Request for an intuitive explanation of why $\sum_{i=1}^{n}p_i\log P_i$, where $\sum p_i=\sum P_i=1$, is maximised by setting $p_i=P_i$ for all i. I'm designing a multiple choice quiz wherein, rather than picking one answer to a question, each participant has 100 percentage points of credence to dish out between the answers. An example question might be: Who won the men's 100m at the 2016 Olympics, A: Tyson Gay, B: Linford Christie, or C: Usain Bolt? and a participant might answer A: 20%, B: 0%, C: 80%, indicating that they believe there's a 20% chance it was Gay and an 80% chance it was Bolt.
I want a scoring system that incentivises responses that genuinely reflect a participant's degree of credence in each answer. Simply giving a participant the number of percentage points they allocated to the correct answer as a score won't do, since under that system my expected score is $p_{A}P_A+p_BP_B+p_CP_C$ (where $p_i$ is my genuine credence in answer i, and $P_i$ the number of points I allocate it), a function which is maximised by allocating 100 points to whichever answer I'm most confident in.
I've seen other people using logarithmic scoring systems for similar purposes, and sure enough, I was able to prove that $\sum_{i=1}^{n}p_i\log P_i$, where $\sum p_i=\sum P_i=1$, is maximised by setting $p_i=P_i$ for all i. That is, the optimal strategy under logarithmic scoring is to allocate each answer your subjective probability of it being correct.
The thing is, I know I'm going to be called on to explain my scoring system to a bunch of numerate non-mathematicians, and I haven't managed to reach an intuitive understanding of it. It's pretty clear to me why $\sum_{i=1}^{n}p_iP_i$ is maximised by sticking as much weight as possible on the highest $p_i$ - shifting any of that weight to a smaller $p_i$ gets you less bang for your buck. I can see that the same loss of bang in the case of $\sum_{i=1}^{n}p_ilogP_i$ might be made up for by increased logarithmic buck, but I can't for the life of me see an intuitive reason why it peaks it at $p_i=P_i$. A handwavy explanation would be much appreciated. Thanks.
 A: If we take an exponential of your expression (which will not change the maximizer) we get
\begin{aligned}
\exp(\sum_{i=1}^{n}p_i\log P_i) &= \prod_{i=1}^{n} {P_i}^{p_i}.  & (1)\end{aligned}
For simplicity, let's say that the $p_i$ are actually nonnegative integers.  In that case, the above expression has a probabilistic interpretation.  If we have a spinner that has a probability $P_i$ of landing on $i$, and we spin it a fixed number of times, (1) is the probability that we see any particular sequence of outcomes where the spinner landed on $i$ $p_i$ times for each $i$.  For example, if we have $1/3$ chance of landing $1$ and $2/3$ chance of landing $2$, the probability that we see the sequence $1,1,2,1,2$ is $\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{2}{3}\cdot\frac{1}{3}\cdot\frac{2}{3}$.
Now suppose that we observe the outcomes $p_i$ but we do not see the spinner and we don't know the probabilities $P_i$.  Then maximizing (1) is the same as choosing the probabilities $P_i$ that would maximize the likelihood that we would observe the outcomes we did given that spinner.  In other words, we give our best guess as to what the spinner would look like by seeing which spinner assigns the highest probability to our observed outcomes.  If we observe the outcomes $(2,1,1,3,2,1)$, it seems reasonable to choose the estimates $P_1 = 3/6$, $P_2 = 2/6$ and $P_3 = 1/6$. And you have shown that this indeed maximizes the likelihood, as we might expect.
This is equivalent to choosing the distribution that minimizes the cross-entropy between the distribution and the observed outcomes.
A: A basic explanation:
Suppose you wanted to "steal form Peter to give to Paul", that is decrease $P_1$ and increase $P_2$ by same amount. How much impact would it have? Observe that how much adding $D$ to $P$ or subtracting $D$ from $P$ impacts $\log P$ is determined by how big $D$ is relative to $P$; that is, we always get the same effect if $D$ is 10% of $P$, no matter what $D$ and $P$ are. So for this "steal form Peter to give to Paul" to have no effect on $p_1\log P_1+ p_2\log P_2$ on net, we must have that $p_1 D$ is the same fraction of $P_1$ as $p_2D$ is of $P_2$, that,  is $(P_1, P_2)$ is proportional to $(p_1, p_2)$. Since we can only be optimally allocating $P_i$s when no such trade is possible, between any two options, we conclude that the $P$s must be proportional to $p$s; since they both sum up to 1, they must be equal.
Of course, all  that we have done is explain why the derivative of $\log x$ is $\frac{1}{x}$, so the balance equation works out to $\frac{p_1}{P_1}=\frac{p_2}{P_2}$. The text above tries to avoid the language of derivatives. If you plan to explain this to people who know about derivatives and their use in optimization, then it is simpler to just write out the derivatives, and say that all $\frac{p_i}{P_i}$ must be equal, concluding, again, that $p_i=P_i$.

Some "higher-brow" things:
Of course, the "deeper" reason this is true is that you are optimizing is (up to the constant equal to $\sum p_i \log p_i$) the cross-entropy, aka the KL divergence, between the distributions given by the $p_i$s and the $P_i$s. This divergence (like all divergences) is minimal (equal to zero) when the distributions are the same, and is positive then they differ (this follows from Jensen's inequality and convexity of the log).
In addition, the quality $\sum p_i \log P_i$ can be conceptually interpreted as follows (there are other interpretations, this is just one option). Imagine repeatedly sampling one of $n$ outcomes, with $i$th outcome having probability $p_j$. Suppose you record the result by writing it down as a code (or message) of length $\log_k P_i$ (here $P_i$ is the total possible number of messages of size $\log_k P_i$ written using $k$ letters). Then the expected length of your message/code is precisely $\sum p_i \log_k P_i$. Then what we want to say is, that if we require the codes to be uniquely interpretable (this means that no code is a prefix of another), then, to get the shortest codes on average, one should record the $i$th outcome by a code whose length is proportional to $\log p_i$.
For example, suppose you have 4 possible outcomes A, B, C,D, with probabilities 1/2, 1/4, 1/8, 1/8 respectively. Using binary, you'd want to code outcome A by 1 symbol (say, 0), the second outcome B by 2 symbols (say, 10) and C and D by three (110 and 111). This means that if I receive the message "1110101100" I can uniquely decode it as outcomes "D, A, B, C, A".
The point here is that a code of length $l$ cuts out $k^{-l}$ part of the space of possible codes (all the ones that have it as a prefix), and I want that fraction to correspond to the probability of seeing the outcome corresponding to that code. Thus $k^{-l_i}$ is proportional to $p_i$, and so $P_i$ is proportional to $p_i$.
