# P(tomorrow is the end of the world) =?

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.

• 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. Aug 14, 2013 at 15:04
• 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. Aug 14, 2013 at 15:08
• "Problems" like this one give probability a bad name. Aug 14, 2013 at 15:14
• "they both", do you mean "they all"? Aug 14, 2013 at 15:29
• @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. Aug 14, 2013 at 18:42

## 3 Answers

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 $3.375 p$.

• 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 Aug 14, 2013 at 18:32
• @MarcvanLeeuwen: assuming prophets can predict the future somehow seems to contradict causality.
– robjohn
Aug 14, 2013 at 18:50
• @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. Aug 14, 2013 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. Aug 14, 2013 at 22:35

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.

• Are you sure they're all independent?
– Neal
Aug 14, 2013 at 15:02
• @Neal: No, but without that assumption you can't get a number. I'll clarify. Aug 14, 2013 at 15:03
• 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, 2013 at 17:51
• @robjohn: Indeed, that is the equivalent Bayesian prior. Aug 14, 2013 at 18:05
• 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, 2013 at 0:23

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}