I am a bit stuck in computing the expected length of this simple variant of the War card game.
In this variant, the cards are numbered from $[1,\dots,2n]$. After shuffling the cards, each player is given $n$ cards as their initial hand. This game is played in rounds. In each round, the players chose a random card from their hand. They compare the card that they chose. The player with higher card takes both card into their hand. We continue to play until one of the player runs out of card. The question : In expectation, how many rounds are being played until the games end?
Some observations:
- This game is not symmetric, the player who gets the highest card at the beginning of the game will be the winner at the very end.
- If a player holds the $k$ card and all cards higher than $k$, then this $k$ card will not change hand for the rest of the game. This seems to suggests that I should compute the stopping times $T_k$'s equal to the time where the $k$ card is firmly in the winner's hand. But I didn't manage to produce a proof through this route.
- I did a monte carlo simulation of this game. Until $n = 30$, it strongly suggests that the expected value is $n^2$. For me, this is surprising by itself as it is equal to the expected time that a random walk that starts at 0 will hit either $-n$ or $n$. But this shouldn't be as this game is not symmetric.
P.S. : This might be related to the game mentioned in this answer. But I strongly suspect that in their version, the selection of card at each rounds is not random. Which might explain their solution that it is $O(n^2 \log n)$ instead of $n^2$.