mainhoon wrote:
Excellent Ian. I had sort of made up the options to understand the thinking process. What boggles my mind is that 1 and 2 give the same answer - I had heard that was the case but the logic was not clear to me. Let me indulge you for a second. Let's say we have 1000 Red and 1 Black ball. If I want to find the probability of finding the black ball:
1. with replacement: clear answer 1/1001
2. without replacement: lets say in the 60th try = as I draw the balls, it would seem that the probabilty of finding the black ball should improve (because the red ball population is depleting) so the larger number of tries the higher the probability of finding the black ball.
When you pick the 61st ball in your example, there are two possible situations. Either we have not yet picked the black ball, in which case, as you say, it's a bit more likely than 1/1001 that we pick the black ball. But it can also happen that we've already picked the black ball in one of our first 60 selections, in which case there is *no* chance our 61st selection is black. These probabilities will 'balance out' to give the answer 1/1001.
It might be easier to see with a different example. If you have 1001 people, 1 of whom is a man and 1000 of whom are women, and you line them up, what's the probability the 61st person in line is the man? Well, there's nothing special about the 61st place in line; when we pick the 61st person, we are selecting from 1001 people, 1 of whom is a man, so the probability is 1/1001. The same would be true for any other position in the line.
Now, there's a different kind of question that you might be thinking of here - a conditional probability question. We might instead ask: If you have 1001 balls, one of which is black, and you pick 61 balls one at a time without replacement,
if none of the first 60 selections is black what is the probability the 61st ball is black? Notice the difference between this question and the one you asked above - here we *know* that we haven't yet picked the black ball when we make our 61st selection. So we know that we have 1 black ball among the 941 balls we are selecting from (we've removed 60 red balls), so the answer is 1/941.
mainhoon wrote:
On a subtle note, I have an issue with the wording around such problems "draw n balls with replacement" - unless this is done one by one it makes no sense, so why does it not say one by one. In other words this can equally mean that I just grab n balls at once (which is my case 3 above). Is the fact that I add with replacement makes it now a 1 by 1 case?
I'm not sure I quite understand the question here, but you might look at this situation as follows: say you have a bag of marbles, and you need to pick two of them without replacement. You can put both of your hands in the bag and grab two marbles. Now, whether you take both of your hands out of the bag at once, or take your left hand out first and then your right, the results you can get will be identical. That's why it's logically equivalent to pick 'two marbles at once' or to pick one marble, then the next, without replacement.
