MHB Calculating Probability of Getting Two Aces in Texas Hold Em

[Jameson]

You are playing your friend in Texas Hold Em. Each player (you and your friend) gets two cards, face down. Assuming that your friend has only one A, what is the probability that you are dealt two aces?
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[Jameson]

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]