Averages

Solution

Let the three numbers be chosen independently and uniformly from \[ \{1,2,3,\dots,20\}. \] There are \[ 20^3 = 8000 \] ordered triples \((a,b,c)\). We want the probability that one of the numbers is the average of the other two. That means one of the following must hold: \[ 2a=b+c, \qquad 2b=a+c, \qquad 2c=a+b. \] Because the variables are symmetric, it suffices to count the number of solutions to \[ 2a=b+c \] and multiply appropriately. For fixed \(a\), we must have \[ b+c=2a. \] The number of ordered pairs \((b,c)\) with \(1\le b,c\le 20\) and \(b+c=s\) is \[ \min(s-1,\,41-s). \] Since \(s=2a\), this becomes \[ \min(2a-1,\,41-2a). \] Summing for \(a=1,\dots,20\), \[ 1+3+5+\cdots+19+19+17+15+\cdots+1. \] The first ten odd numbers sum to \[ 1+3+\cdots+19=100. \] The decreasing part sums to \[ 19+17+15+13+11+9+7+5+3+1=100. \] Thus the number of ordered triples satisfying \(2a=b+c\) is \[ 200. \] By symmetry the same count holds for \[ 2b=a+c \quad \text{and} \quad 2c=a+b. \] So initially we get \[ 3 \times 200 = 600. \] However, triples with all numbers equal satisfy all three equations. There are \[ 20 \] such triples, and each has been counted three times, but should be counted once. Thus we subtract the extra counts: \[ 600 - 2\cdot20 = 560. \] Therefore the probability is \[ \frac{560}{8000}. \] Simplifying, \[ \frac{560}{8000}=\frac{7}{100}. \] So the probability that one number is the average of the other two is \[ \frac{7}{100}. \]