## The cost of errors in assessing probabilities

What is the cost of an error in assessing a probability? This is a very vague question, and clearly the answer is very context-dependent. To set up the general case, there is an unknown true probability $$p_{true}$$ but we only have an estimated probability $$p_{est}$$. To make sense of a "cost" we must need to take a decision (based on our $$p_{est}$$) with quantitative consequences.

We will envisage the setting where the probabilities are not close to 0 or 1; that is, not the setting of very unlikely events with very large consequences if they happen. To enable us to do a little mathematics we envisage the error $$p_{est} - p_{true}$$ as being small.

The insight here is that, if we make a story that fits this setting, we will usually find that the cost of the error scales as $$(p_{est} - p_{true})^2$$ rather than as $$|p_{est} - p_{true}|$$ . In other words, small errors are rather less costly than one might think. The mathematics is outlined below, for two stories. A conceptual point is a contrast with a familiar fact from freshman Statistics. The accuracy of an opinion poll (or other sampling exercise) involving $$N$$ samples is (under ideal conditions) expressed by saying the error scales as $$N^{-1/2}$$. This is relevant in an election context where we are mainly interested in whether a population percentage is greater than 50%. But in most sampling contexts we are not focussed on determining whether or not the percentage exceed a given threshold. If instead we intend to make some decision, based on a sample percentage, then stories like these suggest that the cost of sampling estimation error typically scales as $$N^{-1}$$ rather than $$N^{-1/2}$$.

### 1. A simple decision setting

Suppose you are planning a wedding, several months ahead, and need to choose now between an outdoor venue A, which you would prefer if you knew it would not be raining, or an indoor venue B, which you would prefer if you knew it would be raining. How to choose? Invoking utility theory (and ignoring the difficulty of actually assigning utilities), there are 4 utilities

• (choose A): utility $$=a$$ if no rain, utility $$=b$$ if rain
• (choose B): utility $$=c$$ if no rain, utility $$=d$$ if rain
where $$a>c$$ and $$d>b$$. Now we calculate the expectation of the utility in terms of the probability $$p$$ of rain:
• (choose A): expected utility $$= pb + (1-p)a$$
• (choose B): expected utility $$= pd + (1-p)c$$
There is a critical value $$p_{crit}$$ where these mean payoffs are equal, and this is the solution of $$\frac{p_{crit}}{1-p_{crit}} = \frac{a-c}{d-b}$$. Note that $$p_{crit}$$ depends only on the utilities. If we knew $$p$$ our best strategy would be $\mbox{ do action A if $$p < p_{crit}$$, do action B if $$p > p_{crit}$$}.$ Instead all we have is our guess $$p_{est}$$, so we use this strategy but based on $$p_{est}$$ instead of $$p$$.

What is the cost of not knowing $$p$$? If $$p_{est}$$ and $$p$$ are on the same side of $$p_{crit}$$ then we take the optimal action and there is zero cost; if they are on opposite sides we take the sub-optimal action and the cost is $\mbox{ |p - p_{crit} | z where z = a- b - c + d > 0.}$ So what happens over many repeated different decision problems of this type? Assume the different utilities are all of order $$1$$ and are independent (over problems) of the probabilities, and hence $$p_{crit}$$ is independent of $$p$$ and $$p_{est}$$. Then the proportion of times that $$p_{crit}$$ happens to be in the interval between $$p$$ and $$p_{est}$$ should be of order $$| p - p_{est} |$$, assuming the latter is small; and when this occurs the mean cost is also of order $$| p - p_{est} |$$.

Combining these two factors, in this particular "decision under uncertainty" context the cost of errors is indeed of order $$( p - p_{est} )^2$$.

### 2. Setting the odds

Imagine you offer to take bets on a future event, on either side at fair odds based on your own estimated probability $$p_{est}$$. If a gambler is more accurate than you at estimating probabilities then you're going to lose money in the long run. How much? To study this let us assume the gambler in fact knows the true probability $$p$$ (a similar argument works if we just assume the gambler is more accurate than you). For each unit bet, your expected loss is proportional to the difference $$| p_{est} - p|$$. But by the Kelly criterion the gambler will make bets of size proportional to this difference $$| p_{est} - p|$$. Again combining these two factors, in this particular setting our loss from our error in estimating probabilities will again scale as order $$( p - p_{est} )^2$$.

Our analysis suggests an interesting, albeit rather abstracted, strategy for bookmakers. One can regard their business as first taking a percentage commission and then offering odds corresponding to probabilities adding up to one. What if they offered those odds based on a probability slightly different from their true assessment? This has (according to our story above) only a second order cost in payouts. But if it increases the amount of bets received, there is a first-order effect from increased commission earned. For a monopoly bookmaker, a small change in odds would likely cause only a small change in amount bet, but competition with other bookmakers could cause a substantial change in amount bet.

The same principle holds when (as often happens) much more money is bet on the favorite than on the underdog. Offering worse odds on the favorite would decrease the amount bet and, even if your offered odds are slightly wrong, this strategy will lower your profit in the long run.

### Footnotes

We are imagining the classical setting of linear utility, so you are not risk-averse. In that case if payoffs are random we can just take their expectations.

Some papers loosely relevant to "setting the odds" are Levitt (2004) and Green-Lee-Rothschild (2019).