On the 24. of September there is a federal election coming up in Germany. A few friends and me are all playing a game, where we try to predict the upcoming results of the election.
As it looks now, there are 6 major parties and one big set of parties that will not get more than $5\%$, which means that they won't get any seats in the parliament. I am now trying to find a system to determine which prediction was the most accurate.
The real result might look like this:
- CDU $40\%$
- SPD $20\%$
- FDP $5\%$
- Grüne $5\%$
- Linke $10\%$
- AFD $10\%$
- Bunch of parties that won't get more than $5\%$ each, remaining percent
where the number is the percentage of votes each party got.
Each person will hand in 6 predictions on a sheet of paper for the share of votes of each big party. Let's say I get $n$ such sheets of paper with predictions written on it. I want the game to be a "winner takes it all" of who was the most accurate. How can I determine who was the most accurate after the election?
I don't know what would be the appropriate tags here, so feel free to edit them
You need to define a distance function on results and then let the "closest" one win. Depending on the function, your winner might differ and you might even be able to define a function to let whoever you want win. Thus, you might consider telling your function to everybody before guesses are handed in. However, note that this might motivate to adjust the guess to the chosen function...
To give you a starting point, here is an example:
We want to compare two lists of numbers, $X = (x_1,x_2,\ldots, x_6)$ and $Y = (y_1,y_2,\ldots, y_6)$. For that, we might take these as vectors in $\mathbb{R}^6$ and compute their distance, e.g. in the 2-norm we would get $$\sqrt{\sum_{i=1}^6 (x_i-y_i)^2}.$$ If you prefere the 1-norm, this would give $$\sum_{i=1}^6 |x_i-y_i|.$$ There are many different distances on $\mathbb{R}^6$ that could be used, so I would suggest to take a look at their pros and cons (e.g. how do they behave if one guess is completely correct/really far of) and decide on one of them.