Gigantic Polynomial Game Harder Version

Euler has a winning strategy, since he has the chance to fill in the last gap. When there is 1 gap left, Euler writes down minus the sum of all other coefficients . This will let the equation $f(x)=0$ have a solution of $x=1$ .

Hence, the probability that Fermat wins is close to 0%.

Euler will definitely win this. Besides the method mentioned by Tran, Euler can play this way:

First, Euler places $0$ in the last term, which is the constant.

Next, Fermat can put anything he wants, anywhere he wants. Euler will counter it this way:

If Fermat places a real number $a$ for a term $x^n$ where $n$ is even, Euler will put a real number $-2a$ in the term $x^{n-1}$

Eg. Fermat places $3$ beside $x^{2016}$ , Euler places $-6$ beside $x^{2015}$

If Fermat places a real number $b$ in a term $x^m$ where $m$ is odd, Euler will put a real number $-\dfrac{b}{2}$ in the term $x^{m+1}$

Eg. Fermat places $8$ beside $x^{2013}$ , Euler places $-4$ beside $x^{2014}$

It doesn't matter what Fermat places. It can be positive, negative, natural, rational or irrational, Euler will counter Fermat's decision with the two methods mentioned above.

The resulting polynomial will be in the form of $f(x) = ax^{2016} - 2ax^{2015} + bx^{2014} - 2bx^{2013} + \ldots + qx^2 - 2qx$ which will have two roots: $x=0$ and $x=2$

Example:

$f(x) = 3x^{2016} - 6x^{2015} -4x^{2014} + 8x^{2013} + \dfrac{3}{4}x^{2012} - \dfrac{3}{2}x^{2011} + \ldots + \sqrt{5}x^2 - 2\sqrt{5}x\\ =3x^{2015}(x-2) -4x^{2013}(x-2) + \dfrac{3}{4}x^{2011}(x - 2) + \ldots + \sqrt{5}x(x-2)$

Even if Fermat puts $0$ for all of his turns, Euler will also do the same, which results in the polynomial $f(x) = 0$ , which has infinitely many solutions.

Therefore, assuming that Euler and Fermat plays optimally, the probability of Fermat winning is $\boxed{0\%}$

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Note: Euler can actually decide to create a polynomial which has a root of $x=3,x=4$ or just about any real root that he wants. We can generalize the steps like this:

Say Euler wants the polynomial to have a real root of $p$

If Fermat places a real number $a$ for a term $x^n$ where $n$ is even, Euler will put a real number $-pa$ in the term $x^{n-1}$

If Fermat places a real number $b$ in a term $x^m$ where $m$ is odd, Euler will put a real number $-\dfrac{b}{p}$ in the term $x^{m+1}$

$\underline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}$

Note note: Euler can also tweak the strategy by placing $0$ as the coefficient of $x^{2016}$ instead. This is what he has to do after that:

If Fermat places a real number $a$ for a term $x^n$ where $n$ is odd , Euler will put a real number $-pa$ in the term $x^{n-1}$

If Fermat places a real number $b$ in a term $x^m$ where $m$ is even , Euler will put a real number $-\dfrac{b}{p}$ in the term $x^{m+1}$