Expected value of $\overline{ABC} \times \overline{DEF}$ Here is a question from HMMT:
https://hmmt-archive.s3.amazonaws.com/tournaments/2013/nov/team/solutions.pdf

The digits $1$, $2$, $3$, $4$, $5$, $6$ are randomly chosen (without replacement) to form the three-digit numbers $M = \overline{ABC}$ and $N = \overline{DEF}$. For example, we could have $M = 413$ and $N = 256$. Find the expected value of $M \cdot N$.

Here's what I did. Each digit on average is going to be $(1 + 2 + 3 + 4 + 5 + 6)/6 = 3.5$, so the expected value is $(100(3.5) + 10(3.5) + 3.5)^2 = 150932.25$. However, the answer at the link above is $143745$. What did I do wrong? Did I overcount something?
 A: You mistake is this: the problem asked you to compute $E[MN]$, and you responded with $E[M]E[N]$. In general these are not equal. They would be equal if $M$ and $N$ were independent, but they are clearly negatively correlated in this case; the larger $M$ is, the more large digits it received, meaning $N$ will typically be smaller.
To solve the problem, write
$$
\begin{align}
E[MN]
  &=E[(100A+10B+C)(100D+10E+F)]
\\&=100^2E[AD]+100\cdot 10E[AE]+\dots+1\cdot1\cdot E[CF]
\\&=E[AD](100+10+1)^2
\end{align}
$$
The last equality follows by realizing $E[AD]=E[AE]=\dots=E[CF]$, and that the remaining terms factorize nicely.
All that remains to compute $E[AD]$. Recall this is not $E[A]E[D]=(3.5)^2$. Instead, you need to average over all $30$ possibilities for the ordered pair $(A,D)$. That is, you need to compute
$$
\frac1{30}(1\cdot 2+1\cdot 3+\dots +2\cdot 1+2\cdot 3+\dots+6\cdot 5)
$$
A: 143745 is correct. Your assumption is wrong. Just see this with the set {3, 4}. Its mean is 3.5, but 34*43 is not equal to the square of (35 + 3.5).
A: The correct answer is 143745.
$$
\frac{1}{6}\cdot\frac{1\cdot(2+3+4+5+6)+2\cdot(1+3+4+5+6)+...+6\cdot(1+2+3+4+5)}{5}\cdot111\cdot111
$$
Your answer is incorrect because you average the numbers evenly
A: Just for curiosity, I did 100 million Monte Carlo simulations to see what does this distribution looks like.
import numpy as np
import pandas as pd
import seaborn as sns
from tqdm import tqdm
from numpy.random import default_rng
rng = default_rng()
digits = np.arange(1,7)
def iteration():
    rng.shuffle(digits)
    return (sum([digits[i]*10**i for i in range(3)])*sum([digits[i]*10**(i-3) for i in range(3,6)]))
N = 100000000

mc_results = []
for i in tqdm(range(N)):
    mc_results.append(iteration())
mc_results = pd.Series(mc_results)

The summary statistics are:
count    10000000.00
mean       143754.39
std         83542.57
min         33210.00
25%         76358.00
50%        129176.00
75%        204048.00
max        342002.00
dtype: float64

And the histogram is:

