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)

Update:

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

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  • 4 weeks ago
    Best 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.

    • 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

  • 4 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.

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