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It's akin to saying that if I give you a $(1/2+\epsilon)$ coin-flip opportunity to double-or-nothing any amount of money, obviously the best expected value from that would be to flip as much money as you can. But you wouldn't flip your entire net worth because your expected utility falls off a cliff - you "die" $1/2-\epsilon$ amount of the time and are "twice as wealthy" $1/2+\epsilon$ of the time, but it would have been better to just bet say 5% of your net worth so you could still be in the same general band of wealth even if you lose.
@NateRiver The reason why that should be obvious as well. It is optimal in regards to maximizing your expectation, not maximizing your utility. The utility often involves something like the expectation minus a constant times the standard deviation of your strategy. Setting that constant to zero gives your question, but if this game were about real money that is meaningful to someone, then simply maximizing the expectation would be a losing strategy with respect to maximizing their utility from the game. But the constant chosen for risk is a personal choice, though $0$ is never the best.
@NateRiver It should be obvious why. When $n$ is odd, you get 1 coin which is indifferent to choosing between heads and tails $(p=0.5)$. But when $n$ is even, every coin prefers exactly one of heads and tails $(p \neq 0.5)$.
@Kostya_I it may be ill posed in a technical sense, but you can still minimize loss with respect to a fourier expansion capped to N=5000, for example.. Even though the problem is ill-posed I still believe there could be something to it, if that makes sense.
But there is sometimes a sense in which it can be well-posed, so maybe this can be an area for people to look into solving special cases for. (Or rather that it has been solved for some special cases, is what I'm really interested in) as some special case might still be useful in some contexts.
