There are $25$ coins with values $1,2,\dots,25$. Two persons, $A$ and $B$, play the following game with these coins. Person $A$ chooses a coin and person $B$ decides whether to keep the coin for himself or to hand over the coin to him. Each player with the most coins must choose the next coin and the other player must decide whether to keep the coin or give it to another player. If the number of coins is equal, the previous state is repeated. The game continues in the same way until all the coins are selected. At the end, the player with the most valuable coins wins the game. Which player has a winning strategy?
I think $B$ has the winning strategy, but I can't prove it. I will be grateful if someone helps me to solve this problem.
Let $X(\mathcal{C}, k)$ be the largest difference in scores achievable by the player with the next move when he has $k \ge 0$ more coins than the other player and the set of remaining coins is $\mathcal{C}$. He will choose some $c\in \mathcal{C}$. The other player may decide to let him keep that coin, giving him a score of $c + X(\mathcal{C}\setminus c, k+1)$. Or the other player may take the coin herself, leading to a score for the first player of $-c + X(\mathcal{C}\setminus c, k-1)$ (for $k\ge 1$). The $k=0$ case is special; in that case, if the second player takes the coin, she then has strictly more coins and moves into the choosing role. She will score $c + X(\mathcal{C}\setminus c, 1)$ with optimal play, so the first player's score is $-c-X(\mathcal{C}\setminus c, 1)$. All told, we have that $$ X(\mathcal{C}, k)=\max_{c\in \mathcal{C}}\min\{c+X(\mathcal{C}\setminus c, k+1),-c+X(\mathcal{C}\setminus c, k-1)\}, $$ with boundary conditions $X(\emptyset, k)\equiv 0$ and $X(\mathcal{C}, -1)\equiv -X(\mathcal{C},1)$.
Playing with this analysis for smaller sets of coins (say, $\{1,2,\ldots,N\}$) suggests the following:
If this pattern (checked through $N=18$) holds, then $B$ has the winning strategy for $N=25$.