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

Question

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?

Answer 1

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 $\frac{4^{13}}{52\choose13}\approx 0.01\%$.

Answer 2

Your reasoning for "total possible number of such poker hands" is wrong. You asked for no pairs, not all being of the same suit!

Answer 3

There are 13C13*(4c1)^13 = 4^13 = 67,108,864 ways to select one suit each of ALL thirteen available ranks. And 52C13 = 635,013,559,600 possible 13-card ("Chinese") poker hands exist. Thus, the probability of a thirteen-card pairless hand (guaranteed a straight, of course) is 67,108,864/635,013,559,600, or about 0.01%.