even or odd card game

Question

There are cards numbered $1,2,\ldots,2n$ where $n$ is odd. The deck is shuffled. The top two cards are drawn. If the sum is even the even player gets a point and if the sum is odd the odd player gets a point. The next two cards are drawn and if the sum is even the even player gets a point and if the sum is odd the odd player gets a point. The game continues until all the cards are drawn. Which player will get the most points? Now for one draw odd is favored because once a even card is drawn there are more odd cards in the deck so it would appear that odd is favored. Here are my results. (Possibly wrong because the math is complicated) $$ \begin{array}{|l|l l|} 6\text{ cards} & \text{even wins }\dfrac{3}{5} & \text{odd wins }\dfrac{2}{5}\\\hline 10\text{ cards} & \text{even wins }\dfrac{67}{315} & \text{odd wins }\dfrac{248}{315}\\\hline 14\text{ cards} & \text{even wins }\dfrac{597}{1001} & \text{odd wins }\dfrac{404}{1001}\\ \end{array} $$
Each of these results was a surprise to me. Is there a simple explanation. What happens as $n$ gets large?

Answer 1

Counting the binary strings with exactly $n$ ones and $n$ zeroes by pairing the digits and counting the orderings of pairs by looking at the number of $11$-pairs, we find the identity $$ \sum_{j=0}^{\lfloor n/2\rfloor}\binom{n}{j}\binom{n-j}{j}2^{n-2j}=\binom{2n}{n} $$ The summands can be interpreted as follows: If $j$ is the number of $11$-pairs in a feasible string, that string also must contain $j$ $00$-pairs and the remaining $n-2j$ pairs must be either $01$ or $10$. We can choose $j$ pairs to be $11$ in $\binom{n}{j}$ ways. From the remaining $n-j$ pairs, we must choose $j$ to be $00$ in $\binom{n-j}{j}$ ways, and for each of the remaining $n-2j$ pairs, we can decide them to be either $01$ or $10$.

For each case that we count this way, player even gets $2j$ points and player odd gets $n-2j$ points. To determine the winner, we have to compare them and we see that player odd wins for $0 \le j < n/4$ and player even wins for $n/4 < j \le n/2$. To compare these, you have to compare "the lower" and "the upper" part of the sum. We see that we have a tie for $j=n/4$ which can only happen if $4\mid n$, so in the other cases, both parts together build the whole sum, so their sum is $\binom{2n}{n}$.

At least I can say that for $n=4k$ and $k \rightarrow\infty$, the probability for "ties" approaches $\sqrt{\frac{2}{\pi k}}$, so it vanishes.