# Math Help - Playing Card Odds

1. ## Playing Card Odds

I know how to solve this problem using the Choose method, but I'm trying to do it this way and running into a problem. Could someone see where I'm going wrong?

Given a deck of 52 cards, what is the probability of drawing 1 ace, 1 two, 1 three, 1 four, and 1 five?

(4/52) * (4/51) * (4/50) * (4/49) *(4/48)
because you have a 4 / 52 chance of drawing the ace, then a 4 / 51 chance of the two, and so on.

The question then goes on:
what is the probability of getting any straight?

I'm not able to figure this out, even using the Choose method.
(8C1)(4C1)(4C1)(4C1)(4C1) / (52 C 5) = 0.000788
though the book's answer is 0.00355
The way I see this one is that you have (52 C 5) possibilities.
For the first card, anything works. For the second, there are eight possibilities. Then four for each subsequent card.
This strategy worked for the first half, but not here.

2. ## Re: Playing Card Odds

Hello, Relmiw!

I know how to solve this problem using the Choose method,
but I'm trying to do it this way and running into a problem.
Could someone see where I'm going wrong?

Given a deck of 52 cards, what is the probability of drawing
. . an Ace, a Two, a Three, a Four, and a Five?

My answer: .(4/52) * (4/51) * (4/50) * (4/49) *(4/48) because you have
a 4/52 chance of drawing the Ace, then a 4/51 chance of the Two, and so on.

You have imparted an order to the cards.
You have insisted that the first card is an Ace, the second is a Two, and so on.

There are $5!$ possible orders to the five cards.

The answer is $5!$ times your answer.

3. ## Re: Playing Card Odds

Originally Posted by Soroban
Hello, Relmiw!

You have imparted an order to the cards.
You have insisted that the first card is an Ace, the second is a Two, and so on.

There are $5!$ possible orders to the five cards.

The answer is $5!$ times your answer.

Ahhhhh, of course. I even knew my answer was 5! times smaller than it should be but i couldn't figure out why.
It makes perfect sense now, thanks