maximize $\tbinom{500}{x}\times\tbinom{500}{900-3x}$

2 fair coins A and B, for A, tossing H gets 3 dollars; for B, tossing H gets 1 dollar. Now toss A and B 500 times each, finally we get 900 dollars. Estimate the most likely number of times of H for A and B (you are not necessary to calculate the exact value, just give the best possible bound).

It is equivalent to find the x to maximize:

$$f(x)=\tbinom{500}{x}\times\tbinom{500}{900-3x},$$ here $134\leq x \leq 500.$ Even I use the fact that the optimal $x^*$ meets: $$f(x^*)\geq f(x^*+1)\ and\ f(x^*)\geq f(x^*-1).$$

I still can hardly solve $x^*.$

Solution 1:

Using algebra, the problem is not too bad.

Go from binomial coefficients to the gamma function. $$\binom{500}{x}\times\binom{500}{900-3x}=\frac{\Gamma (501)^2}{\Gamma (x+1)\,\, \Gamma (501-x)\,\, \Gamma (901-3 x) \,\,\Gamma (3x-399)}$$ which implies that you want to minimize $$g(x)=\Gamma (x+1)\,\, \Gamma (501-x)\,\, \Gamma (901-3 x) \,\,\Gamma (3x-399)$$ or, much better, its logarithm. Using derivative $$\frac{g'(x)}{g(x)}=3 H_{3 x-400}-H_{500-x}-3 H_{900-3 x}+H_x$$ and you want it to be equal to $0$. This is a nice and smooth function and you look for the solution between $134$ and $299$. Make an expansion to $O(x^2)$ around the midpoint and solve for $x$. You will obtain $x=220.0138$ while the "exact" solution given by Newton method is $220.0128$.

I think that we have the good approximation. Just try three points $(219,220,221)$ and you will see which is the one giving the maximum value of the initial expression.