Why are you excluding 6 cards for the third card? You can accept 50/50 for the third card.
Sorry wrong quote - in your original analysis you listed 11/13 for the third card when it should have been 50/50 since its face value was inconsequential.
Chronos is right. Put in terms of binomial coefficients, it’s
(4, 1)(4, 1)(44, 1) / (52, 3) = ~0.031855
for it to be an ace, a king, and something else (not ace or king). AAK and AKK each has probability
(4, 2)*(4, 1) / (52, 3) = ~0.0010860
and adding it all up gives ~0.034027.
Once again I disagree, if you already have your A & K on the first two cards then the third card can be anything (50/50). Remember the problem states having at least one ace and one king and does not exclude the possiblity of having either two aces and king or one ace and two kings. You are determining the probability of obtaining only one ace and one king out of a three card deal. Two different problems.
On a related note, what would you define as the probability of getting **at least one ace ** if being dealt three cards from a well shuffled deck?
Are you disagreeing with me? Because although I calculated the probability of an ace and a king and something else (non-AK) along the way, I then explicitly added to it the probability of getting AAK and AKK to come up with a higher number that is the correct answer.
That’s an easy one. The probability of getting zero aces is the number of ways of picking 3 from 48 divided by the number of ways of picking 3 from 52:
(48, 3) / (52, 3) = ~0.783
so the probability of at least one is about 0.217.
In my third case, I exclude the high cards for the first card because if I had one, it’d be the first or second case. In order to be able to just add the probabilities like I did, the cases need to be mutually exclusive.
Trust me, we’re quite aware that AAK or AKK is acceptable.
I think I see where you are coming from.
For further clarification and understanding:
what is the probability of drawing at least one king in two tries from a conventional deck of 52 cards?
what is the joint probability of drawing both a king and a queen in two tries from a conventional deck of 52 cards?
what is the conditional probability that one of the two cards is a king, given that the other is a queen?
This is like the “at least one ace in three tries”, but with two tries. The probability is
1 - (48, 2) / (52, 2) = 0.14932
Note that this is slightly less than 2/13 (which is the average number of kings you’ll get).
(4, 1)*(4, 1) / (52, 2) = 0.012066
This is slightly more than the 21/131/13 that you’d get if you drew with replacement.
4/51, if you stipulate that a particular card is a queen (as opposed to that one or the other is a queen). Same as if you just said it’s not a king.
I believe I understand your methodolgy as I was just calculating by a more less brute force method similar to the description that Chronos initially offered.
Again I agree and see where you are getting your numbers. I don’t recall being taught Binomial Coefficients but admittedly that was 35 years ago.
Here I’m going to admit that this was a setup and I apologize for any deception. According to Epstein’s The Theory of Gambling and Statistical Logic (pages 16-17) the conditional probability problem is equal to the probability of drawing one king out of two cards plus the probability of drawing one queen out of two cards less the probability of drawing a king-queen combination. Numerically he shows it as being (396+396-32)/(5251)=0.2866 (twice the number in the first example above minus the second example or 20.149-0.012). I would assume the high probability (compared to your value of 0.078) has to do with his statement given that the other is a queen . His explanation contains the following footnote:
Note that either of the two cards may be the postulated queen. A variation asks the probability of one card being the king, given the other to be a non-king. In the latter case the probability is 0.145.
I was looking for an explanation for the last part since I frankly don’t understand where he is coming from. 
That’s definitely not the answer to the question that you asked. In fact it seems to be the answer to a very different question: what’s the probability of at least one king or one queen.
The question that you asked is easy. I deal you one card. It’s a queen. What’s the probability that the next is a king? Four kings left, of 51 cards.
What you said seems to make sense except why do you subtract the probability of getting both a queen and a king since that seems to meet the criteria for getting at least one queen or king?
On your latter issue what about getting the queen on the second card in which case is the overall probability then (2*4/51)?
On a more general question, my understanding is that the binomial coefficient represents the number of permutations possible for a given situation where order is important?
BTW - thanks for sharing your knowledge!
If you just add the probabilites of getting at least one king and at least one queen, you’ve counted the possibility of getting a king and a queen twice. So you have to subtract it once.
As I suggested above, the question is open to multiple interpretations. But I don’t think any version has 2*4/51 as an answer. You can conditioin on there being at least one queen, in which case the probability of a king is roughly twice that, but not exactly. It’s actually a bit lower than 2/13, whereas 8/51 is a bit larger than 2/13.
No, it’s the number of combinations possible, without regard to order (so it’s not about permutations at all).
Mr. (or Ms.) DePlume:
Do you add the probabilities of individual events to determine “joint” probability?
Though I was the who brought it up originally, what actually is the definition (or at least explanation) of “conditional” probability?
One last card related question having to do the fact that I still don’t completely grasp the application of Binomial Coefficients (BC), but what is the distinction, in terms of BC, between saying “one and only one ace and king out of three cards” and “at least one ace and king out of three cards”?
BTW, thanks for your responses.