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?
Probability of getting a royal flush with four wild-cards
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0Be brave and consider all five different cases. If you don't tell them they're complicated they won't be. – 2011-04-14
1 Answers
The probability is $\frac{4 \binom{9}{5}}{\binom{52}{5}} = \frac{3}{15,470}.$
Let's do this two ways.
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.
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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0@Nailo: This approach doesn't entail picking the cards one by one. If you try to count them that way the argument, as you note above, quickly splits into several cases and becomes messy. But the expression does $52(51)(50)(49)(48)$ does show up; it's in the $\binom{52}{5}$ expression. For more examples of counting poker hands the way I'm doing here, see http://en.wikipedia.org/wiki/Poker_probability#Derivation_of_frequencies_of_5-card_poker_hands. – 2011-04-14