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The digits 1, 2, 3, and 4 are randomly arranged to form two two-digit numbers, AB and CD.

For example, we could have $AB = 42$ and $CD = 13$.

What is the expected value of $AB * CD$?

On Brilliant, there is an explanation of this problem, but I don't get it... FIrst of all, shouldn't $E[AC]$ require ${}_4 \mathrm{ P }_2 = 12$ permutations instead of 6?

https://brilliant.org/wiki/linearity-of-expectation/

Mining
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2 Answers2

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Rewriting the expression, foiling it out, and using linearity of expectations: \begin{align} E[AB \cdot CD] &= E[(10 \cdot A + B) \cdot (10 \cdot C + D)] \\ &= E[100 \cdot A \cdot C + 10 \cdot A \cdot D + 10 \cdot B \cdot C + B \cdot D] \\ &= 100 \cdot E[A \cdot C] + 10 \cdot E[A \cdot D] + 10 \cdot D[B \cdot C] + E[B \cdot D] \end{align}

Note by symmetry that all of these expectations are equal to each other, therefore, adding up the coefficients:

$$E[AB \cdot CD] = 121 \cdot E[A \cdot C]$$

Yes, there are 12 different equally likely permutations for the values of $A$ and $C$. However, multiplication doesn't depend on order, so we can just look at the six combinations instead. The average of the six equally likely combinations is:

$$E[A \cdot C] = \frac{1\cdot 2 + 1 \cdot 3 + 1\cdot 4 + 2\cdot 3 + 2\cdot 4 + 3\cdot 4}{6} = \frac{35}{6}$$

Thus, the final answer is:

$$E[AB \cdot CD] = \frac{121 \cdot 35}{6}$$

Sherwin Lott
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Using decimal expansion $AB=A\cdot10+B$, $CD=C\cdot10+D$. $$AB\cdot CD=A\cdot C\cdot 100 + (A\cdot D+C\cdot B)10+B\cdot D$$ HEnce

$$\mathbb{E}[AB\cdot CD]=100 \mathbb{E}[A\cdot C] + 10\mathbb{E}[A\cdot D+C\cdot B] + \mathbb{E}[B\cdot D]$$

$$\begin{align} \mathbb{E}[A\cdot C]&=\frac{1}{4\cdot 3}(1\cdot 2+1\cdot 3+1\cdot 4+2\cdot 1+2\cdot 3+2\cdot 4+3\cdot 1+3\cdot 2+3\cdot 4 +\\ &\qquad 4\cdot 1+4\cdot 2 + 4\cdot 3)\\ &=\frac{1}{6}(2+3+4+6+8+12)=\frac{35}{6} \end{align}$$

$$\begin{align} \mathbb{E}[A\cdot D+C\cdot B]&=\mathbb{E}[A\cdot D]+\mathbb{E}[C\dot B]=2\frac{35}{6} \end{align}$$ since $A\cdot D$ and $C\cdot B$ have the same distribution as $A\cdot C$.

Similarly, $B\cdot D$ has the same distribution as $A\cdot C$ and so,

$\mathbb{E}[\mathbb{E}[AB\cdot CD]=(100+20+1)\frac{35}{6}$

Mittens
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