$Pr(A>B)$, where $A$ and $B$ are sum of Bernoullies Let $X= x_1 + x_2 + \ldots + x_m$, $Y=y_1 + y_2 + y_3 + \ldots + y_n$, and $Y' = y'_1 + y'_2 + \ldots + y'_n$, where 


*

*Each $x_i$ is a Bernoulli variable which takes value $1$ with probability $p_i>0$.

*Each $y_i$ is a Bernoulli variable  which takes value $1$ with  probability $q_i>0$.

*For $i>2$, each $y'_i$  is a Bernoulli variable  which takes value $1$ with  probability $q_i>0$.

*$y'_1$ and $y'_2$ are two Bernoulli variables  which take value $1$ with  probability $(q_1 + q_2)/2$ (Suppose that $q_1 \neq q_2$).

*All the variables are independent.


Furthermore, suppose that $E(X) \geq E(Y)$ and $m>n$. Now, let $A = Pr(Y\geq X)$  and $A' = Pr(Y' \geq X)$. The question is whether $A>A'$ or $A'>A$? 
 A: Both $A>A'$ and $A'>A$ are possible.
Let $S:= Y_3+Y_4+\cdots Y_n$.
Assuming independence, we have
\begin{align}
   \triangle A := A-A' &=
      \mathbb{P}(Y\geq X) - \mathbb{P}(Y'\geq X) \\
   &=
      \sum_a\mathbb{P}(X=a)\big[\mathbb{P}(Y\geq a) - \mathbb{P}(Y'\geq a)\big] \\
   &=
      \sum_a\mathbb{P}(X=a)\Big[\sum_{b\geq a}\mathbb{P}(Y=b)-\sum_{b\geq a}\mathbb{P}(Y'=b)\Big] \\
   &=
      \sum_a\mathbb{P}(X=a)\sum_{b\geq a}\sum_{c=0}^2\mathbb{P}(S=b-c)
         \big[\mathbb{P}(Y_1+Y_2=c) - \mathbb{P}(Y'_1+Y'_2=c)\big] \;.
\end{align}
Let $\bar{q}:=(q_1+q_2)/2$ and $\varepsilon:=|q_1-q_2|/2$.  We can verify that
\begin{align}
   \mathbb{P}(Y_1+Y_2=c) - \mathbb{P}(Y'_1+Y'_2=c) &=
      \begin{cases}
         -\varepsilon^2 &\text{if $c=0$,} \\
         2\varepsilon^2 &\text{if $c=1$,} \\
         -\varepsilon^2 &\text{if $c=2$.}
      \end{cases}
\end{align}
It follows that
\begin{align}
   \triangle A &= \sum_a\mathbb{P}(X=a)\sum_{b\geq a}
      \varepsilon^2\big[
         -\mathbb{P}(S=b) + 2\mathbb{P}(S=b-1) - \mathbb{P}(S=b-2)
      \big] \\
   &=
      \varepsilon^2\sum_a\mathbb{P}(X=a)\Big[
         -\sum_{b\geq a}\mathbb{P}(S=b)
         +2\sum_{b\geq a-1}\mathbb{P}(S=b)
         -\sum_{b\geq a-2}\mathbb{P}(S=b)
      \Big]\\
   &=
      \varepsilon^2\sum_a\mathbb{P}(X=a)\big[
         \mathbb{P}(S=a-1)-\mathbb{P}(S=a-2)
      \big] \;.
\end{align}
Now, let us consider two examples.
Observe that if $n=2$, then $\mathbb{P}(S=0)=1$, and consequently
\begin{align}
   \triangle A &=
      \varepsilon^2\big[\mathbb{P}(X=1)-\mathbb{P}(X=2)\big]
   \tag{$\star$}
\end{align}
Also, note that $\mathbb{E}(X)\geq \mathbb{E}(Y)$ if and only if $p_1+p_2+\cdots+p_m\geq q_1+q_2+\cdots+q_n$.
Example 1 ($A<A'$).
Let $m=4$, $n=2$, $p_1=p_2=p_3=p_4=1/2$, and let $0<q_1<q_2<1$ 
Then, $p_1+p_2+p_3+p_4=2>q_1+q_2$, hence the requirements are satisfied.  From ($\star$) we get
\begin{align}
   \triangle A &=
      \varepsilon^2(1/4)^4\Big[
         \binom{4}{1}-\binom{4}{2}
      \Big] < 0 \;.
\end{align}
Example 2 ($A>A'$).
Let $m=3$, $n=2$, $p_1=1-\delta$, $p_2=p_3=\delta$ where $0<\delta\ll 1$, and assume $q_1,q_2$ are distinct and satisfy $q_1+q_2\leq 1$. Then, $p_1+p_2+p_3=1+\delta>1\geq q_1+q_2$, hence the requirements are satisfied.
From ($\star$) we get
\begin{align}
   \triangle A &=
      \varepsilon^2\Big[
         (1-\delta)^3+2\delta^2(1-\delta) - 2\delta(1-\delta)^2-\delta^3
      \Big] \;.
\end{align}
which is positive (it is $\approx \varepsilon^2$) when $\delta>0$ is sufficiently small.
