A royal flush consists of a certain five cards of the same suit. Assume we are dealt one of these cards. Then the suit is determined, and we now need the remaining four cards of that suit. So, let's take the probability of being dealt one of the certain five cards (10, J, Q, K, A) of *any* suit. Since there are five such cards per suit, and there are four suits (the set of suits is \( \{ \diamondsuit, \heartsuit, \clubsuit, \spadesuit \} \)), then we can get any one of those cards to start building a royal flush. Therefore, the probability of getting a starting card is
\[
P(\text{starting card}) = \frac{20}{52}.
\]
Once we have a starting card of a certain suit, our suit is restricted. That is, we may only get the remaining four cards from the same suit as our starting card. Then the probability of getting one of the four cards from the same suit is
\[
P(\text{one of the four from the same suit}) = \frac{4}{51}.
\]
Likewise, after we get the second card, there are three left to get, so the probabilities are \( 3/50 \), then \( 2/49 \), and then \( 1/48 \). What, then, is the probability that we will get *all* of the necessary cards? It is the product of all these probabilities:
\[
P(\text{all}) = \frac{20}{52} \cdot \frac{4}{51} \cdot \frac{3}{50} \cdot \frac{2}{49} \cdot \frac{1}{48} = \frac{1}{649\,740} = 1.539\ldots \times 10^{-6} = 0.000\,153\,9\ldots \%.
\]
Not very good odds.

## Monday, June 20, 2011

### How Likely Are You to Be Dealt a Royal Flush?

In the game of poker, you have a royal flush when you have a 10, J, Q, K, and A all of the same suit. Assuming a uniform random distribution of cards in the deck, what is the probability of being dealt a royal flush?

Labels:
game theory,
math,
probability,
statistics

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