# What is the probability of getting dealt five cards with five different values?

Five cards are dealt without replacement from a standard deck of 52 cards. What is the probability that they have five different values?

My idea is that the probability is $\frac{13 \choose 5}{4^5}$. Is that correct?

• One way to see that $C(13,5)/4^5$ can't possibly be correct is to compute it: $C(13,5)=1287$ whille $4^5=1024$, which means $C(13,5)/4^5=1287/1024\gt1$. Probabilities must always be between $0$ and $1$. – Barry Cipra Jun 5 '17 at 19:21

$$\frac{52}{52} \cdot \frac{48}{51} \cdot \frac{44}{50} \cdot \frac{40}{49} \cdot \frac{36}{48} \approx 0.507$$
There are $13$ groups of cards with the same value. Each group has $4$ members.
• The total number of possible options is $\binom{52}{5}$.
• To find the number of favorable options, note that in $\binom{13}{5}$ ways a group is selected. In addition, in $4^5$ ways five specific cards are selected from each group.
Hence, the desired probability is $$\frac{\binom{13}{5}\cdot4^5}{\binom{52}{5}}=0.5071$$