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Thread: Pr (10 and a heart) - Am I overthinking this?

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    Pr (10 and a heart) - Am I overthinking this?




    Hi guys,

    This may be way simplistic for this forum, so I apologise in advance.

    Drawing two cards from a standard 52 card deck. What is the possibility of drawing a 10 and a heart?


    So, the 10 of hearts changes things because they are dependent events.

    So it's not simply 4/52*13/51

    The possible draws involve draw 1 = 10 hearts and event 2 is 15/51 10's or hearts

    draw 1 heart (not 10) 12/52 and draw 2: 4/51 10's

    Draw 1 a 10 = 3/52 (already did 10 hearts) and 13/51 (hearts)

    Simply add these probabilities? Is my logic flawed?

    Seriously, ignore all my babbling, can anyone explain this simply for me.

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    Re: Pr (10 and a heart) - Am I overthinking this?

    The total number of way to draw 2 cards from a deck of 52 is {52}\choose{2} = 1326

    There are 51 winning hands that have a 10 of hearts(a ten of hearts and any other card.)

    There are 3*12 winning hands that do not have the ten of hearts. The other 3 tens * the other 12 hearts.

    \frac{51 + 36}{1326} = \frac{87}{1326} \approx  0.0656
    Last edited by asterisk; 11-26-2013 at 09:46 PM. Reason: corrected combination calculatio

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    Re: Pr (10 and a heart) - Am I overthinking this?

    Thank you for your response, but I am still unsure.

    Wouldn't the number of potential hands be 1/52 x 1/51. We can halve that if we do not care about order?

    Therefore 1326?

    Or must this be calculcated with combinations/permutations stuff.
    Last edited by camm91; 11-26-2013 at 04:22 PM. Reason: stupidity

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    Re: Pr (10 and a heart) - Am I overthinking this?


    You are correct {52}\choose{2} = 1326

    Or 2 * (1/52) * (1/51) = 1/1326

    I must have hit a wrong key

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