Two players play a game of chess. When they get home, they use a computer to analyze the game.
For each position of the game, the computer knows what move is best, and it assigns a winning probability to that move $P(c)$. The computer is also able to calculate a winning probability to the move that the player had made in that position $P(p)$. These probabilities are in the range [0.0-1.0].
So, for each position $x$ of the game we can calculate a delta, which represents a drop of winning probability.
$\delta(x) = P(c) - P(p)$
If the player played the best move, then this delta will be $0$. A value of $0.1$ or higher would mean that it is a terrible move. ($10\%$ drop of the winning probability). In the extremely rare case, that the human player surpassed the computer and played something brilliant, the delta could be negative.
Next if we make a sum for the delta's for all moves, we get a sum-of-lost-opportunities. In other words, a low sum will indicate a high level of play, and a high sum will indicate a low level of play.
Unfortunately we cannot use this number to compare the level of play of different games, because not all games have the same number of turns. So, we should divide it by the number of turns.
$Level = \dfrac{\sum \delta}{\# moves}$
The next step, (which is outside the scope of this question) is to scale this Level to an ELO rating (which is an international system for ratings of chess players).
But already I have some doubt about all the above.
One of the things that I have doubts about is the subtracting of the winning probabilities. I think the following is actually not entirely correct.
$\delta(x) = P(c) - P(p)$
It raises the question: What is worst? (a drop of $1$ to $0.95$) vs (a drop of $0.5$ to $0.45$) My gut feeling tells me that the latter is worst.
Could somebody explain to me how I could calculate the delta in a non-linear way?