(related to Problem: The Montenegrin Dice Game)
The players should select the pairs $5$ and $9,$ and $13$ and $15,$ if the chances of winning are to be quite equal. There are $216$ different ways in which the three dice may fall. They may add up $5$ in $6$ different ways and $9$ in $25$ different ways, making $31$ chances out of $216$ for the player who selects these numbers. Also, the dice may add up $13$ in $21$ different ways, and $15$ in $10$ different ways, thus giving the other player also $31$ chances in $216.$
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