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