Andrew and Betty play a game in which they lay out a row of three coins, heads up. They take turns, begining with Andrew, turning over one of the coins at a time. They must not produce a pattern of heads and tails which has already occurred earlier in the game. The first person who cannot make a move is the loser.
1. If they each play as well as possible, who is the winner?
2. If the game were played with four coins instead of three, who would be the winner?
3. If the game is played with three coins but the player who cannot make a move is declared the winner, who wins now?
(In reply to
Solution by vswitchs)
vswitch's reasoning would only be correct if the only way to lose was
to repeat the initial position. If that was the game, then by
parity the only person who could lose is the second player.
However, the loser is the person who first repeats any position, so
either player could lose.
For instance, the first player, if he played very poorly, loses on his
second turn if he flips the same coin that the second player turned.
More realistically. here is a four coin win for the second player.
Start: HHHH (binary 0)
Player 1: HHHT (binary 1)
Player 2: HTHT (5)
Player 1: HTTT (7)
Player 2: HTTH (6)
Player 1: HHTH (2)
Player 2: THTH (10)
Player 1: THTT (11)
Player 2: HHTT (3), wins!
Player 1: loses no matter what he does. Only plays
are 1,2,7 and 11, and all are repeats
I don't know yet
which player has a forced win, but I do know that we can't prove it is
automatically player 1 based on a simple parity argument
Edited on September 14, 2006, 2:55 pm
re: Solution -- not so fast!
