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.
I will show that the first player should infallibly win, if he always
plays in the best possible manner. Examine carefully the following
diagram, No. 1, and all will be clear.
[Illustration: 1]
[Illustration: 2]
The first player must place his first cigar _on end_ in the exact centre
of the table, as indicated by the little circle. Now, whatever the
second player may do throughout, the first player must always repeat it
in an exactly diametrically opposite position. Thus, if the second
player places a cigar at A, I put one at AA; he places one at B, I put
one at BB; he places one at C, I put one at CC; he places one at D, I
put one at DD; he places one at E, I put one at EE; and so on until no
more cigars can be placed without touching. As the cigars are supposed
to be exactly alike in every respect, it is perfectly clear that for
every move that the second player may choose to make, it is possible
exactly to repeat it on a line drawn through the centre of the table.
The second player can always duplicate the first player's move, no
matter where he may place a cigar, or whether he places it on end or on
its side. As the cigars are all alike in every respect, one will
obviously balance over the edge of the table at precisely the same point
as another. Of course, as each player is supposed to play in the best
possible manner, it becomes a matter of theory. It is no valid objection
to say that in actual practice one would not be sufficiently exact to be
sure of winning. If as the first player you did not win, it would be in
consequence of your _not_ having played the best possible.
The second diagram will serve to show why the first cigar must be placed
on end. (And here I will say that the first cigar that I selected from a
box I was able so to stand on end, and I am allowed to assume that all
the other cigars would do the same.) If the first cigar were placed on
its side, as at F, then the second player could place a cigar as at
G--as near as possible, but not actually touching F. Now, in this
position you cannot repeat his play on the opposite side, because the
two ends of the cigar are not alike. It will be seen that GG, when
placed on the opposite side in the same relation to the centre,
intersects, or lies on top of, F, whereas the cigars are not allowed to
touch. You must therefore put the cigar farther away from the centre,
which would result in your having insufficient room between the centre
and the bottom left-hand corner to repeat everything that the other
player would do between G and the top right-hand corner. Therefore the
result would not be a certain win for the first player.