# A Poisson process paradox

post by nerfhammer · 2012-04-06T05:20:01.130Z · LW · GW · Legacy · 11 comments

From Steven Pinker's latest book:

Suppose you live in a place that has a constant chance of being struck by lightning at any time throughout the year. Suppose that the strikes are random: every day the chance of a strike is the same, and the rate works out to one strike a month. Your house is hit by lightning today, Monday. What is the most likely day for the next bolt to strike your house?

.

.

.

.

.

.

The answer is “tomorrow,” Tuesday. That probability, to be sure, is not very high; let’s approximate it at 0.03 (about once a month). Now think about the chance that the next strike will be the day after tomorrow, Wednesday. For that to happen, two things have to take place. First lightning has to strike on Wednesday, a probability of 0.03. Second, lightning can’t have struck on Tuesday, or else Tuesday would have been the day of the next strike, not Wednesday. To calculate that probability, you have to multiply the chance that lightning will not strike on Tuesday (0.97, or 1 minus 0.03) by the chance that lightning will strike on Wednesday (0.03), which is 0.0291, a bit lower than Tuesday’s chances. What about Thursday? For that to be the day, lightning can’t have struck on Tuesday (0.97) or on Wednesday either (0.97 again) but it must strike on Thursday, so the chances are 0.97×0.97×0.03, which is 0.0282. What about Friday? It’s 0.97×0.97×0.97×0.03, or 0.274. With each day, the odds go down (0.0300 … 0.0291 … 0.0282 … 0.0274), because for a given day to be the next day that lightning strikes, all the previous days have to have been strike-free, and the more of these days there are, the lower the chances are that the streak will continue. To be exact, the probability goes down exponentially, accelerating at an accelerating rate. The chance that the next strike will be thirty days from today is 0.97^29x0.03, barely more than 1 percent.

Almost no one gets this right. I gave the question to a hundred Internet users, with the word next italicized so they couldn’t miss it. Sixty-seven picked the option “every day has the same chance.” But that answer, though intuitively compelling, is wrong. If every day were equally likely to be the next one, then a day a thousand years from now would be just as likely as a day a month from now. That would mean that the house would be just as likely to go a thousand years without a strike as to suffer one next month. Of the remaining respondents, nineteen thought that the most likely day was a month from today. Only five of the hundred correctly guessed “tomorrow.”

I wonder if there's a name for this problem. He doesn't cite any prior art.

## 11 comments

Comments sorted by top scores.

comment by orthonormal · 2012-04-07T00:34:43.998Z · LW(p) · GW(p)

Dammit, System 1.

comment by [deleted] · 2012-04-08T03:41:19.436Z · LW(p) · GW(p)

Technically, if the strike happened early enough in the morning, the answer is "today".

comment by VincentYu · 2012-04-06T06:42:39.982Z · LW(p) · GW(p)

This problem is described by the geometric distribution with p = P(no lightning strike on a day), which is a special case of the negative binomial distribution with r = 1.

comment by b1shop · 2012-04-07T17:03:15.992Z · LW(p) · GW(p)

Sounds like the error happened because the problem has a not-immediately-obvious conunction.

comment by Incorrect · 2012-04-06T12:39:36.932Z · LW(p) · GW(p)

If he had said "What is the most likely day for another bolt to strike the house and it be the next strike" I think I would have got it.

comment by Cyan · 2012-04-10T16:53:38.139Z · LW(p) · GW(p)

Got it. I didn't stop to think it through (System 2 is lazy!), so this is more good luck than good management. Apparently my System 1 is trained to associate "exponential distribution" with "Poisson process" and to replace "most likely" with "mode". Of course, "exponential distribution" is not quite right; as the answer shows, given the phrasing of the problem statement the relevant distribution is geometric. If the title hadn't mentioned "Poisson process" I don't know what would have happened.

comment by [deleted] · 2012-04-07T14:46:10.594Z · LW(p) · GW(p)

This only means that the mode is an awfully bad measure of central tendency.

comment by cousin_it · 2012-04-06T10:29:51.952Z · LW(p) · GW(p)

Is it true in general that maximum likelihood estimation is less intuitive than expected value estimation?

comment by N_R · 2012-04-09T19:49:31.700Z · LW(p) · GW(p)

relevant (Benford's law)

Replies from: Anubhav
comment by Anubhav · 2012-04-10T10:57:10.448Z · LW(p) · GW(p)

No it isn't.

comment by Thomas · 2012-04-06T06:36:07.538Z · LW(p) · GW(p)

Looks quite elementary, Still interesting, though.