# High School Statistics Problem 15 (Awarding B.A.)?

The 10, A, J, Q, K, A are removed from a standard deck. Each of the 4 players are dealt 5 cards.

What is the probability that each player gets their own suite of cards?

(ex. player 1 gets diamonds, player 2 gets clovers, player 3 gets hearts, player 4 gets spades)

My Answer: (4!) / (24C5*18C5*12C5*6C5)

I only care about the thought process, not about whether or not I'm right (though, that would be kind of nice too)

### 2 Answers

- Steve ALv 74 weeks agoBest Answer
There would be 32 cards left, correct?

First player draws one card. Does not matter what that is. Say diamonds.

(7/31)(6/30)(5/29)(4/28) for him.

Next player draws a card. Cannot be one of the three remaining diamonds.

(24/27)(7/26)(6/25)(5/24)(4/23)

Next player (16/22)(7/21)(6/20)(5/19)(4/18)

Final player (8/17)(7/16)(6/15)(5/14)(4/13)

Leaving 3 cards of each of four suits.

- PuzzlingLv 74 weeks ago
Here's how I approached it.

There are 4! ways to assign the suits to each player --> 4!

The probability the first player is dealt 5 cards all in their suit would be:

C(8,5) / C(32,5)

That comes from there being C(8,5) ways to pick 5 cards all in their suit. But that's out of 32 cards in the deck where they are being dealt 5.

The probability the second player is dealt 5 cards all in their suit would be:

C(8,5) / C(27,5)

That comes from there being C(8,5) ways to pick 5 cards all in their suit. But that's out of 27 cards remaining in the deck where they are being dealt 5.

The probability the third player is dealt 5 cards all in their suit would be:

C(8,5) / C(22,5)

That comes from there being C(8,5) ways to pick 5 cards all in their suit. But that's out of 22 cards remaining in the deck where they are being dealt 5.

The probability the fourth player is dealt 5 cards all in their suit would be:

C(8,5) / C(17,5)

That comes from there being C(8,5) ways to pick 5 cards all in their suit. But that's out of 22 cards remaining in the deck where they are being dealt 5.

So the final probability is:

4! C(8,5)^4 / [C(32,5), C(27,5), C(22, 5), C(17,5)]

= 448 / 5028338002545

Your answer is of the right magnitude but isn't exactly correct. I'm trying to understand your logic. Maybe if you explained it, I could figure out where yours is different.

You repeated the Ace.

My bad, I admit I terribly worded that question. The removed cards are used, not discarded. That aside, I think I should just rewrite the entire question. Thanks though