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!
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