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?

  • 3
    $\begingroup$ 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 $1$. $\endgroup$ – Barry Cipra Jun 5 '17 at 19:21

There are 52 options for the first card, 48 for the second, 44 for the third, 40 for the fourth and 36 for the fifth. The probability of selecting five different values thus equals:

$$\frac{52}{52} \cdot \frac{48}{51} \cdot \frac{44}{50} \cdot \frac{40}{49} \cdot \frac{36}{48} \approx 0.507$$


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 $$\frac{\binom{13}{5}\cdot4^5}{\binom{52}{5}}=0.5071$$


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