Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expected value of 3 cards dealt

  1. Nov 16, 2009 #1
    If each card on a regular deck has points that corresponds to their number (like 2 of hearts is 2 points, 7 of clubs is 7 points), the Jack, Queen, King each being 10 points...what's the expected value of your opponent's hand if you deal them 3 cards?

    I know the empirical expected value...but I'd like to know -how- to get the theoretical expected value, please : )

    Help please D: I've been stewing over this question for days now. The only way I can think of doing this is by doing a tree diagram to get each probability but that'll have like 1000 end branches -headdesk-
  2. jcsd
  3. Nov 16, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    What's an ace worth? 1, 10, 11, 13?

    The expected value wouldn't be too far off from three times the expected value of one card.
  4. Nov 16, 2009 #3


    User Avatar
    Science Advisor
    Homework Helper

    I'll assume the ace is valued at something other than 10.

    So there are 52 * 51 * 50 ways to choose three cards from a deck. Of those, the breakdown is

    216 ways to draw 3 of the same non-10
    10368 ways to draw 2 of the same and one different, none 10
    32256 ways to draw 3 different non-10s
    5184 ways to draw a 10 and 2 of the same non-10s
    55296 ways to draw a 10 and 2 different non-10s
    25920 ways to draw 2 10s and a non-10
    3360 ways to draw 3 10s

    So calculate the average value for each, multiply, add, and divide.
  5. Nov 16, 2009 #4
    Ah, yes, Ace is worth 1 points : )

    Uhm, sorry, would you happen to know the breakdown for 4 cards? I misread the question and apparently it's the expected value of 4 cards dealt. I tried doing the breakdown myself but I always seem a few hundred thousand short of the total ways.

    Thank you very much for all the help!
  6. Nov 17, 2009 #5
    Note that E(X+Y)=E(X)+E(Y) holds regardless of the dependence between X and Y - so you won't need to work out all 13^4 combinations.
  7. Nov 19, 2009 #6
    To expand on bpet's remark, the expected value of one card is 85/13, assuming an ace is 1 and Jack, Queen, King are 10 each. Let's say the value of the ith card is [tex]X_i[/tex]. Then the expected value of 3 cards is

    [tex]E(X_1 + X_2 + X_3) = E(X_1) + E(X_2) + E(X_3) = 3 \times 85/13[/tex].

    That's all it takes.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook