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Owen comments on Bayesian exercise - Less Wrong Discussion

5 Post author: RolfAndreassen 21 September 2011 09:34PM

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Comment author: Owen 21 September 2011 10:03:39PM *  2 points [-]

I think the error lies in this sentence:

"Presumably your chances of success this time are not affected by the next one being a failure."

I assume you think this is true because there's no causal relationship where the next shuttle launch can affect this one, but their successes can still be correlated, which your probability estimate isn't taking into account.

If you want to update meaningfully, you need to have an alternative hypothesis in mind. (Remember, evidence can only favor one hypothesis over another (if anything); evidence is never "for" or "against" any one theory at a time.) Perhaps the engineers believe that there is a 4% chance that any given shuttle launch will fail (H1), but you estimate a 25% chance that they're wrong and the shuttles are actually foolproof (H2). Then you estimate the probability that the first shuttle launch will fail (F) as

P(F) = P(F|H1) * P(H1) + P(F|H2) * P(H2) = (4%)(75%) + (0%)(25%) = 3%.

But the shuttle launch goes off ok, so now you update your opinion of the two hypotheses with Bayes' rule:

P(H1|~F) = P(~F|H1) * P(H1) / P(~F) = (100% - 4%) * (75%) / (100% - 3%) ≈ 74.2%.

Then your estimate that the next shuttle will fail (F') becomes:

P(F' | ~F) = P(F' | H1, ~F) * P(H1 | ~F) + P(F' | H2, ~F) * P(H2 | ~F)

= (4%) * (74.2%) + (0%) * (100% - 74.2%) ≈ 2.97%.

So the one successful shuttle launch does, in this case, lower your expectation of a failure next time. As the shuttles keep succeeding, you become gradually more and more sure that the shuttles are foolproof and the engineers are wrong. But if the launch ever does fail, you will instantly believe the engineers and assign no credence to the claim that the shuttles never fail. (Try the math to see how that works.)