# What is the probability of getting NO PAIRS in a $13$-card poker game?

What is the probability of getting NO PAIRS in a $13$-card poker game?

Here is my attempt:

The setup for the required poker hand would be: $$ABCDEFGHIJKLM$$ where $A, B, \ldots, M$ are distinct faces.

The total possible number of such poker hands is $${13}\cdot{12}\cdot{11}\cdot{10}\cdot{9}\cdot{8}\cdot{7}\cdot{6}\cdot{5}\cdot{4}\cdot{3}\cdot{2}\cdot{1} = 6227020800.$$

The total possible number of $13$-card poker hands from the standard deck of $52$ playing cards is $${52 \choose 13} = 635013559600.$$

Therefore, the required probability is $$\dfrac{6227020800}{635013559600} = \dfrac{2223936}{226790557} \approx 0.9806\%.$$

My question is:

Is this probability computation correct?

• You must be dealt one card of each value (Ace, King, ... , 2). For each value you have a choice of 4 suits. So $4^{13}$ possible hands. There are ${52\choose 13}$ hands in total, so prob $\frac{4^{13}}{52\choose 13}=0.01\%$ Apr 15 '16 at 9:49
• @almagest, please write that down as an actual answer so that I may accept accordingly. Thank you very much. ^_^ Apr 15 '16 at 9:50

To get no pairs you must be dealt one card of each value (Ace, King, ... , 2). For each value you have a choice of 4 suits, so $4^{13}$ possible hands. There are ${52\choose13}$ hands in total, so the prob of no pairs is $\frac{4^{13}}{52\choose13}\approx 0.01\%$.
• @KashitokikuTeshikiari: I know that, but neither does your question tell us much about how you got $13 \times 12 \times \cdots \times 1$. And that's the problem. Without knowing that information, we can't guess what went wrong with your reasoning, only that it's wrong. Apr 15 '16 at 9:50