Assessing which player "wins" a series of predictions A group of friends makes predictions about the outcome of the elections and we would like to assess who's predictions are closer to the final "outcome". For example, Here is a table of predictions for 4 parties (A, B, C, D) and the final "outcome" of the elections:
| Party  | Andrew | Bob  | Claire | Dominic | Eva  | Outcome |
|--------|--------|------|--------|---------|------|---------|
| A      | 44     | 41   | 44     | 45      | 43   | 44      |
| B      | 36     | 41   | 43     | 40      | 42   | 41      |
| C      | 7      | 8.5  | 5      | 7       | 8    | 7       |
| D      | 8      | 3    | 5      | 5       | 4    | 3       |
| Others | 5      | 6.5  | 3      | 3       | 3    | 5       |
| Sum    | 100%   | 100% | 100%   | 100%    | 100% | 100%    |

It is tempting to use the formula for standard deviation:
$$
\sigma = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \mu)^2} {n}}
$$
to calculate each player's deviation from the final "outcome". For example, Eva's standard deviation is:
$$
\sigma_E = \sqrt{\frac{(43-44)^2+(42-41)^2+(7-7)^2+(4-3)^2+(3-5)^2} {5}} = 1.26
$$
which is the lowest of the lot. However, this is not a correct use of the standard deviation formula since we are not measuring the st. dev. of a sample with a given mean. Furthermore, the sum of predictions for the 4 parties + Other always sums up to 100 so the "free" predictions are really 4, not 5. My questions are:


*

*Is this calculation valid?

*Is there a better way to formulate the problem, without getting into parametric tests?

 A: The quantity you have specified is not a standard deviation, because $\mu$ is not the mean prediction.
It's called a root-mean-square deviation, and it's a common measure of accuracy.
However, it's not necessarily a suitable measure of accuracy in this case.
1) the variance of sample proportions is larger near $\frac{1}{2}$ than near the ends -- in some sense it's "easier" to get within a few percent of a very small proportion than one near $\frac{1}{2}$, so you might want to weight accuracy in different regions according to how hard it would be to predict proportions in that region.
2) in terms of predicting outcomes of elections, the relationship between the proportions and the outcomes may be relatively complex -- and depend on the particular kind of election -- a vote is handled very differently when voting for an American president (where accuracy on anything but the top two matter not at all) compared to voting for a StackExchange moderator election, the Irish house of representatives or the Australian Senate (where accuracy in the "small party" votes may be much more important).
A: You could use a "proper score function", which was intended just for your purpose!  You can read about proper score functions here: Metrics to asses the ability of a model to predict a probability  or at  https://en.wikipedia.org/wiki/Scoring_rule
