= Solution
First take $s_i>0$, the hypothesis required for the printed ratios and their optimization formula to be well-defined. Since $\sum p_i=1$,
$$
\sum_i\frac{p_ix_i}{s_i+x_i}=1-\sum_i\frac{p_is_i}{s_i+x_i}.
$$
The maximizing and minimizing problems therefore have exactly the same solutions.
Let $\tau>0$ and minimize the separable <optimization Lagrangian>
$$
L(x,\tau)=\sum_i\frac{p_is_i}{s_i+x_i}+\tau\left(\sum_i x_i-b\right),\qquad x_i\ge0.
$$
Each summand is convex, and its <derivative> is $-p_is_i/(s_i+x_i)^2+\tau$. Its minimizer is $x_i(\tau)=(\sqrt{p_is_i/\tau}-s_i)^+$. The sum of these coordinates is continuous in $\tau$, decreases from infinity to zero, and is strictly decreasing wherever positive. Thus there is a unique $\tau$ for which the sum is $b$. For every feasible $y$, coordinatewise minimization gives $L(y,\tau)\ge L(x(\tau),\tau)$; both constraint terms vanish. The <Lagrangian sufficiency theorem> proves global optimality, not just a necessary stationary condition.
Put $r_i=p_i/s_i$ in decreasing order. A positive bet occurs exactly when $r_i>\tau$. Hence the positive bets form an initial block. To interpret the maximal index in the question, one may include any subsequent indices tied at $r_i=\tau$: their bets are zero but their marginal-return equality still holds. Take $k=\max\{i:r_i\ge\tau\}$. Then
$$
\boxed{\frac{p_1s_1}{(s_1+x_1)^2}=\cdots=\frac{p_ks_k}{(s_k+x_k)^2}=\tau,\qquad x_i=0\ (i>k).}
$$
For this block, $\sqrt\tau=(\sum_{i\le k}\sqrt{p_is_i})/(b+\sum_{i\le k}s_i)$, with the threshold inequalities determining $k$. Zero-<probability> horses receive zero bets.
Let $H$ be the set attaining $\rho=\max r_i$, and write $S_H=\sum_{i\in H}s_i$, $P_H=\sum_{i\in H}p_i=\rho S_H$. For sufficiently small $b$, the solution uses only $H$, with $x_i=bs_i/S_H$ there and $\tau=\rho[S_H/(S_H+b)]^2$. If the next ratio is $r_*<\rho$, this remains valid when $b\le S_H(\sqrt{\rho/r_*}-1)$; if all outside ratios are zero there is no finite threshold. Put $S=\sum_i s_i$. The fixed pool is $S+b$, so the expected gross payout and net gain are
$$
\boxed{E(\text{payout})=(S+b)\frac{P_Hb}{S_H+b},\qquad E(\text{net gain})=(S+b)\frac{P_Hb}{S_H+b}-b.}
$$
Tied best horses receive bets proportional to their existing stakes; betting all on one tied horse is not generally optimal for finite $b$. This is <pari-mutuel expected-return allocation>.
The literal allowance $s_i=0$ needs qualification. At $s_i=x_i=0$ the printed objective contains $0/0$ and the ordered ratios are undefined. Even defining an unbacked horse's payout to be zero does not always yield an attained optimum. For example, take $b=1$, $p_1=p_2=1/2$, $s_1=0$, $s_2=1$. For $x_1>0$, the normalized expected payout is $1/2+(1/2)(1-x_1)/(2-x_1)$, tending to $3/4$ as $x_1\downarrow0$ but never attaining it; at $x_1=0$ that zero-payout convention gives $1/4$. Thus \b[the proved allocation formula requires positive existing stakes, or an explicitly modified market model]. The zero-stake case cannot be justified by silently dividing by zero.
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