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Math Help - Euchre Questions

  1. #1
    Ari
    Ari is offline
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    Euchre Questions

    Okay so I have a few data management questions relating the game of Euchre.
    the rules of euchre are: the Euchre deck is A,K,Q,J,10,9 of all suits, and 5 cards are dealt to each of the players.

    What is the chance of not being dealt any jacks?
    whats the chance of being dealt atleast 1 jack?
    what is the probability of being dealt atleast 1 jack and 1 ace?
    what is the probability of being dealt neither jacks nor aces?
    what is the probabiltiy of being dealt a jack knowing you are already dealt an ace


    thanks in advance and im more interested in the way the solutions are gotten, this is pre-exam study
    Last edited by CaptainBlack; October 8th 2010 at 01:46 AM. Reason: Restored deleted quest
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  2. #2
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    Hello, Ari!

    The Euchre deck is A ,K, Q, J, 10, 9 in four suits,
    and 5 cards are dealt to each player.

    There are 24 cards in a Euchre deck.

    There are: . \displaystyle {24\choose5} \:=\:\frac{24!}{5!\,19!} \:=\:42,\!504 \text{ possible hands.}




    (a) What is the probability of being dealt no Jacks?

    There are 4 Jacks and 20 Others.

    We want 5 Others: . \displaystyle {20\choose5} \:=\:\frac{20!}{5!\,15!} \;=\;15.\!504\text{ ways.}

    Therefore: . P(\sim\!\text{Jacks}) \;=\;\dfrac{15,\!504}{42,\!504} \;=\;\dfrac{646}{1771}




    (b) What is the probability of being dealt at least 1 Jack?

    This is the exact opposite of (a).

    Therefore: . P(\text{at least 1 Jack}) \;=\;1 - \dfrac{646}{1771} \;=\;\dfrac{1125}{1771}




    (c) What is the probability of being dealt at least 1 Jack and 1 Ace?

    This one requires some Olympic-level gymnastics.


    We use this formula:

    P(\sim\!\text{Jacks} \:\vee \sim\!\text{Aces}) \;=\;P(\sim\!\text{Jacks}) + P(\sim\!\text{Aces}) - P(\sim\!\text{Jacks} \:\wedge \sim\!\text{Aces})


    From (a): . P(\sim\!\text{Jacks}) \:=\:\dfrac{15,\!504}{42,\!504}

    . . .Then: . P(\sim\!\text{Aces}) \:=\:\dfrac{15,\!504}{42,\!504}

    From (d): . P(\sim\!\text{Jacks}\: \wedge \sim\!\text{Aces}) \;=\;\dfrac{4,\!368}{42,\!504}


    We have: . P(\sim\!\text{Jacks}\: \vee \sim\!\text{Aces}) \;=\;\dfrac{15,\!504}{42,\!504} + \dfrac{15,\!504}{42,\!504} - \dfrac{4,\!368}{42,\!504} \;=\;\dfrac{35,\!376}{42,\!504}


    This is the exact opposite of "at least one Jack and at least one Ace".


    Therefore:

    . . P(\text{at least one Jack} \wedge \text{at least one Ace}) \;=\;1 - \dfrac{35,\!376}{42,\!504} \;=\;\dfrac{7,\!128}{42,\!504} \;=\;\dfrac{27}{161}




    (d) What is the probability of being dealt neither Jacks nor Aces?

    There are: 4 Jacks, 4 Aces, and 16 Others.

    We want 5 Others: . \displaystyle {16\choose5} \:=\:\frac{16!}{5!\,11!} \:=\:4,\!368\text{ ways.}

    Therefore: . P(\text{no Jacks and no Aces}) \;=\;\dfrac{4,\!368}{42,\!504} \;=\;\dfrac{182}{1771}



    This is getting too long . . . *yawn*
    I'll let someone else explain part (e)
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