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In Unweaving the Rainbow (page 150) Richard Dawkins mentions the following (famous) reasoning on why it's almost certain that a psychic with a big audience will accurately predict/command rare events like watch-stopping:

We can do a similar calculation for the television guru whose psychic miasma seemed to stop people's watches, but we'll have to use estimates rather than exact figures. Any given watch has a certain low probability of stopping at any moment. I don't know what this probability is, but here's the kind of way in which we could come to an estimate. If we take just digital watches, their battery typically runs out within a year. Approximately, then, a digital watch stops once per year. Presumably clockwork watches stop more often because people forget to wind them and presumably digital watches stop less often because people sometimes remember to renew the battery ahead of time. But both kinds of watches probably stop as often again because they develop faults of one kind or another. So, let our estimate be that any given watch is likely to stop about once a year. It doesn't matter too much how accurate our estimate is. The principle will remain.

If somebody's watch stopped three weeks after the spell was cast, even the most credulous would prefer to put it down to chance. We need to decide how large a delay would have been judged by the audience as sufficiently simultaneous with the psychic's announcement to impress. About five minutes is certainly safe, especially since he can keep talking to each caller for a few minutes before the next call ceases to seem roughly simultaneous. There are about 100,000 five-minute periods in a year. The probability that any given watch, say mine, will stop in a designated five-minute period is about 1 in 100,000. Low odds, but there are 10 million people watching the show. If only half of them are wearing watches, we could expect about 25 of those watches to stop in any given minute. If only a quarter of these ring in to the studio, that is 6 calls, more than enough to dumbfound a naive audience. Especially when you add in the calls from people whose watches stopped the day before, people whose watches didn't stop but whose grandfather clocks did, people who died of heart attacks and their bereaved relatives phoned in to say that their ‘ticker’ gave out, and so on.

I agree with the general reasoning but I'm confused as to how he gets to 25 there in the middle, I think it should be 50. This would NOT change at all the general point he's making but it seems big enough to point out. Can someone please help me see how he got to 25? Is it just a simple mistake on his part or mine?

Here's my reasoning: The minutes in a year are 365 days * 24 hours * 60 min = 525,600 minutes. And that divided by 5, gives 105,120 five-minute periods in a year. So far, consistent with Dawkins account. He then asks us to consider a 10,000,000 audience, only half of whom are wearing watches, so 5,000,000 watch-wearing viewers. If we now multiply these by the probability of stopping in a five-minute period we get, 5,000,000 * (1/104,120) = 47.56.

So, in a 5,000,000 watch-wearing audience ABOUT 50 watches are expected to stop in a 5-minute-period, not 25 as he claims. (Note that his statement is rather ambiguous, if we interpret "in any given minute" as being literally how many watches we expect to stop in ONE minute then the answer is 9.5, so about 10... still not 25.)

Thanks.

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  • $\begingroup$ Maybe he divided by two in the beginning, and then, forgetting he already did, divided by two again (accounting for the half who are wearing watches). $\endgroup$ – Jonny Sep 13 '14 at 0:11
  • $\begingroup$ @Jonny Yeah! So that's exactly what I'm thinking, that he wrongly divided by 2 twice instead of just once. I just wanted to be extra sure I wasn't missing anything. His argument is quoted by David Deutsch for one thing and both him & Dawkins are guys who really sweat the details. $\endgroup$ – Eli Sep 13 '14 at 0:13
  • $\begingroup$ The other thing that catches my eye is the bold text says "any given minute" not "any given five minute period" $\endgroup$ – Jonny Sep 13 '14 at 0:14
  • $\begingroup$ Yes, so he should have said "any given five-minute period" from all his preamble about what a sufficiently simulateneous period would be. He can't mean the probability in just ONE minute for that would be 9.5 as I mention in the question. Maybe what he means, somewhat sloppily, by "any given minute" is 2.5 minutes? $\endgroup$ – Eli Sep 13 '14 at 0:20
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    $\begingroup$ Or he just did some very rough estimating when doing the multiplications? After all he says the accuracy of the estimate isn't too important. $\endgroup$ – Jonny Sep 13 '14 at 0:26
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I think the part of the "story" that isn't covered in your math is this part close to the end of first paragraph:

So, let our estimate be that any given watch is likely to stop about once a year.

If I followed your math correctly, you implicitly translated "likely" to 100% probability of every watch failing once each year with the only uncertainty, when it will fail.

Dawkins might've used another probability. I'm not quite sure, if 50% fits the bill. Can't do the math from the back of my mind.

Drifting Off-topic: I'd be surprised, if he'd get away with a higher probability. Back when I used to wear a watch, I had 2-3 battery changes/repairs in about 10 years. My watch had no backlight or other fancy (energy consuming/breakable) stuff...

Well, he doesn't consider people calling in, because they want to be on TV, because they are assistants to the "psychic", ... either. So I'd be lenient with his estimates.

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  • $\begingroup$ Interesting, yes, I think you're right! I had assumed "likely" to mean a 100% probability but if I use 50% then his math makes sense. Thanks, this was admittedly silly but it kind of obsessed me :) $\endgroup$ – Eli May 10 '15 at 16:35

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