I have encountered the following question and although it is related a problem I have encountered before it is somewhat more complicated.
There are 100 doors. Behind the door number n there are n dollars. I roll a 100-sided dice a hundred times. I then take the dollars behind the door whose number was rolled out but the dollars are not replaced. What's the expected value of my earnings?
Now, this problem is similar to the following one: Expected Value Problem with 100 side dice and 100 doors
I tried the following approach.
For $1≤i≤100$ let $X_{i}$ be a random variable if door number $i$ is opened and $0$ otherwise. Then we want $$E(∑iX_{i})=∑iE(X_{i})$$ by linearity of expectation. But $E(X_{i})$ is just the probability that door $i$ is opened, or 1 minus the probability that it is never opened. This is simply $$(1-(\frac{99}{100})^{100}) \approx (1-\frac{1}{e})$$.
So, we get as the answer $$(1-\frac{1}{e})∑i = 5050(1-\frac{1}{e}) \approx 3192.2$$.
Is my reasoning here correct?
Yes, your reasoning is valid, except that your approximation is poor. The exact value of $$1 - \left(\frac{99}{100}\right)^{100} = 0.63396765872677049507\ldots,$$ whereas $$1 - \frac{1}{e} = 0.63212055882855767840\ldots.$$ The error is on the order of $10^{-3}$, which is large enough that when multiplied by $5050$, affects the expectation, which should be about $3201.5366765701910001\ldots$.