# Two probability questions

1. Jul 19, 2010

### Ryker

I.

1. The problem statement, all variables and given/known data
You have a set of 52 cards, out of which you randomly draw 8. What's the probability of an event that either a) three aces will be drawn or b) three kings will be drawn or c) three aces and three kings will be drawn.

2. Relevant equations
The usual probability mumbo jumbo.

3. The attempt at a solution
This is from a high school textbook and the solution says the said events are going to happen in (4 3)(48 5) + (4 3)(48 5) + (4 3)(4 3)(44 2). I get the first two, but I don't get why there is a plus in front of the third. Shouldn't there be a minus? I mean, the event c) is already covered in both the events a) and b), so if anything you should subtract that event so that it isn't counted twice. If my thinking isn't correct, what am I missing?

II.
1. The problem statement, all variables and given/known data
Out of the same set of 52, you draw 3 cards. What is the probability that at least one of those will be an ace?

2. Relevant equations
Same as above.

3. The attempt at a solution
Again, the solution (which I agree with this time) says that event will happen in (4 1)(48 2) + (4 2)(48 1) + (4 3) cases, which is logical. You figure out how many times you'll get on ace, how many times two aces and how many times three aces.

Where I'm lost here, however, is why can't you simply do (4 1)(51 2)? You secure one of those four aces with the first binomial, and then for the remaining two cards you can draw whichever out of the 51 that are left. But this leads to more events than described above and I just can't rationalize which additional events you get here that you shouldn't get.

2. Jul 19, 2010

### vela

Staff Emeritus
Your reasoning is correct. The solution given is wrong.
You end up double- and triple-counting if you do it the second way. Let's label the aces A, B, C, and D, and I'll just represent non-aces as X. The second calculation counts, for example, (A, BX) and (B, AX) separately, where the first element of the ordered pair denotes one of the (4 1) possibilities and the second element is one of the (51 2) possibilities, whereas the first calculation would count the combination of A, B, and X only once. Similarly, the second calculation counts (A, BC), (B, AC), and (C, AB) separately though they're all the same combination. So you end up over-counting by (4 2)(48 1) + 2x(4 3) = 288+2x4 = 296.

3. Jul 20, 2010

### Ryker

That's great to hear! Although it irks me to death when these things - solutions being wrong - happen, especially when you're not too sure of yourself (yet) and try to figure out for hours where you could've gone wrong.

Ah, makes sense, now I'm kind of mad I missed that :) I again had to take a couple of moments to grasp why it's over-counting by the exact numbers you gave, but I think I get it now. Is there a way other than imagining the two examples that you gave above (for a drawing of two and three aces respectively) to get the (4 2)(48 1) + 2 x (4 3) equation, though? Or do you, in such cases, always need to logically derive how many multiplications of a single solution there are?

4. Jul 20, 2010

### vela

Staff Emeritus
I suppose there might be some identity involving the binomial coefficients that'll give you this result, but I'm not aware of it.

5. Jul 20, 2010

### hgfalling

Also for the "at least one ace" problem, you can calculate the probability that you will draw non-aces on every card and then subtract from one: 1 - (48/52)(47/51)(46/50).

This is of course equal to ((4 1)(48 2) + (4 2)(48 1) + (4 3))/(52 3).

6. Jul 21, 2010

### Ryker

Yeah, you're right, but I guess I had foregone that route as I wanted to understand that other one I was asking about. I know in the end you just have to get the result you want taking the easiest path you know of, but I didn't want to pretend I don't have a problem in understanding some of the material and cop out by going with what I can do :) Thanks for the tip, nonetheless!