The beautiful general solution of this problem is as follows. Express the number in every heap in powers of 2, avoiding repetitions and remembering that 2^0 = 1. Then if you so leave the matches to your opponent that there is an even number of every power, you can win. And if at the start you leave the powers even, you can always continue to do so throughout the game. Take, as example, the last grouping given above--12, 11, 7. Expressed in powers of 2 we have-- 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. 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. 398.--THE CIGAR PUZZLE. Not a single member of the club mastered this puzzle, and yet I shall show that it is so simple that the merest child can understand its solution--when it is pointed out to him! The large majority of my friends expressed their entire bewilderment. Many considered that "the theoretical result, in any case, is determined by the relationship between the table and the cigars;" others, regarding it as a problem in the theory of Probabilities, arrived at the conclusion that the chances are slightly in favour of the first or second player, as the case may be. One man took a table and a cigar of particular dimensions, divided the table into equal sections, and proceeded to make the two players fill up these sections so that the second player should win. But why should the first player be so accommodating? At any stage he has only to throw down a cigar obliquely across several of these sections entirely to upset Mr. 2's calculations! We have to assume that each player plays the best possible; not that one accommodates the other. The theories of some other friends would be quite sound if the shape of the cigar were that of a torpedo--perfectly symmetrical and pointed at both ends.