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Let's say you have 8 balls in front of you, each of them contains a bank note (for example 1\$, 5\$, 10\$ etc...). And let's say you can choose randomly 3 balls among the 8 (you don't know what's inside).

How to compute the expected gain, in function of the value of each notes?

As an example, let's say I have the following available notes: 0\$, 0\$, 0\$, 100\$, 100\$, 100\$, 200\$ and 500\$. The expected value should be 375$ (I did all the combinations by hand, but I want the formulae).

Thanks!

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closed as off-topic by Peter Flom Apr 5 at 13:55

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  • $\begingroup$ Use linearity of expectation: by simply averaging the values of the notes you find the expectation of a single random draw is \$125, whence the expectation of three draws is three times that. $\endgroup$ – whuber Apr 5 at 13:46
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Let's call gains as $x_1,...,x_8$. From a brute force perspective as yours, choosing $x_i,x_j,x_k$ in that specific order has probability $\frac{1}{8\cdot7\cdot6}$. If we think of all the combinations, the expected value can be formulated as

$$E[X]=\sum_{i\neq j\neq k}\frac{1}{8\cdot7\cdot6}(x_i+x_j+x_k)$$ Here, if you fix $i$, you'll actually have $x_i$ a total of $7\times6=42$ times in this sum. This is for a specific $x_i$, e.g. $i=1$. Same situation arises three times when we consider $x_j$ and $x_k$. So, The total sum becomes the following (let $S$ be the sum of your banknotes): $$E[X]=\frac{1}{8}\times 3\times\sum_{i=0}^8 x_i=\frac{3S}{8}$$

It can be generalized to $\frac{kS}{n}$ I suppose (when we select $k$ balls out of $n$).

Another Solution

From a more probabilistic perspective, let $X_1,X_2,X_3$ be the three numbers we get. $E[X_1]$ is clearly $S/8$. Using Law of Iterated Expectations, we can calculate the others:

$$E[X_2]=E[E[X_2|X_1]]=E\left[\frac{S-X_1}{7}\right]=\frac{S}{7}-\frac{E[X_1]}{7}=\frac{S}{7}-\frac{S}{7\cdot8}=\frac{S}{8}$$

$$\begin{align}E[X_3]&=E[E[X_3|X_1,X_2]]=E\left[\frac{S-X_1-X_2}{6}\right]=\frac{S}{6}-\frac{E[X_1]}{6}-\frac{E[X_2]}{6}\\ &= \frac{S}{6}-\frac{S}{8\cdot6}-\frac{S}{8\cdot6}=\frac{S}{8}\end{align}$$

which means $E[X_1+X_2+X_3]=\frac{3S}{8}$.

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    $\begingroup$ While I like the directness of using the law of iterated expectations, what I did intuitively for myself was using symmetry (the order labels do not matter) to argue $E[X_1] = E[X_2] = E[X_3]$. $\endgroup$ – Martijn Weterings Apr 5 at 13:57
  • $\begingroup$ I agree with you actually, I just felt obliged to show in other ways that the expectations of consecutive draws are still $S/8$ whether we do it with or without replacement. $\endgroup$ – gunes Apr 5 at 15:10

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