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Please give me the correct answer to the problem

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  • 07-08-2015 1:30am
    #1
    Doyler92
    Registered Users Posts: 2,664 ✭✭✭


    Hi guys.

    A group of friends and I are having a debate about the percentage chance of the outcome in this situation. Here is how the situation is:

    You have a 52 card deck. One player chooses 3 cards at random (say ace, two and three). What are the percentage chance for any one of these cards (it can be any suit) to come out if three random cards are chosen face down.

    We have came to a conclusion of two differing figures. The first one is ~56% and the second one is ~70%.

    If it's possible, please give a really simplified walk through of how the solution is figured out.

    Many thanks for your time.
    Tagged:


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    #2
    spock.
    Closed Accounts Posts: 8,502 ✭✭✭spock.


    There is a 12/52 chance that card A will be an ace, two or three.
    Therefore there is a 40/52 chance that it is not.

    Given that the first card is not, the probability that the 2nd card is not ace, two or three is 39/51

    3rd card will be 38/50 and so on...

    The probability that none of,the cards are an ace, two or three is
    40/52 * 39/51 * 38/50 = 44.7%

    Therefore the probability that at least one comes up is (100-44.7) 55.3% (or roughly 56% which you iroginally had)


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    #3
    Doyler92
    Registered Users Posts: 2,664 ✭✭✭Doyler92


    Thanks Spock for the answer. It makes a lot more sense now and also brings back the memories of probability in school!


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    #4
    Yakuza
    Registered Users Posts: 5,141 ✭✭✭Yakuza


    For completeness, below is a table for getting one, two or three cards correct.
    Note that the event Two includes getting two twos and something else, and Three could mean three aces. I'm too tired to start working out *exactly one* match etc.
    Total Possible outcomes of 3 from 52	 : 22100	
Matches	Good	Unwanted Total	 %
Zero	1	9880	  9880	44.71%
One	12	780	  9360	42.35%
Two	66	40	  2640	11.95%
Three	220	1	   220	 1.00%
                         22100

    There are 52C3 ("52 choose 3") or 22100 possible ways to deal three cards from a deck of 52.
    Basically, you divide the cards into two piles, the 12 "good" ones and 40 unwanted ones, and read the table as follows:
    There is 1 way to get none of the cards you want from the 12 and 9880 (40C3) ways of getting three cards from the bad pile, leading to 9880 or 9880 / 22100
    Similarly, there are 12 (12C1) ways of getting one of the good cards and 780 (40C2) ways of getting two of the bad cards, leading to 9360 ways to get one good card...etc etc.

    Getting at least one is (as pointed out above) 100% minus getting none, or 55.29%


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