Would shorter block times reduce MEV?

Assumption: blocks are twice as frequent and have half the blockspace each (so blockspace per time remains constant)

Does this reduce MEV per time? (say 2 quicker & smaller vs 1 regular block)

The thought I had:

As a simplification let's assume there are n transactions in a small & quick block and the miner can decide their order (this absolutely simplifies how the mempool and MEV works, maybe this is a critical flaw in my thought?). There are n! possible permutations for how transactions can be ordered combinatorically. In two of the small & quick blocks, miners hence have 2(n!) of options. The regular block holds 2n transactions which allow (2n)! permutations.

As 2(n!) < (2n)!, n>1, the optionality is superadditive, ie bigger, when combining the two smaller & quicker blocks. If MEV is a function of optionality, the smaller blocks would reduce MEV per time.

Any ideas/existing material on this?