This game is based on the concept of minimizing resource wastage.

The Game

  1. Cards which do not have numbers are not used e.g. ace, king, queen, joker. For the remaining cards, regardless of the color and shape, the value of a card is equal to the number on it. The cards are shuffled and distributed between the two players. They get equal number of cards.
  2. In each round, both player pick a card a place it upside down on the table without showing the opponent to complete their moves. After this they show their cards to each other.
  3. The player with the bigger number on the card win the round. The points won by the round winner for this round is equal to the number on card played by the losing players card in this round. E.g if $A$ plays $6$ and $B$ plays $1$ in this round then $A$ wins the round and gets $1$ point.
  4. If the number on both cards are equal than the round is a draw and the game continues until someone wins the round or all the cards are used. The round winner get the points equal to the number on the last card played by the round loser.
  5. After a round is completed, the played cards are discarded and the game continues with the unused cards remaining with each player until all the cards are used
  6. At the end, the player with the greater sum of points win the game.

Minimizing wastage: Clearly we want to win a round by using the smallest card. E.g. if $B$ plays $1$ then $A$ will win the round and get $1$ point if he/she puts any card from $2$ to $10$. But playing a $10$ to win $1$ point is a waste as $10$ can be played to win against bigger numbers later in the match. However before showing, neither player knows what card was played by the opponent.

One may argue that if a player uses up the big cards in the beginning to make small win, it may leave him/her weaker in the later rounds. But using bigger cards in the beginning, leaves the player with smaller cards which leaves the opponent less scope to win more points.

Question: Is there a mathematically optimal strategy that maximizes the chances of winning?

Note: One of the reason why poker is considered a sport as opposed to gambling is because is proven to be a game of strategy and chance and not chance alone.

  • $\begingroup$ If I understand the rules right, we have a great risk to lose the point if we choose a low card. Additionally, an overall draw is still possible, or do I miss something ? And could the game not theoretically end $0:0$ ? Confusing is : We continue until someone wins, but there are only finite many cards for each player. $\endgroup$
    – Peter
    Jun 28 '20 at 13:42
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    $\begingroup$ @Peter Over all draw is still possible if both players play the cards of the same value every time so that the game ends with a score of $0-0$ or if they play a normal game winning or losing rounds and end up with the same total score at the end. $\endgroup$ Jun 28 '20 at 13:45
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    $\begingroup$ Obviously there is no strategy to win the match; if there were, your opponent could play it too. $\endgroup$
    – TonyK
    Jun 28 '20 at 14:15
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    $\begingroup$ There is an optimal strategy, which maximises your chance of winning. But if your opponent plays it too, your chance will be exactly $\frac12$. I suppose it's just a question of what exactly you mean by a "strategy to win the match". $\endgroup$
    – TonyK
    Jun 28 '20 at 14:24
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    $\begingroup$ This is similar to Weighted War - Game of Mind and Probability. $\endgroup$
    – Vepir
    Sep 17 '20 at 9:30

This answer models the game as a simultaneous game, i.e. one where the game is played in one turn where players announce the orders of playing their cards. This may or may not be what you want.

Assuming the players seek to maximize their end-of-game score, there is no optimal pure strategy, since player one can always play the "right-shift" strategy to counter any strategy played by player two. That is, the strategy to completely maximize payoff for player two is to play a $1$ whenever player one plays a $10$, play a $2$ whenever player one plays a $1$, etc.

There are definitely mixed-strategy equilibria, i.e. one where players randomize between the orders they choose. It is clear that there is an equilibrium whenever both players completely randomize the order in which you play the cards (and the expected end result is a draw), as no player has the incentive to resort to another strategy (the expected end result will still be a draw). From this we can conclude that the expected result in all equilibria will be a draw; otherwise, one player will lose in expectation and that player will have the incentive to completely randomize. There may also be other mixed-strategy equilibria, but they must have same result in expectation. Note that they may be difficult to identify, requiring an ad-hoc method to find all of them.

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    $\begingroup$ But this is a very different game! In the OP's game, your strategy will depend on which cards remain in the deck, and on how many points you have accrued. $\endgroup$
    – TonyK
    Jun 28 '20 at 14:32
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    $\begingroup$ @TonyK It's a completely valid criticism. I wanted to post this answer as there may be ways of treating the sequential version similarly to the simultaneous version as there are quite a few similarities in optimal strategies. For example, it does strongly seem like complete randomization is optimal: there is really nothing advantageous one player can do if the other player every turn completely randomizes the card she plays from her hand. $\endgroup$
    – paulinho
    Jun 28 '20 at 14:53
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    $\begingroup$ Thinking about it a bit more, I realise that what you say makes perfect sense: if your opponent plays randomly, you can gain nothing by varying the order of your cards, because the final outcome of the game is independent of the order of the rounds. $\endgroup$
    – TonyK
    Jun 28 '20 at 15:50

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