Suppose we are dealt five cards from an ordinary 52-card deck. What is the probability that we get no pairs (i.e. all cards are different values).
I'm not sure if I've got the right answer on this one. My logic is that we pick 5 cards out of 13 possibilities. Then every card also has 4 possibilities for a colour. Somehow I feel that this is incorrect.
So my solution would look like:
$P(no pairs) = \frac{\binom{13}{5} \cdot 4}{\binom{52}{5}} $
Am I wrong and if so where? Thanks!
Best Answer
The denominator ${52 \choose 5}$ is correct: the number of ways of choosing 5 cards from 52.
If we are going to get no pairs, we can choose any of the 52 for the first card.
But we can only choose 48 for the next card, since 3 of the remaining 51 would give a duplicate. We can only choose 44 for the card after that, 40 after that, and 36 for the last one.
Hence, the probability is $$ \frac{\frac{52*48*44*40*36}{5!}}{52 \choose 5} $$
Which is $$ \frac{4^{5}*{13 \choose 5}}{52 \choose 5} $$
In other words, you're missing a $4^{4}$ in your computation because each of the 5 cards can be any of the 4 suits.