Money Game Puzzle I was recently asked a question.
If you were on a game show and there there is a prize of 1 million dollars. You are faced off against another player. The rules are simple the first player can pick any number say $1\le x\le999999$  then the second player can play any number $1\le y\le2x$  then the first player can play any number $1\le z\le2y$, and so forth. Your plays are added up after each play and the person to play so that the final sum is 1 million wins. You get to play first. Which number should you play to win every time.
Eg.
Player 1 can play any value in the range $1-999999$ || let's say he/she plays $333333$
Player 2 can now play any value in the range $1-666666$ || let's say he/she plays $666666$ now the total is $999999$
Player 1 can now play any value in the range $1-1333332$ || now player one would win because no matter what he plays he would be the first to 1 million
The closest I've gotten to solving this problem was a number smaller than $333333$.
 A: Thoughts, example, but not yet a solution:
We can assume that neither player will give a number that allows an immediate win by the other player unless forced to do so - which would typically mean that the numbers are small, especially at the end where the players are avoiding giving a direct winning move.
Let's try the inverse game, where we are taking away from a pile of $20$ counters. 
If you are faced with a pile of $3$ counters and the previous move was a single take, you have lost. Taking either $1$ or $2$ will allow your opponent to finish. 
Thus if you are faced with a pile of $4$ counters, you can always win by taking a single counter.
If you are faced with a pile of $5$ counters and the previous move was a take of $1$ or $2$, you will lose to best play.
So if you are faced with a pile of $6$ or $7$ counters, you can leave a pile of $5$ and win.
If you are faced with a pile of $8$ counters (and can't win in one move), you should lose.
So if you are faced with a pile of $9$ or $10$ counters, you can leave a pile of $8$ and win. 
If you have a pile of $11$ counters, and taking $3$ is legal, that also wins. However otherwise you will lose. So a pile of $12$ wins by taking a single counter, and a pile of $13$ should lose, which gives a win for piles of $14$ and $15$. Again a pile of $16$, like $11$, can win if a take of $3$ is legal, and not otherwise. So a pile of $17$ takes one for a win, a pile of $18$ loses if $13$ cannot be reached, and piles of $19$ and $20$ should win by reducing to $18$.
This could be a fun game. 
