Besides this problem I also use the following one to rise the interest of my students on the Probabilities Theory. It's a funny puzzle were the use of simple arithmetic rules doesn't work. We have to workaround a little bit to find the answer.

So here it goes:

The Random Family decided to go on a picnic next Sunday.
So they tried to get some information about the weather and predict if it will be a sunny Sunday.
They went online to weatheronsundays.org and happily saw that Sunday most likely be sunny (the algorithm used on this site forecasts correctly the weather about four in five times).
The youngest daughter also went to the CheckTheWeatherOnNextSundaySoYourFamilyPicnicWillBeASucess application on her smart phone and also saw that next Sunday will be sunny (this application forecasts correctly about three times out of four).

Based on this two predictions, what is the probability of the next Sunday will be sunny?

  • $\begingroup$ Just in case you hadn't noticed it, there can only be one "accepted answer", so when you clicked the green checkmark on my simplified answer, hexomino's earlier and more thorough correct answer lost the tick. It's of course entirely up to you to decide where the tick should go, so as long as you know what you are doing, I won't protest any more than this :-) $\endgroup$ – Bass Jul 30 at 11:52
  • $\begingroup$ I knew! Please check on my comment on hexomino's answer ;) $\endgroup$ – Pspl Jul 30 at 11:54
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    $\begingroup$ As commented on both answers, this question is not well-posed. For the accepted answer to be correct, we would have to know that the two websites' forecasts were independent. For example, it's possible that the first website's algorithm is: * roll a 5-sided die; * on a 1-4, just copy the second website's algorithm; * on a 5, use a time machine to fully accurately predict the weather. Then the approach requiring independence is definitely wrong; and the true answer definitely depends on the actual probability of sunny weather. I fear your students are learning more bad than good from this. $\endgroup$ – Greg Martin Jul 30 at 20:06
  • $\begingroup$ @GregMartin, too much... Please don't question my skills on teaching on such few information. I've been teaching for 22 years now, and my method works very well. Besides, if you keep bringing "what if a time-machine..." like arguments up, there is no well-posed questions on this site whatsoever. $\endgroup$ – Pspl Jul 30 at 22:12
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    $\begingroup$ I understand that my critique was pretty blunt, and text communication is always harsher than intended, so I do apologize. However, the fact that you accepted the answer below does give me some information about the importance you place on the unspoken details. The "time machine" is a red-herring—substitute a vague "predicts perfectly" and the mathematical critique still stands. $\endgroup$ – Greg Martin Jul 30 at 23:18

(After rechecking my numbers which gave me a different answer before, it seems I've written a duplicate of hexomino's quicker answer. Since this one has 2 lines of maths instead of many, I think I'm going push post anyway.)

The key to figuring this out is that

either both forecasts are correct, or they are both wrong.

This allows us to figure out the relative frequency of the two cases:

Both right: $\frac{4}{5} \times \frac{3}{4} = \frac{12}{20}$
Both wrong: $\frac{1}{5} \times \frac{1}{4} = \frac{1}{20}$

From this we get that whenever the two forecasts agree, it is

12 times more likely that they agree because both are correct.

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  • $\begingroup$ Yeah, I understand the motivation of this approach and I see a lot of people do this in real life but it always makes me a bit uneasy because it's never quite explicit what information is thrown away. $\endgroup$ – hexomino Jul 30 at 11:46
  • $\begingroup$ That IS the answer I was looking for. :) To bad you did't do a table to show your approach! $\endgroup$ – Pspl Jul 30 at 11:47
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    $\begingroup$ I hate to be critical but this is indeed a duplicate of @hexomino's answer in that it makes the same two highly questionable assumptions: 1) the forecasts are independent 2) the false positive and false negative rates are identical. $\endgroup$ – Paul Panzer Jul 30 at 11:47
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    $\begingroup$ @PaulPanzer I don't think you actually hate it, and it's perfectly ok to be critical around here :-). I don't think this puzzle can have a meaningful solution the critique points are relevant, so I usually assume that the puzzle creator intended to create a solvable puzzle instead on an unsolvable one. $\endgroup$ – Bass Jul 30 at 12:03
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    $\begingroup$ This puzzle has brought out some fascinating points about hidden assumptions, so easy to make with probability. Conceivably, the sun might actually never come out while "sunny" keeps being forecast once or twice a week. But your last point, @Bass, is my very favorite: (my usual way of saying it:) The primary puzzle of any puzzle is to figure out what the puzzle is. $\endgroup$ – humn Jul 31 at 1:39

For these sorts of problems I like to use

Bayes rule

In this scenario we can apply it as follows

Let SS represent the outcome that both weatheronsundays.org and CheckTheWeather... predict that Sunday will be sunny and Sunny represent the outcome that Sunday is sunny. Then, we have $$ P(Sunny|SS) = \frac{P(SS|Sunny)P(Sunny)}{P(SS)} = \frac{P(SS|Sunny)P(Sunny)}{P(SS|Sunny)P(Sunny) + P(SS|Not Sunny)P(Not Sunny)}$$ Now given that we have no other information we shall assume that $P(Sunny) = P(Not Sunny) = \frac{1}{2}$.
Otherwise, we could use that information to inform us and it would affect the probability. This means that we have $$ P(Sunny|SS) = \frac{P(SS|Sunny)}{P(SS|Sunny) + P(SS|Not Sunny)}$$ Given that the website predicts correctly with probability $\frac{4}{5}$ and the app predicts correctly with probability $\frac{3}{4}$, assuming that the forecasts are independent, (credit to Paul Panzer for bringing this up) we have that $$P(SS|Sunny) = \frac{4}{5}\times\frac{3}{4} = \frac{3}{5}\,\,\,\,,\,\,\,\, P(SS|Not Sunny) = \frac{1}{5}\times\frac{1}{4} = \frac{1}{20} $$ Hence $$P(Sunny|SS) = \frac{\frac{3}{5}}{\frac{3}{5} + \frac{1}{20}} = \frac{12}{13}$$

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  • $\begingroup$ You got the right answer. But could you try and solve it without the Bayes Rule? A simple scheme will solve the problem on a very much easier way :) $\endgroup$ – Pspl Jul 30 at 11:32
  • $\begingroup$ I'm very sorry to change the accepted answer, but Bass's is more elegant :) None the less (and because you answer is right as well) I'm keeping my upvote :) $\endgroup$ – Pspl Jul 30 at 11:50
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    $\begingroup$ @Pspl I understand the green tick on this question but I would caution against teaching your students this method as it deliberately hides some of the information and you risk confusing them. For example, the probability changes if you add in the fact that you live in a country which is sunny 90% of the time but it's not immediately clear how to incorporate that in the quick method. $\endgroup$ – hexomino Jul 30 at 11:53
  • $\begingroup$ Thank you so, so very much, for your advice. I love this, when we all try and help each other. It's really great. But here's the scenario: I do this to my students on an "introduction" to probabilities basis, when they don't even know Bayes Rule. So I use this method to inspire them and then, after a couple of classes, I revisit this problem with your approach. Meanwhile that was my intent here at stack. $\endgroup$ – Pspl Jul 30 at 12:00
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    $\begingroup$ Also, as noted by Paul on the comment thread on the other answer, this assumes that the website (and the app) has the same accuracy in predicting sunny and not sunny. It could be the case that the website is always right when predicting sunny, and when it's not sunny the prediction is only 60% accurate (still consistent with the problem description). Then the probability that Sunday is sunny is 100%. $\endgroup$ – justhalf Jul 30 at 12:07

I've made this community wiki, so please edit away!

This is a very good question to put to students as long as one subsequently hammers home the point that it is ill posed and one works out the common fallacies that contribute to the expected answer.

The question is actually well suited for this didactic exercise because the tacit assumptions OP's preferred "simple" answer makes are quite implausible making it easier for students to appreciate that these assumptions are not a mere technicality but an actual mistake.

Assumption 1: p(sun) = p(no sun)

A good exercise for students to figure out where in the simple answer this assumption is hidden and how it impacts on the answer.

Also good to highlight the flaws of "no info: let's use a 'flat' ground truth". If we had asked in terms of "rain" vs "no rain" does this mean 50% rain probabiliy is a natural assumption? What if we had asked "sunny" "rainy" "neither" is it 33% each now?

Assumption 2: the two forecasts are independent

Again, good exercise to ask students where in the simple answer this assumption is used.

Unless we have a very low opinion of weather forecasting in general the assumption the forecasts are independent is not only unjustified, it is actually implausible as both forcasts are predictions presumably based on similar info and methodology.

Assumption 3: p(forecast correct|actual sun) = p(forecast correct|no sun) = p(forecast correct)

One last time challenge students to find where this assumption is used.

This is perhaps the most tricky thing to fully appreciate so be careful to well explain why there is a distinction between false positive and false negative rates and why they are often not the same.

  • $\begingroup$ Assumption 3 is not what people expect. People expect p(forecast correct|forecast sun) = p(forecast correct|forecast no sun) = p(forecast correct). And this makes assumption 1 unnecessary. $\endgroup$ – Florian F Aug 1 at 10:29
  • $\begingroup$ @FlorianF _" And this makes assumption 1 unnecessary."_that is not correct, you still need to justify the (non-) weighting of the branches. Maybe what confuses you is that Bayesianist mantra of "we don't need an explicit prior as we know the posterior"? If so, check the problem statement: We are not given the posterior which here would be the rate of forecast sun/no sun regardless of whether correct or not. $\endgroup$ – Paul Panzer Aug 1 at 11:35
  • $\begingroup$ My idea was that the probability of sun is already accounted for in the probability for the forecaster to be right. But I checked and you are right. We are comparing positive results from sunny days with false positives from non-sunny days. So the a-priori p(sunny) matters. $\endgroup$ – Florian F 2 days ago

We can just draw a cube like so:

enter image description here

The answer is...


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  • $\begingroup$ Cool answer :) However you could have made a simple double entry table... But it's cool, non the less! $\endgroup$ – Pspl Aug 1 at 11:49
  • $\begingroup$ how can you say a+b+e+f = a+c+e+g = 1 ? Shouldn't it be a+b+c+d+e+f+g+h = 1 ? $\endgroup$ – Florian F 2 days ago
  • $\begingroup$ Both grids indicate all possibilities when only one of the predictions is taken into account, so it makes sense that both sums are equal. You can call it 1, 2 or whatever if you like, but it doesn't change anything. $\endgroup$ – Nautilus 2 days ago

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