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