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Math Help - cards problem

  1. #1
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    cards problem

    well i wane just make sure if my answer is wrong or right,
    the question is "you are dealt 3 cards form a deck of cards ,which more likely to have more suits with even number of cards or more suits with odd numbers of cards"

    my answer is the probability of both are the same since you have the same number of even and odd numbers in a deck of cards so their probability should be the same . is this correct or i am wrong . thx
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  2. #2
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    Hello, spywx!

    I hope I'm reading the problem correctly . . .


    You are dealt 3 cards from a deck of cards.
    Which is more likely?
    To have more suits with even numbers of cards
    or more suits with odd numbers of cards

    There are three possible outcomes . . .

    (1) The three cards are of the same suit.
    . . There is one suit with an odd number of cards.
    . . There are no suits with an even number of cards.
    There are more odd suits.

    (2) The three cards are: two of one suit, one of another.
    . . There is one suit with odd number of cards.
    . . There is one suit with an even number of cards.
    It is a "tie".

    (3) The three cards are of three different suits.
    . . There is one card of the first suit.
    . . There is one card of the second suit.
    . . There is one card of the third suit.
    . . There are three suits with an odd number of each.
    There are more odd suits.


    It is more likely to have more suits with an odd number of cards.

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  3. #3
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    To expand slightly on Soroban's response:

    P(\text{3 suited}) = \frac{{52}}<br />
{{52}} \times \frac{{12}}<br />
{{51}} \times \frac{{11}}<br />
{{50}} = \frac{{22}}<br />
{{425}} Here, you have only an odd number of suited cards.

    P(\text{2 suited, 1 not suited}) =3\left( {\frac{{52}}<br />
{{52}} \times \frac{{12}}<br />
{{51}} \times \frac{{39}}<br />
{{50}}} \right) = \frac{{234}}<br />
{{425}} Here, you have an odd and an even number of suited cards.

    P(\text{0 suited}) = \frac{{52}}<br />
{{52}} \times \frac{{39}}<br />
{{51}} \times \frac{{26}}<br />
{{50}} = \frac{{169}}<br />
{{425}} Here, you have only an odd number of suited cards


    In all 3 cases, you have a at least one suit with an odd number of cards, so the probability of having an odd number of suited cards out of 3 dealt cards is \frac{{22}}{{425}}+\frac{{254}}{{425}}+\frac{{169}  }{{425}}=1, whereas only in the second case do you have a suit with an even number of cards, so the probability of that situation is \frac{{254}}{{425}}\approx.55. It is intuitively obvious though that you will ALWAYS have a suit with an odd number of cards if you are dealt an odd number of cards, because it is impossible to partition a set with odd order into subsets of only even order.

    This leads me to believe you may have not properly described the problem, as the result is rather trivial.
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  4. #4
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    you are reading the problem right, but i have one point that you ignoring the 0 which an even number,for example the 1st case we got 3 cards for the same suit so we got 0 cards of the other suits so we got now 3 suits with even number which is 0 .

    Quote Originally Posted by Soroban View Post
    Hello, spywx!

    I hope I'm reading the problem correctly . . .



    There are three possible outcomes . . .

    (1) The three cards are of the same suit.
    . . There is one suit with an odd number of cards.
    . . There are no suits with an even number of cards.
    There are more odd suits.

    (2) The three cards are: two of one suit, one of another.
    . . There is one suit with odd number of cards.
    . . There is one suit with an even number of cards.
    It is a "tie".

    (3) The three cards are of three different suits.
    . . There is one card of the first suit.
    . . There is one card of the second suit.
    . . There is one card of the third suit.
    . . There are three suits with an odd number of each.
    There are more odd suits.


    It is more likely to have more suits with an odd number of cards.

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