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So if there are two players playing a tile game where there are two sets of matching tiles $a_1, a_2, a_3$ and $b_1, b_2, b_3$, what would the optimal strategy be to maximize winning probability? Go first or second?

A turn consists of flipping over three of the cards to see if they match. If going first, then prob of winning is $1/6$? By going second, I can win $5/6$ of the time since I know that what he flipped over must be 2 from one set and 1 from another. Thus, flip any of the remaining three?

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Player $1$ wins on the first turn with probability $\frac 1{10}$, needing to match the first tile, probability $\frac 25$, then again, probability $\frac 14$. Otherwise, player $2$ wins if his first tile matches the pair of the three that player $1$ flipped, probability $\frac 13$, or if he first flips a pair of player $1$'s singleton, probability $\frac 13$, for a total $\frac 23\cdot \frac 9{10}=\frac 3{5}$. The other $\frac 3{10}$ player $1$ wins on the second turn. So if you are player $1$, don't flip any tiles.

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  • $\begingroup$ I disagree that player 1's probability is $1/6$ on their first turn. How many ways can you pick three tiles from six? $\endgroup$ – Empy2 Sep 16 '13 at 14:11
  • $\begingroup$ @Michael: I had it mashed badly. Fixed. Thanks. $\endgroup$ – Ross Millikan Sep 16 '13 at 14:22
  • $\begingroup$ What if player 1 only turns over two tiles? $\endgroup$ – Empy2 Sep 16 '13 at 14:35
  • $\begingroup$ I read the rule requiring player $1$ to flip three (except for my 'flip' last sentence). If he is allowed to flip only two, that makes two new strategies available-flip two and only flip a third if the first two match or flip exactly two. Is he also allowed to flip only one? Each of these can be calculated the same way, but there are more branches to the tree. $\endgroup$ – Ross Millikan Sep 16 '13 at 15:36

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