into this. Call the numbers A, B, C, and D. If

A - B = 3n for some integer n, and

A - C = 3m for some integer m,

Then B - C = (A - 3n) - (A - 3m) = 3m - 3n = 3(m-n)

So if two of the numbers (B and C) each have differences from another number (A)which are divisible by three, the difference between the two of them must necessarily be divisible by three. If we take this pre-determination into account, maybe for any four numbers the odds of the pair differences all being divisible by three should be (1/3)^3, corresponding the the three degrees of freedom for the four numbers.

Thoughts?

Paul

[Ray Koopman]

Enumeration of the zero-differences among 4 observations,
for each of the 3^4 possible sets of 4 readings mod 3,
shows that your conjecture is correct.

# of 0's 1 2 3 6

count 36 18 24 3

However, I'm not so sure that your argument implies that.
The jump between the last two sentences is a little too big.

[Ray Koopman]

All four observations, taken mod 3, are independent and uniformly
distributed on {0,1,2}. Subtract one of the observations from all
four, taking the result mod 3. We now have one 0 and three independent
random values uniformly distributed on {0,1,2}. The probability that
all three random values are 0, so that all six pairwise differences
among the four values are 0, is (1/3)^3.