Probability of same suit if drawing $5$ cards from deck of $52$ cards

If we draw $$5$$ cards from a standard deck of $$52$$ cards without replacement, find the probability of drawing all cards of the same suit.

I think the answer is $$P(\text{same suit})=\frac{13}{52}\times\frac{12}{51}\times\frac{11}{50}\times\frac{10}{49}\times\frac{9}{48}$$.

Is this at least close to correct?

You are on the right track. You just need to multiply your answer by 4 since there are 4 suits.

• I see, this is because I have to also use 4 choose 1 for the suit I think.
– Joz
Jun 20 '15 at 19:59
• @Joz Yes. The favoured space is the ways to choose 1 of 4 suits, and 5 of the 13 cards from that suit. The total space is all ways to choose any 5 of 52 cards, $$\dbinom{4}{1} \, \dbinom{13}{5} \Big/ \dbinom{52}{5}$$ Jun 21 '15 at 4:15

When learning how to calculate the probability of an event I think it is usually a good idea to begin with the definition of the probability of an event. That is, the probability of an event $P(E)$ is the number of ways event $E$ can occur, call this $N(E)$, divided by the total number of possible events,call this $N(S)$. That gives us the formula

$P(E)=\frac{N(E)}{N(S)}$

In your case, your $N(E)$ is how many 5-card hands have the same suit. Your $N(S)$ is the total number of possible 5-card hands.

In your case, both of these calculations involve counting the number of combinations, since we are not concerned with order.

To get the total number of 5-card hands with the same suit, $N(S)$, you must remember there a 4 suits, and you are choosing 5 cards from a possible 13. I will leave that calculation to you.

The total number of poker hands, $N(S)$ is 52 choose 5, which is 2,598,960

If you do the above calculation correctly you will get the same answer as you suggest if you multiply by 4, as Tim says above. However, in my opinion, the definition approach can allow to more readily answer a wider range of problems. For example, what if the problem had asked for the probability of getting dealt a flush (5 cards of the same suit) in an actual poker game with 6 players? From the definition, we should see that problem is the exactly the same. But many students would not recognize your solution as representative of the situation because they are not drawing 5 cards in a row from a deck of 52.

• $$\frac{52!}{47!.5!}= \frac{52.51.50.49.48}{5.4.3.2}=$$ $=13.17.5.49.48=13.85.49.48=$ $2,598,960$. Nov 29 '20 at 1:59
• Your method gives $\frac{\frac{13!}{5!.8!}}{\frac{52!}{47!.5!}}= \frac{13!.47!.5!}{5!.8!.52!}=\frac{13.12.11.10.9}{52.51.50.49.48} =\frac{11.3}{4.17.5.49.4}$. Similarly, the post gives $\frac{13.12.11.10.9}{52.51.50.49.48}=\frac{11.3}{4.17.5.49.4}$. But, we still need multiply post's answer by $4$. So, your answer too needs multiplication by $4$, in order to get the same answer. It is possible that you have meant the same, as you stated :"If you do the above calculation correctly you will get the same answer as you suggest if you multiply by 4, as Tim says above. ". Nov 29 '20 at 2:31