| 12 | = | 8 | 4 | - | - |
| 11 | = | 8 | - | 2 | 1 |
| 7 | = | - | 4 | 2 | 1 |
| 2 | 2 | 2 | 2 |
As there are thus two of every power, you must win. Say your opponent takes 7 from the 12 heap. He then leaves—
| 5 | = | - | 4 | - | 1 |
| 11 | = | 8 | - | 2 | 1 |
| 7 | = | - | 4 | 2 | 1 |
| 1 | 2 | 2 | 3 |
Here the powers are not all even in number, but by taking 9 from the 11 heap you immediately restore your winning position, thus—
| 5 | = | - | 4 | - | 1 |
| 2 | = | - | - | 2 | - |
| 7 | = | - | 4 | 2 | 1 |
| - | 2 | 2 | 2 |
And so on to the end. This solution is quite general, and applies to any number of matches and any number of heaps. A correspondent informs me that this puzzle game was first propounded by Mr. W.M.F. Mellor, but when or where it was published I have not been able to ascertain.
[397.—THE MONTENEGRIN DICE GAME.—solution]
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.