Conditional Probability Question

[XChamp]

I am trying to figure out how to solve this problem, which is:

Two people, E and W, randomly select 13 cards out of a full deck of 52 cards. If E has 2 aces, what is the probability that W has just one ace?


I thought the answer was: (2 1)*(48 12)/(50 13) = 39.27%, where (x y) is read as " x choose y".

I am being told this is incorrect. Can anybody tell me why?

[gaming_mouse]

Looks right to me. Why are they saying you are wrong?

gm

[XChamp]

No reason given as yet. It's for a little drill just to check understanding for this class I'm taking. I'm starting to think that the prof is wrong and I'm right.

The choices for an answer are A- 0.21, B-0.23 C-0.25 and D-0.27 .

[Cobra]

I may be doing this wrong but I thought you would do it as follows:

E has two aces and 11 non aces.

The deck remaining has 2 aces and 37 non aces.

The Probability of W getting exactly one ace would be

=2*(37 choose 12)/(39 choose 13) = 45.6%

Cobra

[XChamp]

You are exactly right up till the end. The deck has 2 aces left and 48 non-aces because you don't know what the other 11 cards are for E, other than they are not aces. So, you need to keep track of all the combinations involving those.

I just did a rough numerical analysis (by hand, no computer). I took two aces out of the deck and dealt out 13 cards. I did this 10 times. 5 times there was one ace present and 5 times there were 0.

[Ozzzz]

I don't see how the 11 other cards E has matters. What is important is that there are 39 cards left and 2 aces left.

There are a total of 39C13 (8122425444) combinations left for W. Of those, 37C13 (3562467300) do not contain an Ace, and 37C11 (854992152) contain both Aces. Thus, 3704965992 contain exactly one Ace, or 45.6%.

[gaming_mouse]

Let's try this one more time, using the formula for cond. prob:

A = event that first person has two aces
B = event that second person has exactly 1 ace.

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

Using the multinomial formula:

P(A & B) = ((4 choose 2)*2*(48!/(11!*12!*25!)))/(52!/(13!*13!*26!))= 0.0973829532

P(A) = (4 choose 2)*(48 choose 11)/(52 choose 13)= 0.213493397

Thus the answer is:

0.0973829532/0.213493397=0.456140352

So the answer is in fact 45.6%

gm

[EliteNinja]

Damn, I wish I discovered 2+2 when I was taking stats 3 years ago.

I would've aced Stats with your (2+2ers) help.

[XChamp]

thanks for all your replies.


So you are saying that P(B given A) = (2 C 1)*(37 C 12)/(39 C 13) = 0.456

Ok when I think about your method it makes sense, except that I don't understand why my method didn't work.

The denominator of the previous equation represents the total number of 13 card combinations (where order doesn't matter) given that 2 aces are present in the first person's 13 cards. Doesn't this mean that the total possible combinations for the second person's 13 cards is (50 C 13)? The second person does not know any of the first persons 11 other cards. It looks like you are arbitrarily excluding cards when you have no knowledge of whethere they are left in the deck or not.

I am beginning to see this is not so, but I keep thinking about other problems where you can't do that. For example, if I deal 2 cards face down out of a full deck, and then ask someone what the odds are of the next card being an ace, it is 1/13 regardless of the fact that the deck is missing 2 cards. I could express it as (4 C 1)/(52 C 1). The reason why this is true is because the two cards that are missing are random. I could also examine the three different situations (2,1 and 0 aces were dealt out) and weight the probabilities of each to come to the same 1/13 answer. It is for this reason that I don't freak out when my friends and I discover that the deck we played poker with for 2 hours was short one card the entire time, because it is a random card (hopefully!) so didn't change a single thing. The fact that the card is missing has the chance of hurting/helping any of us equally.

In other words, if I tried to list the different combinations of 13 cards the second person could possibly see, I will end up including every card in the deck except 2 aces because I dont know, for example, if the 5h is left in the deck or not. The same is similarly true for the numerator (I think, at least).

Sorry that was so long winded. I am confused about this problem and I sense I may have a fundamental misunderstanding about the situation or even probability in general.

[gaming_mouse]

[ QUOTE ]
The denominator of the previous equation represents the total number of 13 card combinations (where order doesn't matter) given that 2 aces are present in the first person's 13 cards. Doesn't this mean that the total possible combinations for the second person's 13 cards is (50 C 13)? The second person does not know any of the first persons 11 other cards. It looks like you are arbitrarily excluding cards when you have no knowledge of whethere they are left in the deck or not.


[/ QUOTE ]

I was having the same thoughts at first, which is why I did it the long way to convince myself that the method was correct.

It is not a case where you can use the "unseen cards" shortcut, which I at first thought it was and is the reason I agreed with you in my first post. The reason for this is that other player does have 13 cards. He has 2 aces, and 11 non-aces. The point is that it does not matter what his 11 other cards are. He has them. Therefore you are drawing out of only 39. Obviously there are thousands of possible holdings for him. But, no matter what his holding, YOU are always drawing to 39 cards.

Does that make sense?

gm

[XChamp]

So, given the fact that I am drawing from 39 distinct cards, I know there must be (39 C 13) combinations. This makes a lot more sense now. Thank you.

[Siegmund]

It comes down to some hair-splitting over how the original question was worded.

If "East has 2 aces" means "East has exactly two aces (and therefore exactly 11 non-aces)," then the 45% answer is correct. If "East has 2 aces" means simply "East has two aces that we know about, but he may have a third or fourth ace among his 11 unknown cards", then you can legitimately arrive at the 39% answer.

I think most of us understand the question to mean the first, though, and will arrive at 45%.

[the alex]

W has at least one Ace:
13[2/(52-13)]=.6667

W has a second Ace:
(12)[1/(52-14)](.6667)= .2162

W has only one Ace:
.6667- .2162= .4505

I got 45.05%. Where am I off here?

[Kellon]

[ QUOTE ]
(12)[1/(52-14)](.6667)= .2162

[/ QUOTE ]

I get .2105.

.6667- .2105= .4562

[Kellon]

What book are you using for your class?

[the alex]

Kellon,

That makes sense. I didn't read all of the posts to this point and I'm not a math guy, so I guess I was algebraically right.

Thanks,
Alex

[Kellon]

I read the stuff in this forum hoping something will sink in by osmossis. The arithmetic I can handle, but the rest I'm having a really hard time with. What makes it especially tough on the ol' ego is that I aced my stat class back, back, way back in college. That's why I'm hoping that one of these days, one of y'all will come forward with a recommendation of a reasonably decent textbook that covers the math of probability.

[the alex]

What's you're problem, Kellon?

That what can happen and how frequently it will happen and make the relations. That's poker.

[Kellon]

I understand "[t]hat's poker." What I want to learn is the math (the formulas, which ones to use in which situations, and how to compute them) of probability. That's my problem. (My previous post shouldn't have been a Reply to yours. My apologies.)

[XChamp]

I'm using "Probability and Statistical Inference" by Hogg and Tanis. I haven't used it much though since my professor rarely assigns anything from the text so I can't make a recommendation.

[Cheeseweasel]

It appears to me that the solution would be to use the combinatorial formula

P(k out of n) = (n!/(k!*(n-k)!)*(p^k)*(q^(n-k))

where

n = 13
k = 1
p = 2/39
q = 37/39

Plugging in the numbers I get .3544

Best of Luck