- one out of nine chance of rain being in Seattle

They say it rains about 10% of the time, and "A base rate of 10% corresponds to prior odds of 1:9." I found this notation a bit confusing, but I think it corresponds to a one-in-ten chance of rain. For each rainy day, you get nine sunny days.

The difference between probabilities expressed as a value between 0 and 1 and these "odds" if that's what 1:9 is called is likely part of my confusion in following the article.

It was actually Professor Brian who demonstrated how useful a simulation can be, while analyzing this bizarre problem:

- Say you know a family has two children, and further that at least one of them is a girl named Florida. What is the probability that they have two girls?

But a simulator, based on a random number generator, seems like a good way to check our work. I still trust my numbers as long as I feel like I know what I am doing. Say we start with a convenient number of days:

`270 days`

243 sunny

`27 rainy`

Each friend will be expected to give the same ratio of answers. We call Albert first and he responds:

`True "no" on 2/3 of 243 sunny days = 162 days`

False "yes" on 1/3 of 243 sunny days = 81 days

True "yes" on 2/3 of 27 rainy days = 18 days

`False "no" on 1/3 of 27 rainy days = 9 days`

So Albert says "yes" on 99 days, and on 81 days it is sunny and on 18 days it is raining. We increase our expectation of rain from 10% to 18/99, about 18%.

We expect the same ratio of responses from Betty:

`True "no" on 2/3 of 81 sunny days = 54 days`

False "yes" on 1/3 of 81 sunny days = 27 days

True "yes" on 2/3 of 18 rainy days = 12 days

`False "no" on 1/3 of 18 rainy days = 6 days`

So Betty says "yes" on 39 days, and on 27 days it is sunny and on 12 days it is raining. We increase our expectation of rain from 18/99 to 12/39, about 31%.

We expect the same ratio of responses from Charlie:

`True "no" on 2/3 of 27 sunny days = 18 days`

False "yes" on 1/3 of 27 sunny days = 9 days

True "yes" on 2/3 of 12 rainy days = 8 days

`False "no" on 1/3 of 12 rainy days = 4 days`

So Charlie says "yes" on 17 days, and on 9 days it is sunny and on 8 days it is raining. We increase our expectation of rain from 12/39 to 8/17, about 47%.

This seems very clear and agrees with the given answer. But I still haven't used Bayes' theorem.

Using odds instead of percentage makes this problem simple.

Our estimate of rain in Seattle is 1:9 (one rainy day for every nine sunny days).

Our friends tell us the truth with odds of 2:1 (two truthful reports for each false report).

So after calling one friend and hearing "yes it's raining" we multiply 1:9 by 2:1 and get odds of 2:9 as our new expectation of rain. That's 2/11, about 18%, so we are more confident of rain.

The second friend says "yes" and we multiply 2:9 by another 2:1 to get 4:9, or 4/13 which is about 31%.

The third friend says "yes" and we multiply 4:9 by 2:1 to get 8:9 which is 8/17, the final answer of about 47%.

I have found a way to force Bayes in, but don't know if it's correct. Replacing the A's and B's with more descriptive terms, Bayes' Theorem is

odds of rain, given a "yes" = odds of "yes," given rain × odds of rain / odds of "yes"

So, after hearing the first "yes" we have

odds of rain, given a "yes" = (2:1 the odds we will hear a "yes" on a rainy day) × (1:9 our prior estimate for rain) / (*1*, our certainty of hearing "yes" because we just talked with Albert and he definitely said "yes")

This gives us 2:9 divided by 1, which is the desired 2/11 result. That denominator seems forced and flaky, though, and I am not sure what to do if Albert says "no." Perhaps the odds of rain given a "no" are 1:2 × 1:9 / *1* again because we definitely heard "no" for a total of 1:18 or about 5% chance of rain.

Odds are good b_b will be able to straighten this out.