Here's another nice 'Bayes' type problem. I think it came originally off Facebook (probably originated elsewhere) but I first saw it on the 'MindYourDecisions' YouTube feed from Presh Talwalker I've just anglicised it (being an Anglo-Saxon myself - if you're allowed to admit that these days).

Here's the problem:



I'm thinking of driving up to Manchester (the original in England, UK ). Now there is a saying from that part of the world that goes like this: "If you can see the hills in Manchester it's going to rain - if you can't see them it's already raining". I lived and studied there for five years or so and that's not so far off the mark. Maybe a bit unfair, Met Office records show it rains around 150 days a year (this data will be useful later on).

Anyway, on with the question. I ring three contacts I have in Manchester and ask them if it actually is raining - and they all say 'yes'. The trouble is, I know from experience that they're a bunch of jokers and will only actually tell me the truth 2/3 of the time.

So, the question is, is it actually raining in Manchester?

Let's look at the 2 possible 'truths':

1. It is raining. The probablility in this case (that they're all telling the truth) is:

$${{2} \over {3}} \times {{2} \over {3}} \times {{2} \over {3}} = {{8} \over {27}}$$

2. It's not raining. So what's the probability they're all lying?

$${{1} \over {3}} \times {{1} \over {3}} \times {{1} \over {3}} = {{1} \over {27}}$$

So now we need the probability of each of these probabilities - and for this we need the a-priori probability that it IS actually raining. Now from the Met Office we know this is \({150}\over{365}\) (roughly).

So, the probaility is IS raining is \({150}\over{365}\) and the probability its NOT is \(1 - {{150}\over{365}}\). These figures all get a bit messy so let's say the probaility it IS raining is \(p\) where \({p} = {{150}\over{365}}\) and the probability it's NOR raining is \((1-p)\).

Now we can see that the overall probability of the first case (it Is raining and they all say it is) is \({{8}\over{27}}\times{p}\) and the probability of the second case (it is NOT raining but they all say it is) \({{1}\over{27}}\times(1 - p)\).

So, the probability that it is ACTUALLY Raining is the probability of the first possibility divided by the sum of the probabilities of the first and second possibility. Or:

$$ {{8p}\over{27}}\over{{{8p}\over{27}} + {(1-p)\over{27}}}$$

We can cancel out the 27s and simplify to get:

$${8p}\over{7p+1}$$

Where \(p\) is, of course, \({150}\over{365}\).

If you want to do the maths, this means that the probability that it actually IS raining is 0.85 (85% - which seems about right 

Actually, to nail my colours to the mast, Manchester is one of my favourite cities in the world - and I've been to a few. It may not be pretty but it is a hell of a lot of fun, is full of history, has lots of great places to eat and drink and is really close to some of the best countryside in the Universe. 