No one correctly answered this week's problem.
Solution:
[sp]There are 4 Aces to start with and 48 non-Aces. If the other player has exactly one Ace and one non-Ace then there are 3 Aces remaining and 47 non-Aces. The question asks what is the probability that I am dealt two Aces if we know the above to be true.
For my first card there are 3 Aces to choose from and for my second there are 2 Aces. This can be represented by [math]\binom{3}{1} \binom{2}{1}[/math], which is $3 \times 2=6$ combinations of me being dealt 2 of the 3 remaining Aces.
Now we must divide by the total number of possible hands, keeping in mind that 2 cards have been dealt to my opponent. That means there are $52-2=50$ cards remaining, and we are choosing 2 from this set. That is expressed by [math]\binom{50}{2}[/math].
Putting this together the final answer is [math]\frac{\binom{3}{1} \binom{2}{1}}{\binom{50}{2}} \approx .004898[/math] or $0.49 \% $ [/sp]