This question arose from the game "Poker Shuffle", where one can rearrange 25 cards in a 5x5 grid to maximize the net score of the resulting 10 poker hands. As the royal flush is the highest-scoring hand, I am wondering what the probability of being dealt a royal flush is.

I don't even know where to begin in attacking such a problem—my only insight is that it will likely have to take the form of an inclusion-exclusion formula due to the possibility of multiple royal flushes in one "hand".

Thanks in advance for any thought any of you put into this!

  • $\begingroup$ From how many decks does one take the 25 cards for the grid? $\endgroup$ Dec 24, 2016 at 14:20
  • $\begingroup$ @GuachoPerez I would assume that all cards come from one deck without replacement, so that you never have two equal cards. I have never seen poker hand scoring take into account the possibility of getting two equal cards, or five-of-a-kind. $\endgroup$
    – Arthur
    Dec 24, 2016 at 14:24

1 Answer 1


As mentioned in the OP, this is a job for the Principle of Inclusion/Exclusion (although the probability of more than one royal flush is small).

There are $N = \binom{52}{25}$ possible hands, all of which we assume to be equally likely. Say a hand has "Property $i$" if it includes a royal flush of suit $i$, for $i = 1,2,3,4$. Let $S_i$ be the number of hands with $i$ of the properties. Then we have $$\begin{align} S_1 &= \binom{4}{1}\binom{47}{20}\\ S_2 &= \binom{4}{2}\binom{42}{15}\\ S_3 &= \binom{4}{3}\binom{37}{12}\\ S_4 &= \binom{4}{4}\binom{32}{5} \end{align}$$ By PIE, the number of hands with none of the properties, i.e. the number of hands without a royal flush, is $$N_0 = N - S_1 + S_2 - S_3 + S_4$$ so the probability of not having a royal flush is $$\frac{N_0}{N} \approx 0.919466$$ and the probability of having at least one royal flush is

$$1 - \frac{N_0}{N} \approx 0.080534$$


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