= Solution
Let the win <probabilities> in games 1,2,3 be $u,v,w$. <Independence> makes a prescribed result pattern the product of its three win/loss <probabilities>. Taking a complement for event (a), <inclusion-exclusion> for overlapping pairs in (b) and (c), and disjoint exact-two patterns in (d) and (e), gives
$$
\begin{array}{c|l}
\text{event}&\text{probability}\\\hline
(a)&1-(1-u)(1-v)(1-w)\\
(b)&uv+vw+uw-2uvw\\
(c)&uv+vw-uvw\\
(d)&uv(1-w)+(1-u)vw\\
(e)&uv(1-w)+(1-u)vw+u(1-v)w
\end{array}
$$
For (b), a three-win outcome is counted three times in the pair sum and must be counted once, so subtract twice its <probability>. For (c), only the overlapping pairs 12 and 23 are allowed, so subtract the three-win intersection once. For (d), the two allowed exact patterns are 110 and 011. For (e), add 101 as well.
Substituting $(u,v,w)=(q,p,q)$ and $(p,q,p)$ gives the complete comparison:
$$
\boxed{\begin{array}{c|c|c|c}
& DMD&MDM&\text{maximizing order}\\\hline
(a)&1-(1-q)^2(1-p)&1-(1-p)^2(1-q)&MDM\\
(b)&2pq+q^2-2pq^2&2pq+p^2-2p^2q&MDM\\
(c)&pq(2-q)&pq(2-p)&DMD\\
(d)&2pq(1-q)&2pq(1-p)&DMD\\
(e)&2pq+q^2-3pq^2&2pq+p^2-3p^2q&MDM
\end{array}}
$$
For (a) and (b), the differences $P_{MDM}-P_{DMD}$ are respectively $(p-q)(1-p)(1-q)>0$ and $(p-q)(p+q-2pq)>0$. For (c) and (d), the differences $P_{DMD}-P_{MDM}$ are $pq(p-q)>0$ and $2pq(p-q)>0$. For (e), $P_{MDM}-P_{DMD}=(p-q)(p+q-3pq)>0$ by the additional hypothesis. Thus all maximizing orders follow with strict inequalities.
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