# The probability of being dealt a royal straight flush (ace, king, queen, jack, and ten of same suit)

The probability of being dealt a royal straight flush(ace, king, queen, jack, and ten of same suit is about $$1.3 \times10^{-8}$$ Suppose an avid player sees 100 hands per week for 52 weeks for 20 years....

my question doesn't have to do with solving the question. It has to do with the statement of the probability of the royal flush.

Now going off of what I have learned so far I had figured out the probability of a royal straight flush to be $$\frac{\binom{4} {1} }{\binom{52} {5} }$$

So my question is where did they obtain the probability of $$1.3 \times10^{-8}$$ from amd how is it liked to the combinatorial expression? if at all?

edit: I forgot the choose expression in latex if someone could edit it for me.... thanks.

edit 2: Thanks for the helpful responses and providing me assurance that I am. actually competent enough to think about these things. The book I am using is Mathematical Statistics and Data Analysis 3rd edition by John Rice

• Your latex expression is \binom{n}{k}. Also, you can change $X$ to \cdot :) Apr 2, 2017 at 20:03
• @Winther i was not clear in my question look at the edit Apr 2, 2017 at 20:14
• Who did obtain that probabillity? You got the correct expression and it evaluates to $1.54\cdot 10^{-6}$ not $1.3\cdot 10^{-8}$. Apr 2, 2017 at 20:21
• @Winther the textbook i using quoted that and i found it odd.... Apr 2, 2017 at 20:22

What is your source? The expression $\frac{4}{\binom{52}{5}}$ is correct (you don't need $\binom{4}{1}$ at the top; one easily counts the hands directly with no calculation!).

I suspect your source is wrong. E.g.

gives "odds of getting any one of these royal flushes are 2,598,956 to 4" and since $\binom{52}{5} = 2,598,960$, but "odds" are given as fail to successes or successes to fails, so this is a probability of $\frac{4}{2598956 + 4} = \frac{4}{2598960} = \frac{4}{\binom{52}{5}}$.

Other sources give the same result, and it's the only one that makes mathematical sense.

Im not really sure where they pulled $1.3*10^{-8}$ from. You're probability is accurate. The total amount of ways you can get a royal flush, ${4\choose 1}$, over the total amount of hands, $52 \choose 5$, gives you the correct probability as there is an equal chance for each hand to be dealt. I suggest you bring it up with whomever wrote this question and correct their mistake, whatever it might be.

The way I think of it, somewhat intuitively:

There are $20$ cards of the deck which could be involved in a royal flush ($5$ cards from each of the $4$ suits).

We first need to choose any one of these $20$ with a probability of $\cfrac {20}{52}$. This then determines which royal flush we are 'working towards'.

Thus there are only $4$ cards left in the deck which will complete the royal flush, we must choose any one of these with a probability of $\cfrac{4}{51}$

The next card to complete the flush will have a probability of $\cfrac {3}{50}$, the fourth card of the flush $\cfrac{2}{49}$ and the final card of the flush $\cfrac{1}{48}$

Therefore, the final probability of drawing a roayl flush is equal to:

$$\frac{20}{52}\times\frac{4}{51}\times\frac{3}{50}\times\frac{2}{49}\times\frac{1}{48}=\frac1{649740}\approx1.54\times 10^{-6}$$

In conclusion:

I don't know where the $1.3\times10^{−8}$ came from, so I agree with @Isaac Browne's answer that you should bring it up with whoever wrote or assigned the question