# 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 $\frac{4^{13}}{52\choose 13}=0.01\%$ – almagest 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. ^_^ – Arnie Bebita-Dris 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 $\frac{4^{13}}{52\choose13}\approx 0.01\%$.

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

• In fact, you should intuitively expect a very low probability. Imagine drawing the cards one by one from a shuffled deck and hoping to get no pairs; for the last card alone there are 40 cards in the remaining deck and only 4 of them would work. That's already 1/10. – user21820 Apr 15 '16 at 9:39
• user21820, I apologize, but your answer (plus your comment) does not enlighten me that much. Care to provide more hints / details to your answer, please? =) – Arnie Bebita-Dris Apr 15 '16 at 9:49
• @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. – user21820 Apr 15 '16 at 9:50

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%.

• Please format this using MathJax – Colbi Jun 23 '16 at 19:45
• Please consider formatting your post in order to make it more readable. – Mårten W Jun 23 '16 at 20:27