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Here we have $3$ prophets alpha, beta, gamma, they all predict that tomorrow is the end of the world. It's known that the accuracy of alpha's and beta's prediction is $90\%$, while that of gamma's is $4\%$. What is the probability that tomorrow is the end of the world?

Here's my calculations.

Let $A, B, C$ be alpha's, beta's and gamma's predictions.

P(tomorrow is the end of the world)

$= P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC)$

$= 0.9 + 0.9 + 0.04 - 0.81 - 0.036 - 0.036 + 0.0324$

$= 0.9904$

But this seems a bit too high, is my calculation correct? Thanks.

Dan Rust
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    More assumptions are needed. If beta always repeats what alpha says, then the condition that their accuracies are equal is clearly met, and yet we can totally ignore beta. In that case, gama is more accurate at mispredicting than alpha at predicting, so the world will (probably) not end tomorrow. – Hagen von Eitzen Aug 14 '13 at 15:04
  • @HagenvonEitzen: Amusingly I added just that example to my answer a moment ago (before seeing your comment). I quite agree. – Charles Aug 14 '13 at 15:05
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    How would one induce an accuracy of $0.9$ in the first place? Out of 100 end-of-world-predictions made by alpha in the past, ten were correct? Note that any ability of predicting Super Bowl results may not be transferred to predictions about the end of the world. – Hagen von Eitzen Aug 14 '13 at 15:08
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    "Problems" like this one give probability a bad name. – André Nicolas Aug 14 '13 at 15:14
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    @HagenvonEitzen Let your imagination run wild! For instance, if $\alpha$ were Nate Silver ... or if $\alpha$ were error bars on NASA projections of an extinction-size asteroid headed toward Earth. – Neal Aug 14 '13 at 15:14
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    "they both", do you mean "they all"? – Mike Flynn Aug 14 '13 at 15:29
  • The historical record on end-of-the-world predictions is not good. When I told my (then) seven-year-old about Harold Camping's prediction of the end of the world in 2011, she was impressed and excited. When I mentioned the impending Mayan Apocalypse to her in 2012, she was only disgusted. “Some people think the world will end this week,” I said. “Again?” she replied. Based on the past success of all known apocalyptic predictions, I think the best strategy is to assign all such predictions a probability close to zero, regardless of who makes them. – MJD Aug 14 '13 at 18:02
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    @MJD You are showing survivor bias. Of course all predictions (for previous dates) have turned out to be false. If they weren't false, we would not be here to ridicule them on MathOverflow. – emory Aug 14 '13 at 18:42
  • This seems perilously close to the fallacy of an ontological proof. Suppose I claim that there are nine-legged purple water buffaloes running amok, killing people. You ask why, if this is so, we haven't observed any; I reply that this is survivor bias, since everyone who has observed the buffaloes of doom has (presumably) been fatally gored or trampled. That is one explanation, true, but another explanation is simply that my claim is mistaken. – MJD Aug 14 '13 at 18:46
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    @HagenvonEitzen: You may have subjectivist interpretation of probability. Now you have 3 prophets which predicts end-of-world and your believe in them is 0.9, 0.9 and 0.04. What is you believe that the end-of-world will indeed happen tomorrow (assuming you hold coherent believes)? – Maja Piechotka Aug 14 '13 at 21:44

3 Answers3

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The calculation is completely bogus. The product formula is only true for independent events. But here the three events are far from independent: either it is the end of the world or it isn't, so either they are all right or they are all wrong.

Let's try to put in some of the assumptions that are needed, and make a correct calculation with a Bayesian approach. Suppose the three prophets are asked to predict whether a certain event $E$ will take place, and that event actually has probability $p$. Suppose that whenever prophet $\alpha$ is asked to make a prediction of this sort, his prediction is correct with probability $0.9$ (i.e. if the event actually happens, then with probability $0.9$ he predicted it would happen, and if it doesn't happen, then with probability $0.9$ he predicted it wouldn't). Similarly for $\beta$ and $\gamma$, with probabilities $0.9$ and $0.04$ respectively. Finally, suppose that the four events E, A = ($\alpha$ correct), B = ($\beta$ correct), C = ($\gamma$ correct), are independent.

The probability that the event will happen and all three say it will happen is $p (0.9)^2 (0.04) = .0324 p$. The probability that the event won't happen but all three say it will happen is $(1-p)(0.1)^2 (0.96) = .0096 (1-p)$. So the conditional probability that the event will happen, given that all three say it will happen, is

$$ \dfrac{.0324 p}{.0324 p + .0096 (1-p)}$$

In this case, I happen to think that $p$ is very small, so this is still very small: it is approximately $3.375 p$.

Robert Israel
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    Even though I can appreciate the effort to make some sense out of the "information" provided in the question, I find the interpretation of the given "accuracy of prediction" very doubtful. (1) there is no reason why the probability of correct prediction would be the same whether the event goes one way or the other; in real-life predictions this characteristic is almost impossible to achieve (except for the useless 0.5 accuracy) (2) it suggests that the actual future somehow influences the predictions, contradicting causality (3) 0.04 accuracy would mean reliable prediction of what won't happen – Marc van Leeuwen Aug 14 '13 at 18:32
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    @MarcvanLeeuwen: assuming prophets can predict the future somehow seems to contradict causality. – robjohn Aug 14 '13 at 18:50
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    @robjohn: Prophets may have a acquired a bad reputation lately, but situations where future events (or facts otherwise beyond direct observation) can be predicted based on pertinent but inconclusive evidence are quite common. Think of medical diagnosis. There is no reason the probabilities of mis-predicting one way or the other (false-positive and false-negative) have to be even approximately equal. – Marc van Leeuwen Aug 14 '13 at 19:00
  • The end of the world either comes or does not, but the events that each prophet predicts so given that it happens or given that it does not msy be independent. – dfeuer Aug 14 '13 at 22:35
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I get 27/35, assuming independence. The two possibilities are that all three are right (.9 * .9 * .04) or all three are wrong (.1 * .1 * .96). The former divided by the sum gives 27/35 or about 77.1%.

If you don't assume independence -- for example, if beta just says whatever alpha says -- you can't reduce the problem to a single number without more information.

Charles
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    Are you sure they're all independent? – Neal Aug 14 '13 at 15:02
  • @Neal: No, but without that assumption you can't get a number. I'll clarify. – Charles Aug 14 '13 at 15:03
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    It seems you are starting with a 50/50 chance (1:1 odds) that the world will end today, before accounting for the prophets. – robjohn Aug 14 '13 at 17:51
  • @robjohn: Indeed, that is the equivalent Bayesian prior. – Charles Aug 14 '13 at 18:05
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    Since none of the almost 20000 days that I have experienced have been the end of the world, prior to the prophets' prognostications, I would put the odds that tomorrow was the end of the world a bit lower than $1:1$. That's just my opinion :-) – robjohn Aug 15 '13 at 00:23
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To me, the most sensible way to look at the probability that a prophet's prediction is true, is to look at the ratio of times when their predictions come true over the times when their predictions don't come true: $$ \Lambda(A)=\frac{P(A\mid X)}{P(A\mid\neg X)} $$ Extending the ideas used in this answer, and assuming that the prophets are independent, we get that $$ O(W\mid A\cap B\cap C)=\Lambda(A)\,\Lambda(B)\,\Lambda(C)\,O(W) $$ Where $O(X)$ is the odds of $X$ defined as $$ O(X)=\frac{P(X)}{1-P(X)} $$ We are given that $$ \begin{align} \Lambda(A)=\frac{.9}{.1}&=9\\ \Lambda(B)=\frac{.9}{.1}&=9\\ \Lambda(C)=\frac{.04}{.96}&=\frac1{24}\\ \end{align} $$ Thus, adding the information from the prophets would increase the odds by a factor of $\frac{81}{24}$: $$ O(W\mid A\cap B\cap C)=\frac{27}{8}O(W) $$

This matches Robert Israel's answer if we convert from odds to probabilities: $$ \begin{align} P(W\mid A\cap B\cap C) &=\frac{\frac{27}{8}\frac{P(W)}{1-P(W)}}{\frac{27}{8}\frac{P(W)}{1-P(W)}+1}\\ &=\frac{27P(W)}{27P(W)+8(1-P(W))}\\ &=\frac{.0324P(W)}{.0324P(W)+.0096(1-P(W))} \end{align} $$

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