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### 5 cards are dealt from a perfectly shuffled deck

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Date: 03/16/98 at 00:43:08
From: Nicholas Chu
Subject: Card problems...

Hello,

I have a question from my teacher and I don't understand how to do it,
can you help me?

"5 cards are dealt from a perfectly shuffled deck. What is the
probability of being dealt exactly 2 aces, given that you were
dealt the ace of spades? What is the probability of being dealt
exactly 2 aces, given that you were dealt at least 1 ace? Do the

Since my teacher was talking about the theorem:

P(A|B) = P(A and B) / P(B)

I was thinking,

P(exactly 2 aces and ace of spade) / P (ace of spades), and

P(exactly 2 aces and at least 1 ace) / P (at least 1 ace)

P(exactly 2 aces) = 1/C(52,5) * c(5,2)
P(ace of spades)  = 1/C(52,5) * ?
P(at least 1 ace) = ?

Thank you very much!

Nic
```

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Date: 03/16/98 at 08:59:32
From: Doctor Mitteldorf
Subject: Re: Card problems ......

Dear Nic,

The point of this problem is that your intuition says it doesn't
matter whether the one ace in your hand is the ace of spades or not;
the answer to the two questions should be the same.  But the
calculations along the way, using Baye's Theorem for the two different
cases, will be very different.

In probability analysis even more than in other areas of math, it's
especially important to think in detail and model the situation in
your mind, not just apply a formula. It's so easy to be fooled by
problems that sound as if they ought to be the same but when you think
them through clearly... Let me get you started with some of the
calculations you'll need to do:

Easiest first:

5-card hand? I'm going to calculate what the probability is of
NOT having the ace of spades in the hand, then subtract from 1.

The probability that the first card is not the ace of spades is
51/52. Given that first draw, there are 51 cards left, and 50 of
them are not the ace of spades.

Continuing for all 5 cards, we find the probability that there is

51/52 * 50/51 * 49/50 * 48/49 * 47/48

You can cancel a lot of numerators with denominators here, and
reduce to 47/52. So the probability of having the ace of spades
in your hand is 1-47/52 = 5/52.

Well, that result is kind of expected. After all, you've got 5
cards in your hand, 47 on the table. The chance that the ace of
spades is in your hand ought to be just 5/52, and that it's on the
table ought to be 47/52. This is the correct answer, and there's
nothing wrong with the shortcut.

But now ask, what's the probability that your hand has at least
one ace? Well, let's just take the shortcut here. The probability
of having the ace of spades in your hand is 5/52. The probability
of getting the ace of diamonds is 5/52. For each of the four aces,
the probability is the same, so 4*5/52 or 20/52, right? Let's just
check with the long method:

The probability that the first card you draw is not an ace is
48/52. Now there are 51 cards left, and 47 of them are not aces,
so the probability of the second card not being an ace is 47/51.

Continuing in this way, we find the probability of NO aces in the
hand is

48/52 * 47/51 * 46/50 * 45/49 * 44/48
(We can do a little canceling here, but not much.)

The probability of having at least 1 ace must be one minus the
probability of no aces, or

1 - 48/52 * 47/51 * 46/50 * 45/49 * 44/48.

This is different from 20/52.  Which is right?  The second
calculation is right. Where did our reasoning go wrong in the
other one?  I'll leave that one to you.

Next calculation:

What's the probability of having 2 or more aces in the hand?
It's easier to calculate the probability of 1 ace and no aces.

The probability of having 2 or more aces is just 1 minus the
probability of having no aces minus the probability of having
exactly 1 ace.  The probability of having no aces we calculated
above:

48/52 * 47/51 * 46/50 * 45/49 * 44/48

For the probability of having exactly 1 ace, we can do this:

Say the ace is the first draw. The probability of that is 4/52.
Then the second draw must NOT be an ace. Since there are 51 cards
left in the deck and 48 of them are not aces, the probability for
the second draw is 48/51. Continue in that way and you find the
probability of getting the ace on the first draw and no aces on the
four subsequent draws is:

4/52 * 48/51 * 47/50 * 46/49 * 45/48.

If you apply the same technique to the probability of getting
the ace on the second or the fifth draw, you'll find the same
expression.  So the probability of having exactly 1 ace in your
hand is 5 times the result above:

20/52 * 48/51 * 47/50 * 46/49 * 45/48

And the probability of having two or more aces is

1 - (20/52 * 48/51 * 47/50 * 46/49 * 45/48)
- (48/52 * 47/51 * 46/50 * 45/49 * 44/48).

There's a lot more to do here, but I think you're off to a good
start now, and I'll ask you to take the next steps yourself.

-Doctor Mitteldorf,  The Math Forum
Check out our Web site http://mathforum.org/dr.math/
```

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Date: 03/16/98 at 11:31:02
From: Doctor Anthony
Subject: Re: Card problems ......

I am using the the notation C(n,r) to mean (n choose r).  It is
a binomial coefficient and can be calculated from:

n!                           10!
C(n,r) = -------      e.g.  C(10,4) = -------  =  210
r!(n-r)!                      4! 6!

C(3,1) x C(48,3)
Prob(2 aces|you have ace of spades) =  -----------------
C(51,4)

51888
=  --------  = 0.207635
249900

Probability of being dealt at least one ace could mean 1, 2, 3, or 4
aces. This is best calculated from 1 - Prob(no aces)

C(48,5)
Prob(no aces)   =    -------    =  0.658842
C(52,5)

Probability of at least one ace =  0.341158

C(4,2) x C(48,3)
Prob(2 aces|at least one ace)   = ------------------
0.341158 x C(52,5)

=  0.11704

Do these results agree with your intuition?

Yes! Because given that you have the ace of spades, it is a lower
probability than the probability of at least one ace.  So the sample
space is less and the probability of 2 aces is proportionally greater.

-Doctor Anthony,  The Math Forum
Check out our Web site http://mathforum.org/dr.math/
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Associated Topics:
High School Probability

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