# Very simple probability question

1. Jun 27, 2005

### Physics_wiz

You have a deck of 50 cards with 4 aces in it. I want to calculate the probability that you will get at least one ace when you draw 3 random cards from the deck without replacement. So, I did the following:
(4/50)+(4/49)+(4/48) = .2449659864

Now I wanted to also calculate the probability that there will be no aces in the 3 random cards using the same way. I could have just subtracted the number I got from 1 but I wanted to do it the same way, so I did:
(46/50)*(45/49)*(44/48) = .7744897959

The only problem is that they don't add up to 1. When I even string the whole thing in my Ti 83+ I get 1.019455782 which is like a 2% difference.

Is that just a calculator significant digits error? Did I do something wrong? I'm thinking the probability of 2 aces out of 3 cards and 3 aces might have to be included in there somehow but isn't that taken care of when I calculate the probability of at least 1 ace in the first step?

Thanks.

2. Jun 28, 2005

### LittleWolf

Probability of at least one ace is ((4C1 * 46C2)+(4C2 * 46C1)+(4C3 * 46C0))/(50C3)=0.225510204082.
Probability of no aces is (46C3)/(50C3)=0.77448979597837. Where nCr=n!/r!/(n-r)!

3. Jun 28, 2005

### Physics_wiz

Shouldn't the probability of at least one ace be (4/50)+(4/49)+(4/48) ?

If that's not the probability of at least one ace then what does (4/50)+(4/49)+(4/48) give the probability of? Is there a way I can find the probability of one ace without using combinations (ie. Just multiplying fractions like I did to find the probability of no aces at all)?

4. Jun 28, 2005

### juvenal

I have no idea what your first calculation is, but it seems to make no sense. You can only add the probability of mutually exclusive outcomes, so you would want:

Probability of at least one ace = Probability of exactly one ace out of 3 cards + probability of exactly two aces out of three cards + probability of exactly 3 aces out of three cards, which looks like it corresponds to what LittleWolf has.

5. Jun 28, 2005

### NateTG

The probability of getting an ace on the first card is
$$\frac{4}{50}$$
The probability of getting an ace on the second card if you did not get an ace on the first card is
$$\frac{4}{49}$$
The probability of getting an ace on the third card if you did not get an ace on either of the first two cards is
$$\frac{4}{48}$$
Then the probability of getting an ace in the first three cards is:

$$\frac{4}{50}+\frac{46}{50} \left(\frac{4}{49}+\frac{45}{49} (\frac{4}{48}) \right)$$

6. Jun 28, 2005

### juvenal

Nate - I think you're missing the scenarios where you get more than one ace, which is important, given that the original question was about >= 1 ace in 3 cards without replacement.

7. Jun 28, 2005

### NateTG

I might be, but the expression accounts for them ;).

The first term is the likelyhood of any scenario where the first card is an ace it covers:
Axx
AAx
AxA
or
AAA

The second term is the likelyhodd of any scenario where the first card is not an ace, but the second is. So the two cases it includes are:
xAx
xAA

The third term is the likelyhood that only the last card is an ace.
xxA

That leaves
xxx
as the other case.

If you do the math, you should be able to see that it adds up to 1 with .774... that Physics_wiz has listed for the probability of no aces.

8. Jun 28, 2005

### juvenal

You're right, sorry.

9. Nov 21, 2011

### PritumD

Why is the probability of getting an ace on the first card 4/50 instead of 4/52?

10. Nov 22, 2011

### PatrickPowers

It is easier to calculate the probability of getting no aces in the first three draws.

46/50 * 45/49 * 44/48

Then subtract that from one to get the prob of one or more.

11. Nov 22, 2011

### PritumD

I thought a deck of cards has 52 cards.

12. Nov 22, 2011

### D H

Staff Emeritus
Not in this case. It is given as 50. Someone removed the two red deuces, for example.