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So our nine year old son comes home from 3rd grade and tells us an amazing thing happened in school today. He was playing a math game with his friend and they got the same score two times in a row!

Here was the game:

Deck of 32 cards, each card is blank on front side and has a fraction $\frac{1}{N}$ on the backside, i.e. $\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}..., \frac{1}{32}\}$. No fraction is repeated. Two players, cards are shuffled, deck is divided into half for each player. It's like War. In every round each player plays his top card (random). Whichever player has a bigger fraction takes both cards and sets them aside. After 16 rounds players count the cards they won. Whoever has more cards wins the game.

Our nine year old thought it was amazing that he and his partner both had 16 cards at the end of the game.

  • The question is, using nothing fancier than algebra, what are the odds that both players end up with the same number of cards after one game (16 hands).

  • The more interesting question is how to explain the answer to a 3rd grader!

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  • $\begingroup$ A short, extremely informal explanation: Two randoms are still a random , that is, the probability of getting a particular card on the top of your deck is still equal for each card. Since every outcome is equally probable and different, the probability of winning a particular play is still $1/2$. So there are $\binom{16}{n}$ equally probable ways to have $n$ wins. Can you work from there? $\endgroup$ – chubakueno Apr 17 '14 at 6:30
  • $\begingroup$ Actually, the hardest part is the second one. $\endgroup$ – Tunk-Fey Apr 17 '14 at 6:59
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The fractions don't matter, only that the cards are ordered in strength and there are no ties. Each player will end with $16$ cards if they each win $8$ rounds. As the rounds are independent, the chance of this is ${16 \choose 8}\frac 1{2^{16}}=\frac {16!}{8!8!2^{16}} \approx 19.64\%$ You could see Wikipedia on the central binomial coefficient and this Alpha calculation of the result.

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    $\begingroup$ This is correct, but not very accessible. Unfortunately, I cannot see any way around it without introducing some equivalently complex computations. $\endgroup$ – M.B. Apr 17 '14 at 17:38
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Use the fact(where events X and Y are defined suitably) that P(X < Y) = 1/2 = P(X > Y) (since here, X=Y is not a possibility, the fractions being distinct).

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