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I am creating a card game AI for an iPhone game and I'm stumped trying to come up with this statistic.

The deck has 4 suits. The cards are numbered 1-14 in each suit and there is one Joker. So, the deck has 57 cards. One player will be dealt 18 cards and this person will also get to decide which suit the Joker is (so there are technically 15 cards of that suit) and calls that suit trump.

What is the average number of trump cards that this person will have? Can you help me understand how I would figure this out? I took a few stats classes in college, and it's all gone!

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  • $\begingroup$ Do you assume that the player always declares the Joker to be of the suit for which she has the most cards? Do you know that such a policy is optimal for your game? $\endgroup$
    – cardinal
    Sep 12, 2011 at 23:10
  • $\begingroup$ Yes. You can assume that the suit with the most cards will be named trump. $\endgroup$
    – dontangg
    Sep 12, 2011 at 23:49

1 Answer 1

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Assuming that the 18-card player always chooses her longest suit to be trumps, then the calculations are possible but lengthy.

There are 1250 different possible patterns of suits for 18 cards. For example one might be: 2 cards from suit A, 7 cards from suit B, 0 cards from suit C, and 9 cards from suit D, so the longest suit has 9 cards. The number of ways of producing this pattern is

$${14 \choose 2}{14 \choose 7}{14 \choose 0}{14 \choose 9} = 91 \times 3432 \times 1 \times 2002 = 625248624$$

(or $18!$ times as many if order matters, but we will ignore that here). Add together all the other ways of having 9 as the number of the longest and you get 3,570,677,566,456 ways. Do the same for the other possible maxima (which you should be able to see are from 5 to 14) and add up all the possible ways to get ${56 \choose 18}$, the number of ways of choosing 18 cards from 56. Divide this into the number of ways for each possible maximum length and you get the following probabilities (rounded at the last decimal place)

Max Probability
5   0.168508745
6   0.461233813
7   0.267882939
8   0.083137716
9   0.016816801
10  0.002225768
11  0.000184999
12  0.000008993
13  0.000000224
14  0.000000002

This is enough to work out the expected number in the longest suit, namely about $6.326$.

But that was for 18 cards. Add the joker and the expected number in the longest suit becomes about $7.326$.

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    $\begingroup$ This is great! Thanks a ton. However, I think that your logic at the end there is assuming that this person will always get the joker. It's just another card in the deck that he may or may not get. Your explanation was good enough though, that I think I can figure it out from here! $\endgroup$
    – dontangg
    Sep 13, 2011 at 2:47
  • $\begingroup$ @dontangg: There are 57 cards so each player gets 19 cards. One of these is the joker, so one player gets 18 other cards plus the joker, and so she chooses trumps and extends the number of trumps she has by one. $\endgroup$
    – Henry
    Sep 13, 2011 at 6:57
  • $\begingroup$ Actually, in this game, 3 players get 13 cards and one person gets 18. I know... It's weird. $\endgroup$
    – dontangg
    Sep 13, 2011 at 13:06

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