Consider $2009$ cards, each having one gold side and one black side, lying on parallel on a long table. Initially all cards show their gold sides. Two player, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of $50$ consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins.
(a) Does the game necessarily end?
(b) Does there exist a winning strategy for the starting player?
Sol – a) Take the standard binary interpretation with gold=$1$ and black=$0$. Note that the total must decrease with each step, and thus must terminate.
b) No. In fact, the starting player loses no matter what moves are made. To see this, let $a_i=1$ if the $i$th card from the left is gold, and $0$ otherwise, and let $S = a_1+a_{51}+\dots+a_{2001}$, and remark that the game can only terminate if $S=0$. But on each move, $S$ changes by $1$, so during the second player's turn, $S$ will always be odd and thus never $0$.
Why game can only terminate if $S=0$ ?
If $2001$th card is $1$ and all previous are $0$'s, game can still end ?
thankyou
Best Answer
The given solution is incorrect, for the reason you mentioned; if $a_{2001}=1$ and all other cards are zero, then $S=1$, yet the game is over. The correct solution is to let $T=a_{1960}+a_{1910}+a_{1860}+\dots+a_{10}$, and apply the same logic to $T$. Indeed, if $T>0$, then any nonzero term of $T$ gives a legal move.
They made two mistakes. First of all, they confused left and right, and meant to say that $a_i$ is the $i^{th}$ card from the right end. Second, they seemed to be thinking that the rules were as follows: you can select any card which is gold, and then flip it and the $49$ cards to its right to the other side. If there are fewer then $49$ such cards, then you just flip all that you can. With these two changes, their value of $S$ works.