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Two players, Player A goes first, Player B goes second, are playing Concentration game. Every cards face down, we have 2 green cards, 2 blue cards and 1 red card. On player's turn, she selects one card and turn it over, then select another card and turn it over. If they match, then she keeps those cards (called a set) and gets another turn. If they don't match, she turns the cards back over and another player goes.

My question is: If you are playing this game, is it better to go first or second by finding the following probabilities:

  1. Player A (first player) gets two sets and wins
  2. Player B (second player) gets two sets and wins
  3. Each player gets one set, then they are tie

For this first question, I have 5!/2!2!1! = 30 ways for player A to pick up the order of the cards, and there are two ways to get the sets consecutively: BBGGR, GGBBR, so P(A>=2) = 2/30

But I'm not sure how to solve for 2 and 3.

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  • $\begingroup$ You are writing this as if you are ordering me to do something. I don't like that. $\endgroup$
    – Gerry Myerson
    Commented Mar 14, 2019 at 1:58
  • $\begingroup$ I'm voting to close this question as off-topic because I don't like being ordered around $\endgroup$
    – Gerry Myerson
    Commented Mar 14, 2019 at 1:58
  • $\begingroup$ @GerryMyerson No one here is ordering you to do something, I'm kindly asking. If you don't like it just move on. $\endgroup$
    – Neung Chung
    Commented Mar 14, 2019 at 2:06
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    $\begingroup$ What have you tried? Where did you have difficulties? (By the way, this is not a site for us to do your homework for you.) $\endgroup$
    – David G. Stork
    Commented Mar 14, 2019 at 2:10
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    $\begingroup$ Are you taking into account the possibilities for strategic play? $\endgroup$
    – saulspatz
    Commented Mar 14, 2019 at 2:24

1 Answer 1

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Make a tree (by continuing this one):

enter image description here

Then count end conditions to compute probabilities.

Assume perfect play: i.e., that if a player knows the locations of some cards, he'll only turn them over if they complete a pair.

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  • $\begingroup$ Is there a better way to calculate as formula rather than drawing a tree? what if there are 10 pairs and 1 single cards? $\endgroup$
    – Neung Chung
    Commented Mar 14, 2019 at 2:33
  • $\begingroup$ I view this as a computational (not mathematical) problem, much like the Travelling Salesman Problem. I'd write a simple program (equivalent to creating my tree, above), and automatically count leaves and edges appropriately. The benefit is that it can generalize to cases where there are 37 blue, 92 red, 15 green, 7 orange, and three players. (Try that mathematically!) $\endgroup$
    – David G. Stork
    Commented Mar 14, 2019 at 4:11
  • $\begingroup$ (1) I don't think there is a formula, or at least, you'd have to derive that formula from drawing a tree. However, the tree can be considerably simplified due to symmetries, e.g. the first turn being GG and BB are equivalent situations. (2) See the article cited by @saulpatz ... There are situations in the general Concentration game where the best strategy (i.e. "perfect play") is to turn over a known card even if it achieves nothing but skipping half of your turn (or even doing this twice to skip your entire turn). It might not happen in this RGGBB version, but that has to be proven. $\endgroup$
    – antkam
    Commented Mar 14, 2019 at 4:40

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