Minimising maximum probabilty over a subset of independent events Something is off with my intuition regarding a relatively simple (high school level) probability question. I'll speak out the problem as a [somewhat contrived] example. Where is my mistake?

You have a week off and really like boat trips. However, you can only afford to go on boat trips on three of your seven days. You get seasick easily, so -- given that you know the probability of bad weather on each day -- what is the minimum probability of experiencing bad weather, while maximising your boat-time?
Let $D = \{ \textrm{Monday}, \textrm{Tuesday}, \textrm{Wednesday}, \textrm{Thursday}, \textrm{Friday}, \textrm{Saturday}, \textrm{Sunday}\}$ and $p_d$ be the probability of bad weather on day $d\in D$; modelled as independent events.
Method 1: The probability of good weather over some subset $D' \subseteq D$ is given by the "and rule":
$$\prod_{d\in D'}(1-p_d)$$
Therefore, the minimum probability of experiencing bad weather over $n$ (not necessarily consecutive) days would be:
$$1 - \max_{D'\in D_n}\left( \prod_{d\in D'}\left(1-p_d\right) \right)$$
...where $D_n$ is the set of subsets of $D$ with cardinality $n\in\{1,\ldots,7\}$. In our example, $n=3$.
Method 2: Using the same notation, the maximum probability of bad weather over some $D'$ would be:
$$\max_{d\in D'} p_d$$
Therefore, intuitively, the minimum probability of experiencing bad weather over some $n$ days would be:
$$\min_{D'\in D_n} \left( \max_{d\in D'} p_d \right)$$
Problem: In general, I assert that:
$$1 - \max_{D'\in D_n}\left( \prod_{d\in D'}\left(1-p_d\right) \right) \neq \min_{D'\in D_n} \left( \max_{d\in D'} p_d \right)$$
What is the mistake in the intuitive method?
 A: Set $A=1 - {\displaystyle \max_{D'\in D_n}}\left( {\displaystyle \prod_{d\in D'}}\left(1-p_d\right) \right)$. You called $A$ "the minimum probability of experiencing bad weather over $n$ (not necessarily consecutive) days", but "over" is a little ambiguous here.
To be precise, $A$ is the minimum probability (across all choices of $n$ days) of experiencing at least some bad weather. This includes bad weather every day, bad weather on just one day, and everything in between.

the maximum probability of bad weather over some $D′$ would be: ${\displaystyle \max_{d\in D'}}\,p_d$

I think this is the main part of where you go wrong. The quantity ${\displaystyle \max_{d\in D'}}\,p_d$ is just the highest chance of bad weather in any particular day in the set $D'$. But  that's not the chance of having bad weather at all during the span $D'$.
"The chance of having bad weather at all during the span $D'$" would be a sum over the cases of "bad weather on one day and good weather all other days", "bad weather on precisely two days",... But it's easiest to exploit the fact that "good weather every day" is the only case left out, so we have the probability: $1 - {\displaystyle \prod_{d\in D'}}\left(1-p_d\right)$.
Note that ${\displaystyle \min_{D'\in D_n}} \left(1 - {\displaystyle \prod_{d\in D'}}\left(1-p_d\right)\right)=1+{\displaystyle \min_{D'\in D_n}} \left(- {\displaystyle \prod_{d\in D'}}\left(1-p_d\right)\right)=1-{\displaystyle \max_{D'\in D_n}} \left({\displaystyle \prod_{d\in D'}}\left(1-p_d\right)\right)=A\checkmark$
