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The waiting time paradox: why is my bus always late? (2018)

(jakevdp.github.io)
219 points skadamat | 3 comments | 20 Aug 24 13:55 UTC | HN request time: 0s | source
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xnorswap ◴[20 Aug 24 15:50 UTC] No.41301135[source]▶
My favourite corollary of this is that even if you win the lottery jackpot, then you win less than the average lottery winner.

Average Jackpot prize is JackpotPool/Average winners.

Average Jackpot prize given you win is JackpotPool/(1+Average winners).

The number of expected other winners on the date you win is the same as the average number of winners. Your winning ticket doesn't affect the average number of winners.

This is similar to the classroom paradox where there are more winners when the prize is poorly split, so the average observed jackpot prize is less than the average jackpot prize averaged over events.

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pif ◴[20 Aug 24 15:58 UTC] No.41301213[source]▶
> The number of expected other winners on the date you win is the same as the average number of winners.

Sorry, but no! The total number of expected winners (including you) is the same as the average number of winners.

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xnorswap ◴[20 Aug 24 16:02 UTC] No.41301245[source]▶
No, it's 1+average number of winners.

If the odds of winning are 1 in 14 million, and 28 million tickets are sold, then you expect there to be 2 winners.

If look at your ticket and see you've won the lottery, then the odds of winners are still 1 in 14 million, and out of the 27,999,999 other tickets sold, you expect 2 other winners, and now expect 3 winners total, given you have won.

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1. adastra22 ◴[20 Aug 24 17:42 UTC] No.41302155{3}[source]▶
This is a variant of the Monty Hall problem, and the trick that makes it unintuitive is that you’ve snuck in the conditional of assuming you have already won.

If you have a ticket and haven’t checked that it is winning, you should expect two winners, regardless of whether you end up being one of them.

If you play the lottery trillions of times (always with the same odds, for simplicity) and build up a frequentist sample of winning events, you will on average be part of a pool of two winners in those instances where you win.

You snuck in the assumption that the first ticket checked (yours) is a winner, which screws up the statistics.

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2. xnorswap ◴[20 Aug 24 18:21 UTC] No.41302539[source]▶
I haven't "snuck in" anything, I very explicitly stated that it's the conditional expectation I'm talking about.

Expected(Number of winners | Pop size N and You win) = 1 + Expected(Number winners | Pop size N-1)

For small p, large N, that's ~1 + Expected(number of winners).

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3. kgwgk ◴[20 Aug 24 19:24 UTC] No.41303124[source]▶
The main problem with your argument is that the "average lottery winner" doesn't win JackpotPool/Average winners.