Anyone interested in a probability problem? I'm stumped.
A deck of 52 cards contains 4 aces. If the cards are shuffled and distributed in a random manner to four players so that each player receives 13 cards, what is the probability that all 4 aces will be received by the same player?
The answer is (4 * 13C4) / 52C4 (using nCr notation for "n choose r"), but I'm really having trouble interpreting that answer and seeing how to get it. I understand that it's written as (# of favorable outcomes) / (total # of outcomes) but I would think that the total number of outcomes would be 52!/(13!)^4, the number of ways to deal 52 cards to 4 people in groups of 13. I don't get where 52C4 comes from. If anyone has any advice or a hint or something I'd love to hear it.
They did write it a weird way. One thing to note is because of how choose is defined, 52c4 = 52c48
So think of it this way
we have 4 guys, hence 4* in front
now we want out of 13 cards, to have 4 cards be aces: (13c4)
now we assumed the 4 cards are acces, so that means we got 4 aces and then delt 48 more cards, so all the ways we can deal 48 cards is.. (52c48). This is weird to me as well, in fact I hate statistics, but... I will explain (try to anyway) how this gives the right answer as well:
So because choose does not depend on order, we can't find the ways to choose 52 cards from 52, which is 1 way, so we have to be more clever. We want a way to choose 48 cards, because we are asumming one of those ways leaves 4 aces left over (which are all together in someones hand). Now thats where (52c48) comes from. Now here is the weird part. Since order doesn't matter with choose, we need to make it matter by being clever. We want out of every 13 cards, the way to choose 4 cards. That way someone is going to have 4 aces (that's what we're choosing, the 4 aces). From here we only have to multiply the answer by 4 because there are 4 chances for this to happen.
Its very counter intuitive and probability is lame, but they yield the same answer.