Expected number of boxes I need to check before finding ball This is a really basic question, but I'm out of practice and have bad intuition.
If I know that there is exactly one ball hidden in a row of $n$ boxes, how many boxes do I need to check successively (check first box, then next one, etc. until I find the ball), on average, before finding the ball? 
Intuitively, I know that the answer is $n/2$. 
But, why can't I think of this as a geometric distribution? (which would lead to a different, incorrect, answer) i.e. probability that the ball is under any given box is $p = 1/n$; probability that it's not is $1 - (1/n)$; therefore, the expected number of boxes I need to check before finding the ball (first success) is $(1-p)/p$
I'm assuming that the probability that the ball is under a given box is independent from the results from previous boxes being checked. I'm confused about what assumptions I'm making that are incorrect.
Thanks!
 A: It's not a geometric distribution because (presumably) you won't check the same box twice. Consequently the chances are not independent - if the first $n-1$ boxes don't have it, it must be in the last one (with probability $1$, not $1/n$). And to a lesser extent this works earlier on. If you don't find it in the first box, then there are $n-1$ boxes left it could be in, so when you look in one you now have a $\frac1{n-1}$ chance of finding it.
If you just picked a random box every time, going with the random box even if it was the same box you just checked, then it would be a geometric distribution and it would take time $n$. But obviously this is less efficient than always trying a new box, so it's not surprising that it takes much longer on average.
A: In the first step, $p_{1}=\frac{1}{n}$. But in the second step, $p_{2} = \frac{1}{n-1}$, ..., $p_{i} = \frac{1}{n-i+1}$,...,$p_{n}=1$.
Thus, each of trial is not a Bernoulli trial. Therefore the distribution is not a geometric distribution.
