Welcome to our community

Be a part of something great, join today!

Problem of the week #25 - September 17th, 2012

Status
Not open for further replies.
  • Thread starter
  • Admin
  • #1

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
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?
--------------------
Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Admin
  • #2

Jameson

Administrator
Staff member
Jan 26, 2012
4,040
No one correctly answered this week's problem.

Solution:

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 \(\displaystyle \binom{3}{1} \binom{2}{1}\), 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 \% $
 
Status
Not open for further replies.