1. ## Probability

I have been working on this question for a while now, and am having trouble with it.

Assume you are playing poker, are dealt five cards in a row from a new deck of cards. What is the probability of having the Ace, King, Queen, Jack, and 10 of Hearts?

I know that the probability of recieveing a single card in a deck would be 1 in 52, and am confident the second card would be 1 in 51 and so on. I am not getting a good answer to this question though.

I don't know if I am right, but I think I need to calculate the total number of possible 5 card combinations in order to get the right answer. I have never done this before and am lost.

2. Originally Posted by MikeMcC
I have been working on this question for a while now, and am having trouble with it.

Assume you are playing poker, are dealt five cards in a row from a new deck of cards. What is the probability of having the Ace, King, Queen, Jack, and 10 of Hearts?

I know that the probability of recieveing a single card in a deck would be 1 in 52, and am confident the second card would be 1 in 51 and so on. I am not getting a good answer to this question though.

I don't know if I am right, but I think I need to calculate the total number of possible 5 card combinations in order to get the right answer. I have never done this before and am lost.

For the first card, you don't care whether you get an Ace, King, Queen, Jack or 10 of hearts. So you can get any of those cards. So you can get any of these 5 cards

Then on the second card, you can only accept 4 cards. Lets say you got a 10. Well then for the second card you need an Ace, King, Queen, or Jack of hearts. So thats 4 possible. Since you already have 1 card though, there are only 51 cards remaining. This pattern repeats

Then for the third card, you can only accept 3 out of the possible 50. For the 4th 3 possible and for the 5th card, 1 possible.

So

$\displaystyle \frac{5}{52}\cdot\frac{4}{51}\cdot\frac{3}{50}\cdo t\frac{2}{49}\cdot\frac{1}{48}=$

3. There are four kinds of Royal Flushes. This is one of those four.

This is just one particular hand out of $\displaystyle \binom{52}{5}=2,598,960$ possible hands.