1. Originally Posted by meymathis
My apologies for some of the confusion (I'm new at the forums and didn't see your "thanks"). I thought people were arguing for the binomial distribution. Looking more carefully (which I should have done before) at awkward's example, the main thrust of my argument still holds, I think. Probably the most succinct way of saying it is this:
$Pr(X_i=0) = (1-1/n)^k \text{ for } i = 1,2, \dots ,n$
doesn't work because the number of people on the bus is not constantly $k$.

$\mathbf{P}(X_2=0|X_1=1)\neq$
$\mathbf{P}(X_2=0|X_1=0)$ because there are no longer k people on the bus, then
$Pr(X_2=0|X_1=1) = (1-1/n)^{k-l}$
where $l$ are the number of people that got off at the first stop.

Again, sorry for not being clearer before.
I'm going to disagree with you; I still think that

$Pr(X_i=0) = (1-1/n)^k \text{ for } i = 1,2, \dots ,n$.

The reason is that we are simply dealing with the unconditional probability that $X_i = 0$. It is true that the probability changes when people get off; but then we would be talking about the probability that $X_i = 0$ given that j passengers had gotten off, or something like that.

It is true that $Pr(X_2 = 0 | X1 = 1) \neq Pr(X_2 = 0)$. That is just another way of saying that the $X_i$'s are not independent. But the solution I gave does not require their independence.

I say again: $E(X+Y) = E(X) + E(Y)$ even if $X$ and $Y$ are not independent. It's hard to swallow, I know.

2. Originally Posted by awkward
I'm going to disagree with you; I still think that

$Pr(X_i=0) = (1-1/n)^k \text{ for } i = 1,2, \dots ,n$.

The reason is that we are simply dealing with the unconditional probability that $X_i = 0$. It is true that the probability changes when people get off; but then we would be talking about the probability that $X_i = 0$ given that j passengers had gotten off, or something like that.

It is true that $Pr(X_2 = 0 | X1 = 1) \neq Pr(X_2 = 0)$. That is just another way of saying that the $X_i$'s are not independent. But the solution I gave does not require their independence.

I say again: $E(X+Y) = E(X) + E(Y)$ even if $X$ and $Y$ are not independent. It's hard to swallow, I know.
I completely disagree(d) with you but I am WRONG.

I had my argument all lined up and was just sitting down to write as a proof, but then I thought, hey I'll just do a test case $(k=4,n=6)$ just to prove it. I computed all the different possible cases, calculated the number of stops for each case, and calculated the expected value. Lo and behold your formula nailed it.

Not that you probably care, but where I think I goofed up was thinking of this more like a random process. Instead of thinking that each passenger randomly picks a stop at the beginning, I was thinking of each passenger going to a stop and deciding whether to get off or not, with probability $1/n$. But that is not a good model of what is happening. For one thing, to work it this way, the probabilities of getting off would have to increase to 1, and not stay constant at $1/n$ ("last stop, everybody off). I interpreted
$Pr(X_i=0) = (1-1/n)^k$ as the probability that of k people on the bus that they don't get off at that stop. And my thought went that there were only k people on the bus near the beginning and certainly not that many toward the end. So it was this formula that I was questioning.

I humbly submit that you did this problem correctly.

But darn it, you should see the formula I had come up with! It is correct, but it is ugly.

And no, it's not hard to swallow:
a) I am quite happy with $E(X+Y) = E(X) + E(Y)$. In fact it is very useful as this case points out. I never questioned it.
b) I absolutely don't mind being wrong, I just didn't believe I was wrong!

3. ## thanks all!

thanks all your posts! the discussion about correct use case of binomial distribution was really directive.

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