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

(jakevdp.github.io)
219 points skadamat | 2 comments | 20 Aug 24 13:55 UTC | HN request time: 0.411s | 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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mecsred ◴[20 Aug 24 18:10 UTC] No.41302432[source]▶
> Your winning ticket doesn't affect the average number of winners.

I think this is a good hint that the conclusion isn't true. Just think about what it would mean if this were true for a sample of lotto winners. For a winner, if they win, their average number of winners is higher than the global average. Repeat this logic for each individual winner... And every winner wins with a higher number of winners than the average. Which is clearly impossible.

It would be true if you were guaranteed to win, since that's the assumption you have conditioned the probability on, but that's not a lottery then. If you want to get the actual expected value across all samples you need to take a weighted sum including the expected value when you don't win.

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1. xnorswap ◴[20 Aug 24 18:26 UTC] No.41302584[source]▶
> Which is clearly impossible.

It's just like the average pupil being in a larger than average class size, it's not impossible!

Take a situation where you have 499 lotteries with zero winners, and 1 lottery with 1000 winners.

There are on average 2 winners per lottery.

From the perspective of all the winners, there was an average of 1000 winners.

That's the very basis of the paradox in the article.

Now, in that case, the lottery would be investigated for fraud. But the paradox plays out in a much gentler sense.

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2. mecsred ◴[20 Aug 24 20:29 UTC] No.41303746[source]▶
Well, I simulated it and the numbers seem to agree with you. The example is interesting. I still have trouble seeing why my original reasoning doesn't hold though. I'll give an example, if anyone can clear up the issue that would be appreciated.

1/10 odds, 10 entrants, one winner expected on average.

Given a particular winner: Expect: 0.9 + 1 winners

Given the same particular loser: Expect: 0.9 winners

Over all cases we see: 0.1(1.9) + 0.9(0.9) = 1 winners

Checks out, but if the numbers are correct then any winner should be able to calculate the higher average and be right knowing only that there is at least one winner. So in cases where there is at least one winner: P(winners>=1)=1-(9/10)^10=~65% The expectation should work out to 1.9. The rest of the time we expect zero winners. However if I use those numbers I get an overall expected number of winners as 1.237, which has increased the overall number of winners across all cases. In order for that number to work out to one, the expected winners when there is at least one winner would have to be ~1.535. Which suggests that the expected outcome is different depending on if you check your own ticket, or someone else's, even if you see the same thing?

Am I just not on for math today? I thought the solution to the paradox would be that the higher expectation discounts outcomes with zero winners.