I'm pretty sure you are.

 

3 Possible combos:

AH-AC
AH-K
AC-K

If he says he has an ace, all 3 hands are open - meaning there is a 1 in 3 chance he has a second ace, 2 in 3 he has the king. If he says he has an ace of clubs, only 2 hands are open - meaning there is an equal chance he has either the ace or king. AC version is 50%, A(s) version is 33%.

Full deck variety. 1326 2-card combos, but only 198 meaningful (those with an Ace):

AH-AC
AH-AD
AH-AS
AC-AD
AC-AS
AD-AS
AH-x (48 possible x values)
AC-x (48 possible x values)
AD-x (48 possible x values)
AS-x (48 possible x values)

AC has only 51 possibilities, while A(s) has all 198. So the AC odds are 3 in 51 (5.9%), while the A(s) odds are 6 in 198 (3.0%).


I think.

 

After all the work, I hope you got it right.

 

This one's correct.

This one needs some work.

There are two problems here. First, you need to count all possible hands for each example listed, not simply the combination of two of the thirteen cards. Second, a hand that holds three or four aces qualifies as having a second ace.

In short, it's more complicated than what you've done.

Note that in the two-card case B = 1.5 * A. In the 13-card case the same is true: B = 1.5 * A. I calculated them a long time ago, and, as I recall, A is in the neighborhood of 36%, and B is around 54%.

 

Err... I did my math based on a 2 card hand in both cases.

I'll trust your calculation.

 

Let's see if I can do this again, and explain it as I go along.

To compute A we need the number of hands that contain at least one ace and the number that contain at least two aces.

The easy way to get the number of hands containing at least one ace is to count all the hands and subtract the number of hands containing no aces.

The total number of 13-card hands is 52C13 = 635,013,559,600.

The number of 13-card hands with no aces is 48C13 = 192,928,249,296.

Thus, the number of hands with at least one ace is 635,013,559,600 - 192,928,249,296 = 442,085,310,304.

The number of hands with at least two aces is the number with at least one ace minus the number with exactly one ace.

The number of 13-card hands with exactly one ace is (4C1) * (48C12) = 278,674,137,872.

The number of hands with at least two aces is 442,085,310,304 - 278,674,137,872 = 163,411,172,432.

Finally, the probability of having a second ace given that he has an ace is 163,411,172,432 / 442,085,310,304 = 36.96%. This is A.

Now, for B.

The number of 13-card hands that contain the ace of clubs is 51C12 = 158,753,389,900.

The number of 13-card hands that contain the ace of clubs and no other ace is 48C12 = 69,668,534,468.

Therefore, the number of hands that contain the ace of clubs and another ace is 158,753,389,900 - 69,668,534,468 = 89,084,855,432.

The probability of having a second ace given that he has the ace of clubs is 89,084,855,432 / 158,753,389,900 = 56.12%. This is B.

 

I'd say 54% and 36% were pretty good estimates.

 

What makes the probability change when the suit is specified is the same for both problems: the reduction in the number of possible hands.

In the 2-card problem, specifying the suit reduces the number of possible hands from 3 to 2.

In the 13-card problem it reduces the number from 442 billion to 159 billion.

 

Next.

 

I think I know what may be tripping up all of us that thought A = B, which I did until I read WestSideBilly and magician's posts and remembered a few things from Prob Stats that I mostly slept through! The key to this problem is that all of the cards are already dealt and we're figuring the overall probably of having a particular hand, and not the probably of a particular future event happening...let me try to explain what the heck I'm talking about.

Let's say there are three cards, AH, AC, and K, face down on a table. Your friend picks up one card, looks at it, tells you it's an Ace, and throws it away. Now there are two cards left on the table. Here are the possibilities:

Case #1: AH thrown away with AC and K on the table

OR

Case #2: AC thrown away with AH and K on the table


Either way, when you go to pick up the next card to see if it's an Ace, you have to choose between two cards and one of them is an Ace. The probability is 1/2 (50/50) or A = B.

If you friend says "I have the Ace of clubs" and then throws it away, you know for a fact that you're dealing with Case #2. You know there are two cards on the table, and one of them is the AH. The probability that you pick up the AH is 1/2 (50/50) again.

Am I making any sense here?