The math of probabilities is not always intuitive, and here is an example. I was looking at this forecast for rainfall in New York City.

I was mainly interested in the estimates for rainfall as I had dinner plans. The forecasts are given in terms of probabilities. In the primary display, which is every three hours, the forecasters say the chance of rain is between 10% and 40%. I also see the probabilities peaking around 3 pm then moderating.

Then, the display includes a daily estimate shown in the row below. In that box, the probability of rainfall was listed at 60%.

Wait a minute, if the individual probabilities are all between 0% and 40%, where did the 60% come from?

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It turns out that when we aggregate probabilities, there are different ways of doing it, and what one should do depends on how one will use the aggregated number.

One explanation of why the daily aggregated probability is 60%, significantly higher than any of the individual 3-hour estimates, is that the forecaster asks this question: what is the chance of any rainfall during [insert time period]?

For any 3-hour period, the forecaster asks: what is the chance of rainfall during these three hours?

For a full day, the question becomes: what is the chance of rainfall during these 24 hours?

The relationship between these two questions is seen when we break down the 24 hours into eight contiguous but non-overlapping three-hour windows. The aggregated question is really asking what is the chance that there is rainfall in *any of* these eight three-hour blocks.

This is an "OR" question, rather than an "AND" question. They are not saying there will be rain in *all of* the eight three-hour blocks. It only takes rain in one or more of those blocks to count as rain during that day!

Therefore, it is not surprising that while the probability of rain is below 40% in a specific eight-hour block, it is 60% across all eight 3-hour blocks. If our intuition says otherwise, it is wrong! In fact, the aggregated number must be higher than the maximum probability in any three-hour block because the aggregated probability is that probability plus the probability that rain falls in other three-hour blocks.

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As I said above, the formula for aggregating the eight time blocks depends on what we plan on using the aggregate forecast to do.

A different way of aggregation is to take the average of the eight individual forecasts. If we did that, the aggregated average forecast of rainfall will surely be below 40% as the eight numbers being averaged range from 0% to 40%.

Imagine someone who is deciding to go to the corner store to buy something during the day. Since this shopping trip happens during a specific three-hour time window, the aggregated daily forecast is actually not useful. If she knows which time window (e.g. 12-3 pm), then the appropriate probability is the disaggregated number.

Let's say she's flexible. Then the number is determined by the store's opening hours. Let's say the store opens at 9am and closes at 6 pm, which spans the first 3 blocks of the three-hour forecasts. The estimates were 30%, 30% and 40%.

If we average these three values, we get 33.3%. If we solve for the probability of rain in any of these three time blocks, then we get a number above 40%.

There's more. We can ask for the minimum of the three values, which is 30%.

In order to pick one of these methods, we need more context around the decision. If she wants to pick the best time to go out, defined as the least chance of being rained on, then she should take the minimum, which is to say go out before 3 pm.

If she can't quite control when she could go out but she is definitely going to the store that day, then taking the average makes sense. When it happens, the trip will take place in one of the three three-hour blocks.

If she will be outside for the entire 9 hours, then she should go with the forecaster's aggregate.

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