Betting Business

Let $q_1, q_2, \ldots, q_n$ be the optimal values such that the expected amount of dollars you pay back is minimized. The probability that team $i$ wins is $p_i$, and if they win we have to pay back $B_i \over q_i$ dollars. Thus our total expected payback will be $S = \sum_{i=1}^n \frac{p_iB_i}{q_i}$ under the constraint that $q_1 + …. + q_n = 1$.

We will find a relationship between $q_i$ and $q_j$ for each $i, j$. Consider what happens when we only vary $q_i$ and $q_j$ and fix the rest. All terms of $S$stays constant except $\frac{p_i B_i}{q_i} +\frac{p_j B_j}{q_j}$. SInce $S$ is minimum so is this term. Let $q_i+q_j = k$. Then $\frac{p_i B_i}{x} +\frac{p_j B_j}{k-x}$must be minimum at $x = q_i$. Remember that the first derivative is 0 at minima. Thus differentiating at $x = q_i$,
we have $- \frac{p_i B_i}{x^2} + \frac{p_jB_j}{(k-x)^2} = 0 \implies \frac{x}{k-x} = \frac {\sqrt{p_i B_i}}{\sqrt{ p_jB_j}} \implies \frac{q_i}{q_j} = \frac{\sqrt{p_i, b_i}}{\sqrt{p_j, b_j}}$ for all $i, j$

In other words, $q_i$ is proportional to $\sqrt{p_i b_i}$. Let $a_i = \sqrt{p_i b_i}.$Then it is a matter of simple algebra to show that

$q_i = a_i/\sum{a_i}$ and $S = \left(\sum a_i\right)^2$