3 Card Mindgame

Rellated to my idea of a card-based RPG, and general thoughts on what direction I want it to take, I’ve been doing a lot of pondering considering these types of games.

Consider the following case:  both players have a 1, a 2, and a 3 card.  Simultaneously, they choose a card and play it. Both cards are discarded, and whoever had the higher card gets a point.  On a tie, neither player gets a point.  After three rounds, whoever has the highest score wins.

Before reading any comments, take five minutes or so to consider what strategy you might take in this game.  What are games between two players of equal familiarity of the game likely to look like?  What common game is this effectively equivalent to?  What changes if each player has five cards, numbered one through five, and the game lasts five rounds?

(I promise that I am going somewhere cool with this, but I can’t get there without YOUR help.)

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8 thoughts on “3 Card Mindgame

  1. josiah says:

    If you’re thinking it’s like paper-rock-scissors, it’s not, because in paper-rock-scissors there’s no mechanism to prevent using symbols which you’ve already used.

    Outcomes: If you choose the same card as your opponent, the final result is ALWAYS a tie. If you choose a different card, you will either not be able to win or not be able to lose, but in either case you can still tie.

    0-1
    Loss 1-2, 2-0
    Tie 1-0, 2-2
    0-2
    Win 1-0, 2-1
    Tie 1-1, 2-0
    1-2
    Loss 0-1, 2-0
    Tie 0-0, 2-1

    So, for any card you pick on first draw you have a 1/3 chance of winning, tying, or losing.
    If that doesn’t force a tie, you have a 1/2 chance of winning or losing.

    Best strategy = random.

    The only way to win at a better than average percentage is if your opponent used a strategy which wasn’t random and you were able to figure it out.

  2. josiah says:

    At higher numbers of cards, the odds of “not tying” increases.

    2 cards you always tie (4/4).
    3 cards 2/3 (24/36) tie.
    4 cards 7/12 (336/576) tie.

    The denominator is just (n!)^2. The numerator is a little trickier and I guess I’ll probably have to calculate 5 cards to find a pattern:

    2^2
    2^3 * 3
    2^4 * 3 * 7

    looks like a pattern is forming, but who knows.

    Note: I think random is still the best strategy with higher number of cards.

  3. josiah says:

    hmm, i wonder if 5 cards is (8064/14400) that being 2^5 * 3 * 7 * 12. oh, man, i don’t want to find out the long way.

  4. josiah says:

    so i guess the numerator is just multiplying itself by (n-2)*(n+3) every time. damn i spent way too much time on that.

    but if it’s true it means my guess was right.

  5. josiah says:

    of course i used my guess to calculate the formula so likely neither is correct. just doing 5 cards by hand got boring quickly.

  6. Willow says:

    The 3 card game is almost equivalent to one game of RPS- except, as you noted, the chance of tying is higher than 1/3.

    The five card game is (roughly) equivalent to ‘best two out of three’ RPS.

    You get a cookie.

  7. Rahvin says:

    I’ve seen some material equate cards to dice in such a way that cards can essentially replace dice, but grant players an opportunity to select their own fate.

    In your “n Card Mindgame”, for example, a higher card can be used to “rescue the princess” if that’s important to the player. If I don’t particularly care about the princess, I’ll use a low card and keep the high card until I need to “slay a dragon” or some such thing.

    Essentially, any area where a traditional game would use dice could be converted to a card system and, wola! instant priority system.

    It also creates really neat situations such as governing card play between players, or the idea of special abilities allowing you to combine cards, or dealing with suites, or any other damn thing you can think of.

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