Expected value of Multiple Values?

Question

In a bucket there are 5 envelopes, where envelope i has value i written on top of it ie. 1,2,3,4,5.

We randomly pick 3 envelopes (without returning them). let X be the sum of values written on those 3 envelopes. Calculate E(X).

How can I approach this question? I cant write X as num of indicators.

we have 10 different options to pick:

1,2,3 = 6 1,2,4 = 7 1,2,5 = 8 1,3,4 = 8 1,3,5 = 9 2,3,4 = 9 1,4,5 = 10 2,3,5 = 10 2,4,5 = 11 3,4,5 = 12

Answer 1

Converting comments:

Let us pick the envelopes one at a time. Let $X_1$ be the value of the first envelope. Let $X_2$ be the value of the second, similarly for $X_3$.

Recognize that $X=X_1+X_2+X_3$. You say "I can't write $X$ as sum of indicators" but clearly you can.

We have that $E[X]=E[X_1+X_2+X_3] = E[X_1]+E[X_2]+E[X_3]$, each of which is simply going to be the average value which is $3$ and so the final answer is $3+3+3=9$. We get to completely ignore the dependence of these random variables here thanks to the linearity of expectation.

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As per Chris's suggestion, if we were to write out the sum of everything so we can take the average of the sums, this would be $(1+2+3)+(1+2+4)+(1+2+5)+(1+3+4)+\dots$ and is a bit annoying and tedious to calculate. You could certainly do it here, however it is by no means recommended (especially if we were to increase the size of the problem).

Instead, if we were to organize our thoughts as performing the sum such that each parenthetical phrase lies on its own row, and we space things out so that all the same digits occur in their own respective columns:

$$\begin{array}{rlllll}&1+&2+&3\\+&1+&2+&&4\\+&1+&2+&&&5\\+&1+&&3+&4\\+&1+&&3+&&5\\+&1+&&&4+&5\\+&&2+&3+&4+\\\vdots&&\vdots&&\vdots\\\hline&6(1)+&6(2)+&(6)3+&(6)4+&(6)5\end{array}$$

You can see that there are six copies of $1$ being added, six copies of $2$, etc... giving the overall sum as $6(1+2+3+4+5)$ making the final average $$\dfrac{6(1+2+3+4+5)}{10}=9$$

If it is difficult to see why there are six copies of each, that can be done with a bit of combinatorics as well, there being $\binom{4}{2}$ ways to pick two other digits than the digit in the question to be added to it. If it is difficult to see why there are ten outcomes possible, that is simply $\binom{5}{3}$.

Answer 2

You can take advantage of symmetry to get a solution that works for a generalized version of the problem, choosing three envelopes numbered $1,2,...,y-1$

Divide the possible combinations of $3$ into three groups:

1. Combinations that don't include the number $\frac y2$ (the middle number, if $y$ is even; if $y$ is odd, all combinations fall in group $1$)
2. Combinations where one envelope is $\frac y2$ and the others sum to $y$
3. Combinations where one envelope is $\frac y2$ and the others don't sum to $y$

We can show that for each of these groups, the expected total given that the combination is a member of that group

For group 1: For every combination $(a,b,c)$ in group 1, combination $(y-a,y-b,y-c)$ is a different combination also in group 1. (Left to the reader to prove). As a result of symmetry, the expected value given the combination is in group 1 is the average of $X(a,b,c)$ and $X(y-a,y-b,y-c)$, which is $\frac{a+b+c+y-a+1-b+y-c}{2}=3\frac{y}{2}$.

For every ,member of group $2$, the total value $X$ is by definition $3\frac{y}{2}$.

For group $3$, similar to what we did for group $1$, for every combination $(a,b,\frac{y}{2})$ in group $3$, the combination $(y-a,y-b,\frac{y}{2})$ is a different combination also in group $3$ (Left to the viewer to prove). As a result of symmetry, the expected value given the combination is in group $3$ is the average of $X(a,b,\frac{y}{2})$ and $X(y-a,y-b,\frac{y}{2})$, which is $\frac{a+b+\frac y2+y-a+y-b+\frac y2}{2}=3\frac y2$.

Answer 3

As per my comment, we're after the average of all possible arrangements. My suggestion was to consider what you might notice if you wrote all arrangements out.

There are $\binom{5}{3}=10$ different ways to choose the three envelopes. Each arrangement is a list of three envelopes; so the full list has $30$.

By symmetry, each envelope will appear the same number of times in the full list; there are five different envelopes, so each is counted $30\div 5=6$ times.

So the total of all numbers in the full list is $(1+2+3+4+5)\times 6=90$.

The average is then simply $90\div 10=\boxed{9}$.

Note that you could easily extend this method as much as you like to more envelopes and more choices.