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This is not a solution but some background to the question.

Define $$S(k,p,x) = \sum_{i=0}^k {k \choose i} p^i (1-p)^{k-i} (x_i-p)^2.$$ Define $$f(k) = \mathrm{argmin}_x \max_p S(k,p,x).$$ Then $f(k)$ is the minimax square-loss solution to trying to estimate the win rate of a random process by observing $k$ results (Wald wrote on this). The neat thing is: we can show if there is a real solution $x$ in $[0,1]^{k+1}$ to $S(k,p,x) = x_0^2$ then $x=f(k)$. Meaning we avoided two nasty quantifiers. See this file for some experimental examples. Also, a change of variables $z = p/(1-p)$ makes collecting terms easier.

This is not a solution but some background to the question.

Define $$S(k,p,x) = \sum_{i=0}^k {k \choose i} p^i (1-p)^{k-i} (x_i-p)^2.$$ Define $$f(k) = \mathrm{argmin}_x \max_p S(k,p,x).$$ Then $f(k)$ is the minimax square-loss solution to trying to estimate the win rate of a random process by observing $k$ results (Wald wrote on this). The neat thing is: we can show if there is a real solution $x$ in the interior of $[0,1]^{k+1}$ to $S(k,p,x) = x_0^2$ then $x=f(k)$. Meaning we avoided two nasty quantifiers. See this file for some experimental examples. Also, a change of variables $z = p/(1-p)$ makes collecting terms easier.

This is not a solution but some background to the question.

Define $$S(k,p,x) = \sum_{i=0}^k {k \choose i} p^i (1-p)^{k-i} (x_i-p)^2.$$ Define $$f(k) = \mathrm{argmin}_x \max_p S(k,p,x).$$ Then $f(k)$ is the minimax square-loss solution to trying to estimate the win rate of a random process by observing $k$ results (Wald wrote on this). The neat thing is: we can show if there is a real solution in $[0,1]^{k+1}$ to $S(,)$ then it is $f(k)$. Meaning we avoided two nasty quantifiers. See this file for some experimental examples. Also, a change of variables $z = p/(1-p)$ makes collecting terms easier.

This is not a solution but some background to the question.

Define $$S(k,p,x) = \sum_{i=0}^k {k \choose i} p^i (1-p)^{k-i} (x_i-p)^2.$$ Define $$f(k) = \mathrm{argmin}_x \max_p S(k,p,x).$$ Then $f(k)$ is the minimax square-loss solution to trying to estimate the win rate of a random process by observing $k$ results (Wald wrote on this). The neat thing is: we can show if there is a real solution $x$ in $[0,1]^{k+1}$ to $S(k,p,x) = x_0^2$ then $x=f(k)$. Meaning we avoided two nasty quantifiers. See this file for some experimental examples. Also, a change of variables $z = p/(1-p)$ makes collecting terms easier.

This is not a solution but some background to the question.

Define $$S(k,p,x) = \sum_{i=0}^k {k \choose i} p^i (1-p)^{k-i} (x_i-p)^2.$$ Define $$f(k) = \mathrm{argmin}_x \max(p,S(k,p,x)).$$ Then $f(k)$ is the minimax square-loss solution to trying to estimate the win rate of a random process by observing $k$ results (Wald wrote on this). The neat thing is: we can show if there is a real solution in $[0,1]^{k+1}$ to $S(,)$ then it is $f(k)$. Meaning we avoided two nasty quantifiers. See this file for some experimental examples. Also change of variables $z = p/(1-p)$ makes collecting terms much easier.

This is not a solution but some background to the question.

Define $$S(k,p,x) = \sum_{i=0}^k {k \choose i} p^i (1-p)^{k-i} (x_i-p)^2.$$ Define $$f(k) = \mathrm{argmin}_x \max_p S(k,p,x).$$ Then $f(k)$ is the minimax square-loss solution to trying to estimate the win rate of a random process by observing $k$ results (Wald wrote on this). The neat thing is: we can show if there is a real solution in $[0,1]^{k+1}$ to $S(,)$ then it is $f(k)$. Meaning we avoided two nasty quantifiers. See this file for some experimental examples. Also, a change of variables $z = p/(1-p)$ makes collecting terms easier.