Optimality of the uniform distribution for a load balancing problem For some $n ∈ ℕ$, you have $n$ buckets with given capacities $c_1,…,c_n ∈ ℕ$ and given average loads $λ_1,…,λ_n ∈ ℝ_{≥ 0}$. You may choose probabilities $p_1,…,p_n ∈ [0,1]$.
Now, for $i ∈ [n]$ independently, we put a Poisson distributed number $X_i \sim \mathrm{Po}(λ_i)$ of balls into bucket $i$, with each ball independently coloured red with probability $p_i$ and blue otherwise.
We say a bucket $i ∈ [n]$ is overflowing if $X_i > c_i$. The total overflow is $T = \sum_{i ∈ [n]} \max\{0,X_i-c_i\}$.
You lose if


*

*the total overflow is $T = 2$ and

*all balls in overflowing buckets have the same colour.


Question: Is the probability to lose minimised by $p_1 = p_2 = … = p_n = \frac{1}{2}$, regardless of $(λ_i)_{i ∈ [n]}$ and $(c_i)_{i ∈ [n]}$?
Note: If 1. holds, then either one bucket overflows by two balls or two buckets overflow by one ball. We may zoom in on these cases since the colours of the balls (and thus $(p_i)_{i ∈ [n]}$) are irrelevant if 1. does not hold.
PS: We suspect that a positive answer could imply a more difficult (but also more natural) claim: Namely, that the probability of a $k$-uniform hypergraph with $n$ vertices and $m$ hyperedges to be orientable is maximised if the $m$ edges of the hypergraph are generated according to the uniform distribution on $\binom{[n]}{k}$ (as opposed to being generated by any other distribution).
 A: This may be true, and comes down to doing some algebra. I haven't had time to finish up this answer in a clean way, but it is too long for a comment, so I post it here in case it may be useful. I only treat the case $n = 2$, and have only checked some cases numerically.
Let $B_i$ be the event that all the balls in bucket $i$ are the same color, and $B_{12}$ the event that the balls in both buckets are the same color. The event of interest is the union of three disjoint events, $E = E_1 \cup E_2 \cup E_3.$
\begin{align*}
E_1 &= \{ X_1 = c_1+2, X_2 \leq c_2, B_1\} \\
E_2 &= \{X_2 = c_2+2, X_1 \leq c_1, B_2\}\\
E_3 &= \{X_1 = c_1+1, X_2=c_2+1, B_{12}\}.
\end{align*}
The probabilities for these events are easy to write down. For instance,
\begin{align*}
P(E_1) &= P(X_1 = c_1+2, B_1) P(X_2 \leq c_2) \\
&=(p_1^{c_1+2}+(1-p_1)^{c_1+2}) \frac{e^{-\lambda_1} \lambda_1^{c_1+2}}{(c_1+2)!}\sum_{k \leq c_2} \frac{e^{-\lambda_2}\lambda_2^k}{k!},
\end{align*}
and similarly
$$P(E_2) = (p_2^{c_2+2}+(1-p_2)^{c_2+2}) \frac{e^{-\lambda_2} \lambda_2^{c_2+2}}{(c_2+2)!}\sum_{k \leq c_1} \frac{e^{-\lambda_1}\lambda_1^k}{k!}.$$
For $E_3$,
\begin{align*}
P(E_3) &= P(B_{12}|X_1,X_2)P(X_1=c_1+1,X_2=c_2+1)\\
&= (p_1^{c_1+1}p_2^{c_2+1}+(1-p_1)^{c_1+1}(1-p_2)^{c_2+1}) \frac{e^{-\lambda_1}\lambda_1^{c_1+1}}{(c_1+1)!} \frac{e^{-\lambda_1}\lambda_1^{c_2+1}}{(c_2+1)!}.
\end{align*}
Thus we can write
$$P(E) = P(E_1)+P(E_2)+P(E_3) = a_1 f_1(p_1) + a_2 f_2(p_2) + a_3 f_3(p_1,p_2).$$
From here it comes down to taking derivatives w.r.t. $p_1, p_2$ and solving the system of equations. Alternatively, one should be able to reduce it to whether
$$ a_1 f_1(0.5) + a_2 f_2(0.5)  + a_3 f_3(0.5, 0.5) \leq a_1 f(0) + a_2 f(1) + a_3 f_3 (0, 1),$$
where the $a_j$ depend on $c_i, \lambda_i$,
and whether the same inequality holds true for $(p_1,p_2) = (1, 0)$. If it does, then $(1/2, 1/2)$ is the optimal selection of $p_i$'s in this case. 
