What are the Odds? Part 2

There are ${52 \choose 7}=133,784,560$ possibilities for the 7 cards dealt.

There are 4 possible suits for the 5 cards that make up the royal flush, and $47 \choose 2$ possibilities for the other two cards. Multiply 4 by $47 \choose 2$ to find the total number of combinations that make up a royal flush. $4 \times {47 \choose 2}=4324$ .

Therefore, the probability of getting dealt a royal flush from 7 cards of a deck is $\frac{4324}{133,784,560}$ ; however, what the problem is asking for is $\frac{1}{\frac{4324}{133,784,560}}$ , which is $\frac{133,784,560}{4324}=\boxed {30,940}$ .

A royal flush consists of an ace, king, queen, jack, and ten, all of the same suit. You are drawing 7 cards.

1) The probability that the first card is in a royal flush is $\frac{20}{52}$ . There are four suits and five cards in a royal flush, so there are $20$ possible cards that can be in a royal flush. There are $52$ total cards.

2) The probability that the next card is in the same royal flush as the first card is $\frac{4}{51}$ . The suit has been chosen by the first card, so there is one possible suit and four possible cards this card can be. One card has already been chosen, so now there are only $51$ total cards.

3) The probability that the next card is in the same royal flush as the first and second card is $\frac{3}{50}$ . There is one possible suit and three possible cards, and a total of $50$ cards.

4) Following the same logic, the fourth card has probability $\frac{2}{49}$

5) and the fifth card has probability $\frac{1}{48}$ .

Then, there are still two more cards, and these cards can be anything you want them to be, so there is a probability of $1$ for each.

However, the first five cards in your set of seven don't have to be the ones with a royal flush. It could be the last five, or the middle five, etc. There are ${7 \choose 5}$ different combinations the royal flush could be in your 7 cards, and so the number of different ways is ${7 \choose 5} = 21$ .

So the final answer is $\frac{20}{52} \times \frac{4}{51} \times \frac{3}{50} \times \frac{2}{49} \times \frac{1}{48} \times 1 \times 1 \times 21 = \frac{1}{30940}$ .