# Scores FAQ

Below are Frequently Asked Questions (and answers!) about scores. The general FAQ is here, and the medals FAQ is here.

Contents:

# Scores

## What is a scoring rule?

A scoring rule is a mathematical function which, given a prediction and an outcome, gives a score in the form of a number.

A naive scoring rule could be: "you score equals the probability you gave to the correct outcome". So, for example, if you predict 80% and the question resolves Yes, your score would be 0.8 (and 0.2 if the question resolved No). At first glance this seems like a good scoring rule: forecasters who gave predictions closer to the truth get higher scores.

Unfortunately this scoring rule is not "proper", as we'll see in the next section.

## What is a proper scoring rule?

Proper scoring rules have a very special property: the only way to optimize your score on average is to predict your sincere beliefs.

How do we know that the naive scoring rule from the previous section is not proper? An example should be illuminating: consider the question “Will I roll a 6 on this fair die?”. Since the die is fair, your belief is “1/6” or about 17%. Now consider three possibilities: you could either predict your true belief (17%), predict something more extreme, like 5%, or predict something less extreme, like 30%. Here’s a table of the scores you expect for each possible die roll:

 outcome die roll naive score of p=5% naive score of p=17% naive score of p=30% 1 0.95 0.83 0.7 2 0.95 0.83 0.7 3 0.95 0.83 0.7 4 0.95 0.83 0.7 5 0.95 0.83 0.7 6 0.05 0.17 0.3 average 0.8 0.72 0.63

Which means you get a better score on average if you predict 5% than 17%. In other words, this naive score incentivizes you to predict something other than the true probability. This is very bad!

Proper scoring rules do not have this problem: your score is best when you predict the true probability. The log score, which underpins all Metaculus scores, is a proper score (see What is the log score?). We can compare the scores you get in the previous example:

 outcome die roll log score of p=5% log score of p=17% log score of p=30% 1 -0.05 -0.19 -0.37 2 -0.05 -0.19 -0.37 3 -0.05 -0.19 -0.37 4 -0.05 -0.19 -0.37 5 -0.05 -0.19 -0.37 6 -3 -1.77 -1.2 average -0.54 -0.45 -0.51

With the log score, you do get a higher (better) score if you predict the true probability of 17%.

## What is the log score?

The logarithmic scoring rule, or "log score" for short, is defined as:

$\text{log score} = \ln(P(outcome))$

Where $$\ln$$ is the natural logarithm and $$P(outcome)$$ is the probability predicted for the outcome that actually happened.

Higher scores are better:

• If you predicted 0% on the correct outcome, your score will be $$-\infty$$ (minus infinity).
• If you predict 100% on the correct outcome, your score will be 0.

This means that the log score is always negative. This has proved unintuitive, which is one reason why Metaculus uses the Baseline and Peer scores, which are based on the log score but can be positive.

The log score is proper (see What is a proper scoring rule?). This means that to maximize your score you should predict your true beliefs (see Can I get better scores by predicting extreme values?).

One interesting property of the log score: it is much more punitive of extreme wrong predictions than it is rewarding of extreme right predictions. Consider the scores you get for predicting 99% or 99.9%:

 99% Yes, 1% No 99.9% Yes, 0.1% No Score if outcome = Yes -0.01 -0.001 Score if outcome = No -4.6 -6.9

Going from 99% to 99.9% only gives you a tiny advantage if you are correct (+0.009), but a huge penalty if you are wrong (-2.3). So be careful, and only use extreme probabilities when you're sure they're appropriate!

## What is the log score for continuous questions?

Since the domain of possible outcomes for continuous questions is (drum roll) continuous, any outcome has mathematically 0 chance of happening. Thankfully we can adapt the log score in the form:

$\text{log score} = \ln(\operatorname{pdf}(outcome))$

Where $$\ln$$ is the natural logarithm and $$\operatorname{pdf}(outcome)$$ is the value of the predicted probability density function at the outcome. Note that on Metaculus, all pdfs have a uniform distribution of height 0.01 added to them. This prevents extreme log scores.

This is also a proper scoring rule, and behaves in somewhat similar ways to the log score described above. One difference is that, contrary to probabilities that are always between 0 and 1, $$\operatorname{pdf}$$ values can be greater than 1. This means that the continuous log score can be greater than 0: in theory it has no maximum value, but in practice Metaculus restricts how sharp pdfs can get (see the maximum scores tabulated below).

## What is the Baseline score?

The Baseline score compares a prediction to a fixed "chance" baseline. If it is positive, the prediction was better than chance. If it is negative, it was worse than chance.

That "chance" baseline gives the same probability to all outcomes. For binary questions, this is a prediction of 50%. For continuous questions this is a uniform (flat) distribution.

The Baseline score is derived from the log score, rescaled so that:

• Predicting the same probability on all outcomes gives a score of 0.
• Predicting perfectly on a binary question gives a score of +100.
• The average scores of binary and continuous questions roughly match.

Here are some notable values for the Baseline score:

 Binary questions Continuous questions Best possible Baseline score on Metaculus +99.9 +183 Worst possible Baseline score on Metaculus -897 -230 Median Baseline empirical score +17 +14 Average Baseline empirical score +13 +13

Theoretically, binary scores can be infinitely negative, and continuous scores can be both infinitely positive and infinitely negative. In practice, Metaculus restricts binary predictions to be between 0.1% and 99.9%, and continuous pdfs to be between 0.01 and ~35, leading to the scores above.

The empirical scores are based on all scores observed on all resolved Metaculus questions, as of November 2023.

Note that the above describes the Baseline score at a single point in time. Metaculus scores are time-averaged over the lifetime of the question, see Do all my predictions on a question count toward my score?.

You can expand the section below for more details and maths.

The Baseline scores are rescaled log scores, with the general form:

$\text{Baseline score} = 100 \times \frac{ \operatorname{log\ score}(prediction) - \operatorname{log\ score}(baseline) }{ \text{scale} }$

For binary questions, the $$scale$$ is chosen so that a perfect prediction ($$P(outcome) = 100\%$$) gives a score of +100. The final formula is:

$\text{binary Baseline score} = 100 \times \frac{ \ln(P(outcome)) - \ln(50\%) }{ \ln(2)}$

Note that you can rearrange this formula into: $$100 \times(\log_2(P(outcome)) + 1)$$.

For continuous questions, the $$scale$$ was chosen empirically so that continuous scores have roughly the same average as binary scores. The final formula is:

$\text{continuous Baseline score} = 100 \times \frac{ \ln(\operatorname{pdf}(outcome)) - \ln(baseline) }{ 2 }$

Where $$\ln$$ is the natural logarithm, $$P(outcome)$$ is the probability predicted for the outcome that actually happened, and $$\operatorname{pdf}(outcome)$$ is the value of the predicted probability density function at the outcome.

The continuous $$baseline$$ depends on whether the question has open or closed bounds:

• If both bounds are closed, the $$baseline$$ is 1, corresponding to a uniform distribution in range.
• If one bound is open, the $$baseline$$ is 0.95, corresponding to a uniform distribution in range + 5% probability out of the open bound.
• If both bounds are open, the $$baseline$$ is 0.9, corresponding to a uniform distribution in range + 5% probability out of each open bound.

## What is the Peer score?

The Peer score compares a prediction to all the other predictions made on the same question. If it is positive, the prediction was (on average) better than others. If it is negative it was worse than others.

The Peer score is derived from the log score: it is the average difference between a prediction's log score, and the log scores of all other predictions on that question. Like the Baseline score, the Peer score is multiplied by 100.

One interesting property of the Peer score is that, on any given question, the sum of all participants' Peer scores is always 0. This is because each forecaster's score is their average difference with every other: when you add all the scores, all the differences cancel out and the result is 0. Here's a quick example: imagine a binary question that resolved Yes, with three forecasters having predicted:

 Forecaster log score Peer score Alex $$-1$$ $$\frac{(A-B)+(A-C)}{2} = \frac{(-1-1)+(-1-2)}{2} = -2.5$$ Bailey $$1$$ $$\frac{(B-A)+(B-C)}{2} = \frac{(1-(-1))+(1-2)}{2} = 0.5$$ Cory $$2$$ $$\frac{(C-A)+(C-B)}{2} = \frac{(2-(-1))+(2-1)}{2} = 2$$ sum $$-2.5+0.5+2 = 0$$

Here are some notable values for the Peer score:

 Binary questions Continuous questions Best possible Peer score on Metaculus +996 +408 Worst possible Peer score on Metaculus -996 -408 Median Peer empirical score +2 +3 Average Peer empirical score 0* 0*

*The average Peer score is 0 by definition.

Theoretically, binary scores can be infinitely negative, and continuous scores can be both infinitely positive and infinitely negative. In practice, Metaculus restricts binary predictions to be between 0.1% and 99.9%, and continuous pdfs to be between 0.01 and ~35, leading to the scores above.

The "empirical scores" are based on all scores observed on all resolved Metaculus questions, as of November 2023.

Note that the above describes the Peer score at a single point in time. Metaculus scores are time-averaged over the lifetime of the question, see Do all my predictions on a question count toward my score?.

You can expand the section below for more details and maths.

The Peer scores are built on log scores, with the general form:

$\text{Peer score} = 100 \times \frac{1}{N} \sum_{i = 1}^N \operatorname{log\ score}(p) - \operatorname{log\ score}(p_i)$

Where $$p$$ is the scored prediction, $$N$$ is the number of other predictions and $$p_i$$ is the i-th other prediction.

Note that this can be rearranged into:

$\text{Peer score} = 100 \times (\ln(p) - \ln(\operatorname{GM}(p_i)))$

Where $$\operatorname{GM}(p_i)$$ is the geometric mean of all other predictions.

As before, for binary questions, $$p$$ is the probability given to the correct outcome (Yes or No), and for continuous questions it is the value of the predicted pdf at the outcome.

## Do all my predictions on a question count toward my score?

Yes. Metaculus uses time-averaged scores, so all your predictions count, proportional to how long they were standing. An example goes a long way (we will use the Baseline score for simplicity, but the same logic applies to any score):

A binary question is open 5 days, then closes and resolves Yes. You start predicting on the second day, make these predictions, and get those scores:

 Day 1 Day 2 Day 3 Day 4 Day 5 Average Prediction 40% 70% 80% N/A Baseline score 0 -32 +49 +49 +68 +27

Some things to note:

• Before you predict, your score is considered to be 0 (this is true for all scores based on the log score). This means that if you believe you can do better than 0, you should predict as early as possible.
• You have a score for Day 4, despite not having predicted that day. This is because your predictions stay standing until you update them, so on Day 4 you were scored on your Day 3 prediction. On Day 5 you updated to 80%, so you were scored on that.
• This example uses days, but your Metaculus scores are based on exact timestamped predictions, so a prediction left standing for 1 hour will count for 1/24th of a prediction left standing for a day, etc.

## Can I get better scores by predicting extreme values?

Metaculus uses proper scores (see What is a proper scoring rule?), so you cannot get a better score (on average) by making predictions more extreme than your beliefs. On any question, if you want to maximize your expected score, you should predict exactly what you believe.

Let's walk through a simple example using the Baseline score. Suppose you are considering predicting a binary question. After some thought, you conclude that the question has 80% chance to resolve Yes.

If you predict 80%, you will get a score of +68 if the question resolves Yes, and -132 if it resolves No. Since you think there is an 80% chance it resolves yes, you expect on average a score of

80% × 68 + 20% × -132 = +28

If you predict 90%, you will get a score of +85 if the question resolves Yes, and -232 if it resolves No. Since you think there is an 80% chance it resolves yes, you expect on average a score of

80% × 85 + 20% × -232 = +21

So by predicting a more extreme value, you actually lower the score you expect to get (on average!).

Here are some more values from the same example, tabulated:

 Prediction Score if Yes Score if No Expected score 70% +48 -74 +24 80% +68 -132 +28 90% +85 -232 +21 99% +99 -564 -34

The 99% prediction gets the highest score when the question resolves Yes, but it also gets the lowest score when it resolves No. This is why, on average, the strategy that maximises your score is to predict what you believe. This is one of the reasons why looking at scores on individual questions is not very informative, only aggregate over many questions are interesting!

## Why did I get a small score when I was right?

To make sure incentives are aligned, Metaculus needs to ensure that our scores are proper. We also time-average scores.

This has a counter-intuitive consequence: when a question resolves before its intended close date, the times between resolution and close date need to count in the time-average, with scores of 0. We call this "score truncation".

An example is best: imagine the question "Will a new human land on the Moon before 2030?". It can either resolve Yes before 2030 (because someone landed on the Moon), or it can resolve No in 2030. If we did not truncate scores, you could game this question by predicting close to 100% in the beginning (since it can only resolve positive early), and lower later (since it can only resolve negative at the end).

Another way to think about this is that if a question lasts a year, then each day (or in fact each second) is scored as a separate question. To preserve properness, it is imperative that each day is weighted the same in the final average (or at least that the weights be decided in advance). From this perspective, not doing truncation is equivalent to retroactively giving much more weights to days before the question resolves, which is not proper.

You can read a worked example with maths by expanding the section below.

This example uses the Baseline score, which will be noted $$S$$, but results would be equivalent with any proper score.

Alex wants to predict if they will be fired this year. They have a performance review scheduled this week. They estimate there is a $$20\%$$ chance they fail it, and if so they will be fired on the spot. If they don’t fail this week, there is still a $$5\%$$ chance they will be fired at the end of the year. A proper scoring rule ensures that the best strategy on this question is to predict $$p=(20\%+80\% \times 5\%)=24\%$$ this week, and then $$5\%$$ for the other 51 weeks (if they weren’t fired).

Without truncation

Without truncation, this honest strategy gives Baseline scores of:

• $$S(24\%)≃−106$$ in the $$20\%$$ of cases they are fired this week.
• $$\frac{1}{52}S(24\%) + \frac{51}{52}S(5\%) = -328$$ in the $$80\% \times 5\% = 4\%$$ of cases they are fired at the end of the year.
• $$\frac{1}{52}S(76\%) + \frac{51}{52}S(95\%) = +92$$ in the $$80\% \times 95\% = 76\%$$ of cases they are not fired.

For an average score of $$20\% \times -106 + 4\% \times -327 + 76\% \times +92 = +36$$ in expectation.

But the strategy of “predicting close to 100% in the beginning and lower later”, let’s say 99% today, then 5% the other 6 days, without truncation gives Baseline scores of:

• $$S(99\%) ≃ +99$$ in the $$20\%$$ of cases they are fired this week.
• $$\frac{1}{52}S(99\%) + \frac{51}{52}S(5\%) = -324$$ in the $$80\% \times 5\% = 4\%$$ of cases they are fired at the end of the year.
• $$\frac{1}{52}S(1\%) + \frac{51}{52}S(95\%) = +80$$ in the $$80\% \times 95\% = 76\%$$ of cases they are not fired.

For an average score of $$20\% \times +99 + 4\% \times -324 + 76\% \times +80 = +68$$ in expectation.

Notice that $$+68 > +36$$, so without truncation, the gaming strategy gives you a score almost twice as high in expectation! It is really not proper.

With truncation

With truncation, the honest strategy gives Baseline scores of:

• $$\frac{1}{52}S(24\%) ≃ −2$$ in the $$20\%$$ of cases they are fired this week.
• $$\frac{1}{52}S(24\%) + \frac{51}{52}S(5\%) = -328$$ in the $$80\% \times 5\% = 4\%$$ of cases they are fired at the end of the year.
• $$\frac{1}{52}S(76\%) + \frac{51}{52}S(95\%) = +92$$ in the $$80\% \times 95\% = 76\%$$ of cases they are not fired.

For an average score of $$20\% \times -2 + 4\% \times -327 + 76\% \times +92 = +56$$ in expectation.

While the gaming strategy gives:

• $$\frac{1}{52}S(99\%) ≃ +2$$ in the $$20\%$$ of cases they are fired this week.
• $$\frac{1}{52}S(99\%) + \frac{51}{52}S(5\%) = -324$$ in the $$80\% \times 5\% = 4\%$$ of cases they are fired at the end of the year.
• $$\frac{1}{52}S(1\%) + \frac{51}{52}S(95\%) = +80$$ in the $$80\% \times 95\% = 76\%$$ of cases they are not fired.

For an average score of $$20\% \times +2 + 4\% \times -324 + 76\% \times +80 = +48$$ in expectation.

This time, $$+56 > +48$$, so without truncation, the gaming strategy gives you a worse score than the honest strategy! Which is proper.

## What are the legacy scores?

### What is the Relative score?

The Relative score compares a prediction to the median of all other predictions on the same question. If it is positive, the prediction was (on average) better than the median. If it is negative it was worse than the median.

It is based on the log score, with the formula:

$\text{Relative score} = \log_2(p) - \log_2(m)$

Where $$p$$ is the prediction being scored and $$m$$ is the median of all other predictions on that question.

As of late 2023, the Relative score is in the process of being replaced by the Peer score, but it is still used for many open tournaments.

### What is the coverage?

The Coverage measures for what proportion of a question's lifetime you had a prediction standing.

If you make your first prediction right when the question opens, your coverage will be 100%. If you make your first prediction one second before the question closes, your coverage will be very close to 0%.

The Coverage is used in tournaments, to incentivize early predictions.

### What are Metaculus points?

Metaculus points were used as the main score on Metaculus until late 2023.

You can still find the rankings based on points here.

They are a proper score, based on the log score. They are a mixture of a Baseline-like score, and a Peer-like score, so they reward both beating an impartial baseline, and beating other forecasters.

For full mathematical details, expand the section below.

Your score $$S(T,o)$$ at any given time $$T$$ is the sum of an "absolute" component and a "relative" component:

$S(T,o) = a(N) \times L(p,o) + b(N) \times B(p,o)$

where:

• $$o$$ is the outcome of the question: 1 if the question resolves positive, 0 if it resolves negative.
• $$N$$ is the number of forecasters on the question.
• $$L(p,o)$$ is the log score relative to a 50% prior, defined as:
• $L(p, o) = \begin{cases} \log_2 \left ( \frac{p}{0.5} \right ) & \text{if } o = 1 \\ \log_2 \left ( \frac{1 - p}{0.5} \right ) & \text{if } o = 0 \end{cases}$
• $$2 < B(p,o) < 2$$ is the betting score and represents a bet placed against every other forecaster. It is described under "constant pool scoring" on the Metaculus scoring demo (but with a modification that for computational efficiency, the full distribution of other forecaster predictions is replaced by a fitted beta distribution).
• $$a(N)$$ and $$b(N)$$ depend on $$N$$ only and define how the points scale with the number of forecasters.
• \begin{align*} A(N) &= 45 + 15 \log_2(1 + N/30) \\ B(N) &= 30 \log_2(1 + N/30). \end{align*}

Note that $$B$$, $$N$$, and $$p$$ can all depend on $$T$$ and contribute to the time-dependence of $$S(T, o)$$.

Your final score is given by the integral of $$S(T, o)$$ over $$T$$:

$S = \frac{1}{t_c-t_o} \int_{t_o}^{t_c} S(T, o) \, dT$

where $$t_o$$ and $$t_c$$ are the opening and closing times. (Note that $$S(T) = 0$$ between the opening time and your first prediction, and is also zero after question resolution but before question close, in the case when a question resolves early.)

Before May 2022, there was also a 50% point bonus given at the time the question closes, but it was discontinued and the points multiplied by 1.5 henceforth.

# Tournaments

## What Are Metaculus Tournaments?

Metaculus tournaments are venues for forecasts and analyses organized around a central topic or theme. Tournaments are often collaborations between Metaculus and various nonprofits, government agencies, or other organizations seeking to use forecasting to support effective decision making. Cash prize pools and other rewards are provided for accurate forecasts and other valuable contributions.

To participate in a tournament you must be logged into your Metaculus account. Some tournaments require registration in order to compete.

To begin competing, forecast on Open questions that appear on the tournament page. After at least one question has resolved, a Leaderboard will appear on the tournament page displaying current scores and rankings. A personal score board ("My Score") will also appear, detailing your performance for each question (see How are my tournament Score, Coverage, Take, Prize and Rank calculated?).

After the end of a tournament, the prize pool is divided among forecasters according to their scores. The more accurate you are relative to other participants, the greater proportion of the prize pool you receive.

If you have outstanding tournament winnings, Metaculus is happy to facilitate donations to various non-profits, regranting organizations, and funds. You can find the list of organizations we facilitate payments to here.

## How are my tournament Score, Coverage, Take, Prize and Rank calculated?

Your tournament Score is the sum of your Relative scores over all questions in the tournament. If, on average, you were better than the Community Prediction, then it will be positive, otherwise it will be negative.

Your tournament Coverage is the average of your coverage on each question. If you predicted all questions when they opened, your Coverage will be 100%. If you predicted all questions halfway through, or if you predicted half the questions when they opened, your Coverage will be 50%.

Your tournament Take is the exponential of your Score, times your Coverage: $$\text{Take} = e^\text{Score} \times \text{Coverage}$$.

Your Prize is how much money you earned on that tournament. It is proportional to your take, and is equal to your Take divided by the sum of all competing forecasters' Takes.

Your Rank is simply how high you were in the leaderboard, sorted by Prize.

The higher your Score and Coverage, the higher your Take will be. The higher your Take, the more Prize you'll receive, and the higher your Rank will be.

## What are the Hidden Period and Hidden Coverage Weights?

The Community Prediction is on average much better than most forecasters. This means that you could get decent scores by just copying the Community Prediction at all times. To prevent this, many tournament questions have a significant period of time at the beginning when the Community Prediction is hidden. We call this time the Hidden Period.

To incentivize forecasting during the hidden period, questions sometimes are also set up so that the coverage you accrue during the Hidden Period counts more. For example the Hidden Period could count for 50% of the question coverage, or even 100%. We call this percentage the Hidden Period Coverage Weight.

If the Hidden Period Coverage Weight is 50%, then if you don't forecast during the hidden period your coverage will be at most 50%, regardless of how long the hidden period lasted.