What is the Probability of Getting an Ace in a Standard Deck of Cards?

[mynameisfunk]

Homework Statement

1. All 52 cards of a standard deck of cards are divided equally among 4 players.
(a) What is the probability that each player gets an ace?
(b) And that one player get all the spades?

Homework Equations

The Attempt at a Solution

part (a) http://www.mathhelpforum.com/math-help/attachments/f8/19068d1285520980-probabilititty-1a.gif

part (b) 1/(52 choose 13)

 

[hgfalling]

Could you elucidate on how you got your expression for part a? It's clearly wrong, as it evaluates to about 10^-18. I mean, deal out a few bridge hands with a deck of cards and you'll get a hand where everyone has an ace.

Now the way the question for part b reads to me is "what's the probability that everyone will get an ace AND someone will get all the spades?" If it's just asking a different question, "what is the probability that someone will get dealt all the spades?" your answer is close to right, but remember that there are 4 people who could possibly get all the spades.

 

[mynameisfunk]

for part (a) i figured the bottom would be all possible ways to deal out the cards. and the top, 4 choose 1 ways for the first guy to get an ace and then 48 choose 12 ways to get ...wait a minute here now..

Ok, I think I meant to put (4 choose 1)(51 choose 12) +(3 choose 1)(48 choose... etc for the numerator. But it turns out that is still really small.

OK, well there must be something wrong with my denominator. I'm really unsure of how to calculate the total number of possible ways to deal out the cards.

 

[hgfalling]

Here's a straightforward method. Somebody gets the first ace, always. Now figure out the probability that a different guy gets the next ace. Then do it again. And again. Now multiply all those numbers together.

Figuring out how many possible deals there are is kind of overkill. (It's (52 13)*(39 13)*(26 13) by the way).

EDIT: Oh, I looked back at your work. You did do the denominator right. Multiply your numerator things together instead of adding them.