Hi guys want to understand if my process on the following problem is correct or not.
Digits $1$, $2$, $3$ are randomly arranged to form a one-digit number and two-digit number. What is the expected value of the product of the two numbers?
I first knew that there are $6$ distinct ways to create a one-digit number and two-digit number based on the three available numbers ($3!$). I calculated the products of all of these distinct possibilities.
$$(1, 23): 1 × 23 = 23\\
(1, 32): 1 × 32 = 32\\
(2, 13): 2 × 13 = 26\\
(2, 31): 2 × 31 = 62\\
(3, 12): 3 × 12 = 36\\
(3, 21): 3 × 21 = 63$$
So I calculated the expected value in the following way:
$$E[\text{Product}]=\frac16×23+\frac16×32+\frac16×26+\frac16×62+ \frac16×36+\frac16×63=40.33.$$
Is my idea like this correct?
Best Answer
A more formal way to perform this computation is to observe that if $(a,b,c)$ represents a permutation of the set $\{1,2,3\}$, then take $f(a,b,c) = (10a + b)c$ as the product formed. Then we observe $$f(a,b,c) + f(b,c,a) + f(c,a,b) = (10a + b)c + (10b + c)a + (10c + a)b = 11(ab + bc + ca),$$ and because this expression is symmetric in $a, b, c$, the sum of the other three permutations is also equal to this value. Therefore, the expectation is simply $$\frac{22}{6}(ab+bc+ca)$$ and without loss of generality we may take $a = 1$, $b = 2$, $c = 3$ to obtain $$\frac{22}{6}(2 + 6 + 3) = \frac{121}{3}.$$
It is worth noting that this solution applies to any choice of three distinct decimal digits $1$ through $9$.