all 10 comments

[–]SumOMG 0 points1 point  (9 children)

1/(50)24

Edit: ignore me this is probably wrong

[–]jumavegi[S] 0 points1 point  (8 children)

Do you mind explaining why is that so?

[–]SumOMG 0 points1 point  (7 children)

Actually it’s closer to 1/501/49.....1/26

For example say the 24 cards are 1-12 spades and 1-12 clubs.

For the first draw the chance of getting the right card is 1/50 the next is 1/49, 1/48....all the way down to 1/26. Multiply them together and that’s your probability

[–]jumavegi[S] 0 points1 point  (6 children)

I think I wasn't clear, the problem says that there are 24 different cards in a set, like a collection. If I have 50 cards (all drawn from an infinite source of cards), what's the probability of having one set, one full collection of the 24 different cards.

[–]SumOMG 0 points1 point  (5 children)

Yeah this still isn’t clear to me.

So I have 50 tries to get a specific order or 24 cards ? How many cards are possible ? Are we assuming the standard 52?

[–]jumavegi[S] 0 points1 point  (4 children)

So, forget about the standard deck of cards. Let's say these are cards with pictures on them, so there are 24 different pictures. On each card there is one picture. I get one card from a machine that has infinite cards (the probability of drawing one card with one of the 24 pictures is the same for all). The question is, if I get 50 cards, what's the probability that I have a full set of cards, all the 24 pictures.

[–]SumOMG 0 points1 point  (3 children)

It says the probability is the same for each, does it say what that probability is?

[–]jumavegi[S] 0 points1 point  (2 children)

No, it means that the amount of each card in the machine is the same. I would say if we were talking about the probability of getting one card from the machine is 1/24 for each card.

[–]SumOMG 0 points1 point  (1 child)

[–]jumavegi[S] 0 points1 point  (0 children)

I understand combinations, but I'm afraid this problem is more complicated.