Reasonable method for evaluation of bets

  • A
  • Thread starter tom.stoer
  • Start date
  • #1
tom.stoer
Science Advisor
5,766
161
A couple of friends ##n = 1 \ldots N## of mine bet on the result of the next general election in Germany. We select the six most important political parties ##p = 1 \ldots 6##. For each party ##p## and each friend ##n## we have the forecast ##x_{pn}## and the official election result ##x_p##.

Now we calculate

$$D_n = \sum_p w(x_p) \, d(x_{pn} - x_p)$$

with a weight-function ##w## and a deviation-function ##d##. The winner is the guy with smallest ##D_p##.

My question is, what are the most reasonable functions?

It seems natural to set

$$d(x) = x^2$$
$$w(x) = 1$$

which corresponds to the Euclidean distance with equal weight for each party.

But of course other choices are conceivable, e.g.

$$w(x) = x^c$$

weighting bigger parties more than smaller parties.

Are there any reasonable arguments and choices for the weight-function ##w## and the deviation-function ##d##?
 
Last edited:

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,643
What you are predicting is essentially the expected percentage of votes. I would rank contestants according to the multinomial distribution. The prediction that gave the largest probability for the end result wins.

Edit: Since the number of votes will be large, you will probably want to use a lot of approximations in the computations of the factorials ...
 
  • Like
Likes mfb
  • #3
34,673
10,809
Predicting a larger party with 1% accuracy is harder than predicting a smaller party with that accuracy (as extreme case, imagine you would have included all parties, even those that don't manage to get 0.1%).

You could interpret the party results as probabilities (that a randomly picked person will vote for them) and then give a penalty based on how much additional information would have been needed to get the prediction right (in the sense of Bayesian updates). As an example, predicting 2% instead of 1% would be roughly as bad as predicting 60% instead of 30%. Hmm... not sure if that is a good approach.

I would suggest giving a slightly smaller weight to larger parties. 1/sqrt(x_p) or something similar.
 
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,643
Actually, since the number of votes is large, you may want to approximate the multinomial distribution with a multivariate Gaussian with the appropriate covariance matrix and expectation values. Just beware of the fact that the covariance is singular (you should add a party called ”other” so that the sum of all votes goes to 100%, the singularity stems from the outcomes being constrained to the surface of 100% votes).
 
  • #5
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,643
Predicting a larger party with 1% accuracy is harder than predicting a smaller party with that accuracy (as extreme case, imagine you would have included all parties, even those that don't manage to get 0.1%).
You mean predicting a larger party within 1 pp accuracy? Parties around 50% of the votes will be easier to predict with relative accuracy although more difficult in absolute terms. Consider the special case of only two parties, one very big and the other very small. In this case they have the same absolute error, but the smaller party will have a much larger relative spread.

In the multinomial distribution, the variance in outcome i is ##V_i = np_i(1-p_i)## (the same as in the binomial). The relative accuracy should therefore be ##\sqrt{V_i}/(np_i) = (1-p_i)/\sqrt{np_i}##. This clearly decreases as ##p_i## grows. However, the absolute accuracy ##\sqrt{V_i}## would take its maximal value at ##p_i = 0.5##.
 
  • #6
tom.stoer
Science Advisor
5,766
161
I am sorry, but I don't get it.

Can you do me a favor and translate your notation into mine for the functions ##w## and ##d##?
 
  • #7
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,643
No, it does not have that form. You may be able to put it on a relatively similar form if you use the multivariate gaussian approximation and use the log of the likelihood instead of the likelihood itself.

The point is that the multinomial distribution is what you would have if you had ##n## votes where each vote has a probability ##p_i## of being for party ##i##. Of course, ##n## is a very large number in the case of the German elections.
 
  • #8
tom.stoer
Science Advisor
5,766
161
seems to be not very pragmatic; my idea was to put in in an excel sheet, not to write a paper
 
  • #9
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,643
I don’t see any problem with putting it into an Excel sheet. For practical purposes, you may want to use the Dirichlet distribution instead with the appropriate scaling of parameters (and using the log likelihood to avoid numerical issues with extremely small numbers).
 
  • #10
34,673
10,809
(you should add a party called ”other” so that the sum of all votes goes to 100%, the singularity stems from the outcomes being constrained to the surface of 100% votes).
That "other" party exists anyway, as the 6 big parties won't get all votes. They get all seats (as you need 5% of the votes to get seats) but that is a different point.
You mean predicting a larger party within 1 pp accuracy? Parties around 50% of the votes will be easier to predict with relative accuracy although more difficult in absolute terms. Consider the special case of only two parties, one very big and the other very small. In this case they have the same absolute error, but the smaller party will have a much larger relative spread.
I meant the absolute difference. Above 50% that would get easier as well but no party will get more than 50%, even 40% looks unlikely.
 
  • #11
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,643
That "other" party exists anyway, as the 6 big parties won't get all votes.
Yes, my point was that the relevant multinomial distribution is a distribution with 7 bins, not one with 6 bins.
 

Related Threads on Reasonable method for evaluation of bets

Replies
2
Views
1K
  • Last Post
Replies
3
Views
337
Replies
3
Views
967
Replies
2
Views
1K
Replies
4
Views
615
Replies
3
Views
1K
Replies
0
Views
820
  • Last Post
Replies
3
Views
2K
Replies
7
Views
235
  • Last Post
Replies
5
Views
2K
Top