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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 $333333$.

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    Do you mean the player who chooses a number, which takes the inclusive sum of all plays thus far to be greater than $1,000,000$ wins? – 2017-02-26
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    yes. Sorry I should have made that more clear – 2017-02-26
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    Also, if player A chooses n, then player B can play a number k, with $1\leq k\leq 2^{n}$ - is that correct? – 2017-02-26
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    B can play 1<=k<=2xn. – 2017-02-26
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    I have an answer, will write it up after eating – 2017-02-26
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    See [this link](http://puzzling.stackexchange.com/questions/24591/pop-the-last-bubble) on the puzzling community. – 2017-02-26

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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.

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    an excellent observation I still can find a solution either but I think your on to somthing – 2017-02-26
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    Meh, I'm still working on this, I have the first player winning now :-) – 2017-02-26