Probability of Drawing an Ace

Date: 5/31/96 at 19:48:45
From: Anonymous
Subject: Probability

If I have 40 cards and there are 4 aces in the deck, the probability 
of drawing an ace is 1/10. If I were to draw 2 cards at the same time
it would be 2/10. What happens if I draw 10 cards? It can't be 10/10 -
or can it? 

Date: 5/31/96 at 22:32:34
From: Doctor Pete
Subject: Re: Probability

Think about it this way: Suppose you have ten cards, and only one is 
an ace.  Then the probability of drawing the ace is 1/10. The 
probability of drawing the ace with two draws is 2/10, and if you 
simultaneously pick n cards out of the 10, your chances of one of 
them being the ace is n/10. In particular, if you draw *all* the 
cards, it is certain that one will be the ace.

Now, say you have two aces in those 10 cards. If you pick one card, 
the probability it will be one of those two aces is 2/10 (= 1/5). If 
you draw two cards, there are 3 possible outcomes:

      neither is an ace;
      exactly one of them is an ace;
      both are aces.

If you want to find the probability that *at least* one (it could be 
both) of the two cards you draw is an ace, then it is 1 minus the 
probability of drawing no aces.  So how do you count this?  Well, find 
out how many ways you can pick two cards out of 10 (this is the number 
of combinations of 10 things taken two at a time, or "10 choose 2", or 
10!/(2!*8!), where n! = n*(n-1)*(n-2)*...*2*1.)  Then find out how 
many ways you can pick two cards without either of them being an ace.  
Since there are two aces, this is the number of ways of picking two 
cards out of 8 (= 10 cards - 2 Aces).  

This is 8 choose 2 = 8!/(2!*6!), so our probability is:

            8! 2! 8!         8*7         28
      1 - ----------- = 1 - ----- = 1 - ---- = 17/45.
           2! 6! 10!         9*10        45

So you see, it's not quite so simple to find this kind of probability; 
as a general rule, probabilities don't simply add together. You're 
assuming that drawing two cards makes the probability twice as likely 
as drawing one card, but this is not true because you're always 
drawing two distinct cards, and so these events are not independent.

Let's go back to the example with two aces in 10 cards, and two draws, 
except this time, you draw a card, put it back, shuffle, and then draw 
again.  What is the probability that the card you draw on both 
occasions will be an ace?

Well, each event (drawing of a single card) is independent of the 
other. Since you're doing the same thing twice, each with probability 
2/10 = 1/5 of drawing an ace, the total probability is 

  1/5 * 1/5 = 1/25. 

In this case, the probabilities do multiply, because the occurrence 
of one event does not affect the other.

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/