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(a) Note that Alex knows how many blue cubes are left at any point in the process. We let $f(n, a)$ be the largest factor by which Alex can increase his money, provided that there are $n$ cubes left and $a$ of them are blue. We seek to find $f(100, 1).$ It's apparent that $f(n, 0) = 2^n$ and $f(1, 1) = 2$, as for these situations Alex can go all-in every time. Claim. $f(a, b)$ is the harmonic mean of $f(a-1, b)$ and $f(a-1, b-1)$, whenever $0 < b < a.$ Proof. It's clear that the maximum amount of money that Alex can guarantee is $\max_{x \in (-1, 1)} {\min((1-x)f(a-1, b), (1+x)f(a-1, b-1))}.$ To maximize this, Alex should select $x$ so that $(1-x) f(a-1, b) = (1+x) f(a-1, b-1),$ and then we'd have $f(a, b) = (1-x)f(a-1, b) = (1+x) f(a-1, b-1).$ But then we have that $\frac{f(a, b)}{f(a-1, b)} + \frac{f(a, b)}{f(a-1, b-1)} = 2$, implying the claim. $\blacksquare$ From this and induction, we can deduce that $f(n, 1) = \frac{2^n}{n}$ for all $n \in \mathbb{Z}_{\ge 0},$ and the answer is $f(100, 1) = \frac{2^{100}}{100} = \boxed{\frac{2^{98}}{25}}.$ (b) This part is solved similarly. Let $t(n, a)$ denote the maximum factor by which Alex can increase his money, provided that there are $n$ cubes left and at least $a$ of them are blue. We seek to find $t(100, 2).$ In this case, we have that $t(n, 0) = 1$ because Alex has no information (so he has no way of guaranteeing even an increase in money because he could lose every time). It's also clear that $t(1, 1) = 2$ and $t(2, 2) = 4$ because Alex knows all cubes are blue. From here, we can finish in the same fashion as in part (a). Claim. $t(n, a)$ is the harmonic mean of $t(n-1, a-1)$ and $t(n-1, a)$, whenever $0 < a < n.$ Proof. This claim is proven in the same way as the claim in part (a). $\blacksquare$ This allows us to prove by induction that $t(n, 1) = \frac{2^n}{2^n-1}$ for all $n \in \mathbb{N}$ and $t(n, 2) = \frac{2^n}{2^n - n - 1}$ for all $n \in \mathbb{N}_{> 1}.$ Our answer is simply $\boxed{\frac{2^{100}}{2^{100} - 101}}.$ $\square$