Maybe related to this question
In the comments of this question they say that it gets easier if the variables are identically and independently distributed.
But i don't see how because in my case the variable is discrete
Here is my problem :
I toss 4 dice and keep the 3 best results. What is the expected value of the result ?
I think tossing 4 dice and keep the 3 best is like tossing 4 dice and removing the minimum.
- Let X be the result of a standard die.
- Let Y be tossing 4 dice and keeping the 3 best
Is that correct : $E(Y) = 4*E(X) - E(min)$ ?
So how calculate E(min) ?
I know if the variable was uniform on [0,1] I could have started with $F_Y = 1 - ( 1-F_X )^p$ where p is the number of dice I toss, but here the variable is discrete so i don't know where to start.
Generalization :
How to calculate the expected value of k realizations of a discrete random variable in [0-n]?
It's been a while since i studied probability, so my basic calculation may be wrong. Also,
English is not my mother tongue, so please forgive my mistakes.
edit : spelling mistakes
Answer
For clarity, suppose that the dice have ID numbers $1,2,3,4$. Let $X_i$ be the result on die $i$. Let $Y$ be the sum of the three largest of the $X_i$, and let $W$ be the minimum of the $X_i$.
Then $Y=X_1+X_2+X_3+X_4-W$. By the linearity of expectation, it follows that
$$E(Y)=E(X_1)+E(X_2)+E(X_3)+E(X_4)-E(W).$$
The linearity of expectation is a very useful result. Note that linearity always holds: independence is not required.
The expectation of the minimum can be calculated by first finding the distribution of the minimum $W$.
The minimum is $1$ unless the dice all show a number $\ge 2$. The probability of this is $1-\left(\frac{5}{6}\right)^4$. We rewrite this as $\frac{6^4-5^4}{6^4}$.
The minimum is $2$ if all the dice are $\ge 2$ but not all are $\ge 3$. The probability of this is $\frac{5^4-4^4}{6^4}$/
The minimum is $3$ if all results are $\ge 3$ but not all are $\ge 4$. This has probability $\frac{4^4-3^4}{6^4}$.
And so on. Now use the ordinary formula for expectation. We get that the expectation of $W$ is
$$\frac{1}{6^4}\left(1(6^4-5^4)+ 2(5^4-4^4)+3(4^4-3^4)+4(3^4-2^4)+5(2^4-1^4)+6(1^4-0^4) \right).$$
We leave you the task of computing. Before computing, simplify!
Generalization: Suppose we toss $k$ "fair" $(n+1)$-sided dice, with the numbers $0$ to $n$ written on them. For $i=1$ to $k$, let $X_i$ be the number showing on the $i$-th die. Let $S$ be the sum of the dice. Then $S=X_1+\cdots+X_k$. The expectation of $X_i$ is $\frac{0+1+\cdots +n}{n+1}$. By the usual expression for the sum of consecutive integers, $E(X_i)=\frac{n}{2}$ and therefore $E(S)=\frac{kn}{2}$.
The analysis of the minimum $W$ goes along the same lines as the earlier one. The probability that the minimum is $j$ is $\frac{(n+1-j)^k -(n-j)^k}{(n+1)^k}$. If we use the ordinary formula for expectation, and simplify, we find that
$$E(W)=\frac{1^k+2^k+\cdots+n^k}{(n+1)^k}.$$
A nice way to find $E(W)$: The following is a useful general result. Let $X$ be a random variable that only takes non-negative integer values. Then
$$E(X)=\sum_{i=1}^\infty \Pr(X\ge i).$$
We apply that to the case of the random variable $W$ which is the minimum of $X_1,\dots,X_4$. The probability that $W\ge i$ in that case is $\frac{(7-i)^k}{6^k}$.
The same procedure works for the more general situation you asked about.
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