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- #1

- Jan 26, 2012

- 4,040

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Remember to read the POTW submission guidelines to find out how to submit your answers!

- Thread starter Jameson
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- Thread starter
- Admin
- #1

- Jan 26, 2012

- 4,040

--------------------

Remember to read the POTW submission guidelines to find out how to submit your answers!

- Thread starter
- Admin
- #2

- Jan 26, 2012

- 4,040

Solution:

For my first card there are 3 Aces to choose from and for my second there are 2 Aces. This can be represented by \(\displaystyle \binom{3}{1} \binom{2}{1}\), 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 \(\displaystyle \binom{50}{2}\).

Putting this together the final answer is \(\displaystyle \frac{\binom{3}{1} \binom{2}{1}}{\binom{50}{2}} \approx .004898\) or $0.49 \% $

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