Here's the question: How many dice do you need to throw so that the probability of one of each face showing (i.e. getting at least one dice to show '1', and at least one to show '2', and... etc.) is at least 0.5?
My answer was 44, which is seriously wrong because the correct answer is 17. I would like to know how to arrive at the correct answer. Anyway, here's how I got my answer.
The possible number of configurations our n dice can have is (n+5) choose 5, i.e. C(n+5, 5) is the notation I will use. I get this by saying: use five dividers to partition the dice into six groups, corresponding to the numbers 1, 2, 3, 4, 5, 6, the partitions telling us what the dice will show on its face.
The number of ways to get at least one of each number showing is to take 6 dice and assume they have the numbers 1,2,...,6 showing. Then configure the remaining n-6 as you please. There are C(n-1, 5) ways to do this.
Therefore for n dice the probability of getting one of each face is C(n-1, 5)/C(n+5, 5). To get this greater than or equal to 0.5 we need n = 44.
As I said, my answer is seriously wrong. I would like to know what is the right way to arrive at the answer, and also, if possible, where I went wrong. Thank you for any help.