Why does the probability of something being picked after the nth attempt stay the same? There are $n$ balls in a basket, one is blue and the others are red. Each time we pick out a ball, we take it out of the basket.
The question is: What is the probability of grabbing the blue ball from the basket after an arbitrary number of tries, $k$? Meaning I draw the blue ball on the $k$th attempt.
So my result is: $P=\frac 1n$ for any $k<n$ where $k$ is the number of attempts.
My question here is why does this make sense? Assuming I didn't make a mistake in my calculations, of course. If I did, tell me why. I find it interesting but cannot figure out why this would makes sense.
Interestingly enough, if we divide all the balls into $g$ groups of $b$ balls (let's suppose $n$ divides perfectly into $b$). Then we first pick a group and then pick the ball from the group, we get a similar result ($P=\frac 1{gb}=\frac 1{n}$). Does this mean, in terms of finding the blue ball, both ways of holding the ball are equivalent?
 A: (Your original question is somewhat ambiguously worded; I'm interpreting your result to mean: "For any $k \in \{1,\ldots, n\}$, the probability that my $k$th draw, without replacement, is the one that draws the blue ball". Note that this is off-by-one from your use of the letter $k$, since by restricting to $k < n$, you appear to use it as "the number of draws before the one that gets the blue ball".)
Let's draw the balls a different way: I'll set up an opaque screen between us and randomly order the balls from left to right (all orderings equally likely). Whenever you ask for a ball, I will give you the leftmost ball still available.
From your perspective, this is the same as just drawing the balls out of the bag the way we were doing before (since the initial ordering was random). However, it's now clear that you will get the blue ball on your $k$th pull if and only if I happened to put the blue ball into the $k$th place in my initial ordering, and it's intuitively easier to see that that event has probability $1/n$.
A: The reason your answer is incorrect is that you don't consider the conditional probabilities - depending on your precise sequence of previous choices, the probability that you draw a blue ball changes.
Here, since you have only one blue ball, if you assume that you only continue to choose if you have not yet chosen the blue ball, then your answer becomes $ \frac{1}{k} $ for $ 0 < k \leq n$, where $ k $ is the number of balls remaining in the basket.
Otherwise, the moment you pick a blue ball, the probability of drawing a blue ball becomes $0$, since there are no blue balls remaining in the basket.
