I have no stats training, so I am asking if I am attacking this simple statistical problem correctly.

What are the chances of flopping a royal flush in Texas hold’em?

My attempt: The first card dealt to the player must be either a $10,J,Q,K,$ or $A$, any suit. So the probability of the first card deal would be $\frac{5}{13}$. Then the second card dealt to the player must be the same suit, and one of the values mentioned above, meaning the probability is $\frac{4}{51}$. It follows that the next three cards on the flop must be the three remaining cards needed to complete the royal flush, with probabilities: $(\frac{3}{50})(\frac{2}{49})(\frac{1}{48})$. Therefore I believe the odds of flopping a royal flush is: $\frac{5!}{13\cdot 51\cdot 50\cdot 49\cdot 48}$

I would think that this probability would be independent of how many players are in the game. The odds of the deck the be stacked in the perfect way according to how many players there are to give you the royal flush on the flop seem to be the same.

Have I made any mistakes? Thanks in advance.

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    $\begingroup$ "Have I made any mistakes?" No. For a reference see e.g. en.wikipedia.org/wiki/… $\endgroup$
    – Winther
    Jan 1, 2018 at 20:49
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    $\begingroup$ What does "flopping" mean? $\endgroup$ Jan 1, 2018 at 22:40
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    $\begingroup$ @GerryMyerson The flopped cards are community cards. You (and everyone else) get two face-down cards, there are then three face-up cards that anyone can use. $\endgroup$
    – rogerl
    Jan 1, 2018 at 22:42
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    $\begingroup$ @GerryMyerson in Texas hold’em, the flop is the first three cards that are dealt to the table. So basically you can think of my question as: what is the probability of drawing a 10, jack, queen, king, and ace of all the same suit from 5 random cards $\endgroup$
    – Mason
    Jan 1, 2018 at 22:44
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    $\begingroup$ @MJW: Incidentally, when posting a question like this, its best to include the fully mathematical question derived from the problem -- e.g. the translation you give in the comment above. This makes your question much more accessible to those who aren't familiar with the problem domain (in this case, knowledge poker variants). It also gives the knowledgeable readers a chance to spot if you've made any errors in converting the source problem into a mathematical one. $\endgroup$
    – user14972
    May 23, 2018 at 5:51

2 Answers 2


In the poker game you are describing, you are dealt two cards and there are a further three cards dealt on the "flop". There is no opportunity to exchange cards. So really, this is the same as asking the probability of a Royal Flush when you are dealt five cards from a deck of fifty-two.

There are ${52 \choose 5} = 2,598,960$ different equiprobable hands you can be dealt when you are dealt five cards from a standard deck of fifty-two. There are four different hands that constitute a Royal Flush (i.e., one for each suit). This means that the probability of a Royal Flush is:

$$\mathbb{P}(\text{Royal Flush on the flop}) = \frac{4}{{52 \choose 5}} = \frac{4}{2,598,960} = \frac{1}{649,740}.$$


You are correct.

$\dfrac {5!} {(13 \cdot 51 \cdot 50 \cdot 49 \cdot 48)} = \dfrac {1}{13} \cdot \dfrac {5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{51 \cdot 50 \cdot 49 \cdot 48} = \dfrac {52}{13} \dfrac {5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{51 \cdot 50 \cdot 49 \cdot 48} = \dfrac {4}{C(52,5)} = \dfrac {C(4, 1)} {C(52, 5)} = 0.000001539.$

Most of the time, you'll see the probability in terms of the binomial coefficients, $C(n, k)$.


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