Origin of the Game
The word "Fantam" is often used to designate a variation of the game of Nim, but we use it here in a very specific sense for a quite different game, which starts by being very easy, but which can quickly
become quite complicated.
The game is played with any number of tokens of any kind.
1. The first player decides how many tokens will be used.
2. The second player decides who plays first.
3. Each player may pick up either 1 or 2 tokens.
4. The winner is the one who picks up the last token.
A simple little lesson in sets.
Where "n" represents the size of the sets,
being the addition of the smallest and the largest number which can be picked up,
in this case 3 (1 + 2),
all the "losing positions" which one must leave to the opponent
are always mutiples of n (n, 2n, 3n,...).
1. The first player is at a disadvantage and will always lose if both are good players.
2. The second player counts the tokens chosen by the first player :
(a) if it is a multiple of n, he asks the other to begin, but
(b) if not, he begins and takes the required number of tokens to make it so.
It is evidently preferable to count without it showing, and camouflage the strategy.
Teaching children to play the game
1. When presented with only 1 token, the child takes it and wins.
2. When presented with 2 tokens, the child takes them both and wins.
3. When presented with 3 tokens, the child always loses,
(a) if he takes only 1, the other takes 2,
(b) if he takes 2, the other takes only 1.
This is when to ask the other to start.
4. When presented with 4 tokens, the child takes only 1 to leave the other child 3 tokens.
5. When presented with 5 tokens, the child takes 2 to leave the other child 3 tokens.
6. When presented with 6 tokens, the child asks the other to start.
And so on.
Very young children should count in groups of 3 (1, 2 ,3, 1, 2, 3, ...),
if they come out even, they ask the other to start,
if not, they take what is left over (the last number counted).
Each player could pick up 1, 2, or 3 tokens,
in which the value of n for the losing positions would be "4" (1+3).
Or each player could pick up 1, 2, 3, or 4 tokens,
in which the value of n for the losing positions would be "5" (1+4).
The Mathematics and the Strategy remain very much the same.
More Complicated Rules
If each player can pick up 2 or 3 tokens (but not 1),
the value of n for the losing positions would be "5" (2+3), as expected,
there would now be the possibility of a draw when there is only 1 token left,
which neither player can pick up.
When the second player counts the tokens chosen by the first player :
(a) if it is a multiple of 5, he asks the other to begin,
(b) if it is either 2 or 3 more than a multiple of 5, he begins and takes them,
(c) if it is either 1 or 4 more than a multiple of 5, he may either
take 2 to transform the 1 into 4, or take 3 to transform the 4 into 1,
and pass the problem on to the other, hoping a mistake will be made, or
ask the other to start and still hope for a mistake to be made.
With 2 good players, a draw is inevitable.