# Probability of getting a royal flush with four wild-cards

I am trying to calculate the probability of getting royal flush, if four 5's are wild cards that can be of any suit. I get that the probability of the first card I am picking is $\frac{24}{52}$, but then it seems to be breaking down into many complicated cases (since I can pick which suit the 5 [if I get one] should be of at any time of picking, including the very end). Is there a not complicated way of computing it? Or how can this be calculated?

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Be brave and consider all five different cases. If you don't tell them they're complicated they won't be. –  Yuval Filmus Apr 14 '11 at 4:22

The probability is $$\frac{4 \binom{9}{5}}{\binom{52}{5}} = \frac{3}{15,470}.$$

Let's do this two ways.

1. First, choose the suit. There are four ways to do this. Then, to obtain a royal flush in, say, spades, you can choose any five cards from A, K, Q, J, 10 and the four wild cards. There are $\binom{9}{5} = 126$ ways to do this. Thus there are $4(126) = 504$ total ways to obtain a royal flush. Dividing by $\binom{52}{5}$, the total number of ways to obtain a five-card poker hand, yields the probability.

2. Again, there are four ways to choose the suit. A royal flush in spades can be done with $k$ cards from A, K, Q, J, 10 and $5-k$ wild cards, where $1 \leq k \leq 5$. For fixed $k$, there are $\binom{5}{k}$ ways to choose the non-wild cards and $\binom{4}{5-k}$ ways to choose the wild cards. Summing over $k$, we have that the number of ways to obtain a royal flush with a given suit is $$\sum_{k=1}^5 \binom{5}{k} \binom{4}{5-k} = \binom{9}{5} = 126,$$ where the summation is evaluated using Vandermonde's convolution. As with the first approach, multiplying by $4$ and then dividing by $\binom{52}{5}$ gives the probability of obtaining a royal flush. (Steve D's comment below indicates that he is using a similar approach.)

For those interested, here are some more examples of using combinations to calculate poker hand probabilities.

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There are only 4 wild cards... –  user641 Apr 14 '11 at 4:23
What you want is 31*4, not 32*4. –  user641 Apr 14 '11 at 4:24
@Steve: You're right. Good catch, and thanks! –  Mike Spivey Apr 14 '11 at 4:26
No, this isn't right, either. The wild cards are distinguishable, and I forgot to account for that. Need another fix. –  Mike Spivey Apr 14 '11 at 4:30
On average, we were right: should be 126. –  user641 Apr 14 '11 at 4:34