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

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.

> Average Jackpot prize is JackpotPool/Average winners.

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

That doesn't make a lot of sense.

Maybe you mean that most winners get less than the average prize.

Let's say that there is $1m jackpot and there could be one, two, three or four winners (with equal probability).

To simplify the calculation, let's say that each outcome happens once.

The average prize is $400k (4 x $1m / (1+2+3+4)).

A winner has 40% probability of getting just $250k and 30% probability of getting $333k.

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Edit: Or maybe you tried to say something like the following but didn't get it right because "average winners" means different things when you win and when you don't.

> Average Jackpot prize is JackpotPool/Average winners when there are one or more winners

> Average Jackpot prize given you win is JackpotPool/(1+Average winners when there are zero or more winners).

The key here is that you don't care what happens when you don't win, you don't care how much other people win.

What you care about, is the expected amount you win, given that you have a winning ticket.

Let's say there are N players, and let's say anyone has a 1 in X independent chance to win.

If you don't buy a ticket, there are N/X expected winners.

If you do buy a ticket, it doesn't affect whether other people win or not.

There are still N/X expected other winners.

Your participation doesn't reduce the expected number of people, who are not yourself, that will win.

This isn't a Monty hall problem, because Monty Hall introduced new information.

Buying a ticket doesn't introduce new information.

With Prob of (X-1)/X, you lose, and go home unhappy.

With Prob of 1/X, you win. And now there are 1 + N winners.

Your buying a ticket therefore increased the overall expected number of winners by 1/X. That is correct.

Conditioned on you winning, there are 1+N expected winners.

Conditioned on you losing, there are N expected winners.

Conditioned on you winning, there are 1+N expected winners. The number of winners is always larger than zero [edit: the following is not correct "and the average prize is calculated diving the jackpot by the 1+N expected winners"].

Conditioned on you losing, there are N expected winners. The number of winners can be zero and the average prize cannot be calculated dividing the jackpot by the N expected winners [edit: not that you could before...]. [edit: the following is not correct "You have to divide the jackpot by a higher number: the expected winners conditional on having at least one."] No winner, no prize.

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Let's say there are 2 people playing and the probability of winning is 50%. The number of expected winners is 1. If the jackpot is $1000 the "average lottery winner" doesn't get $1000.

Three outcomes are possible, with the following probabilities:

  1/4 zero winners
  1/2 one winner gets $1000
  1/4 two winners get $500 each

The "average lottery winner" gets less than $1000. The "average lottery winner" gets $750. (Imagine that a lot of draws have happened: for each split jackpot there were two jackpots going to a single winnner. All in all, half the winners got $1000 and the other half got $500.)

Consider now that you are one of the two players and you win. The other person will either win (you get $500) or not (you get $1000) with the same probability. Your expected prize? $750

What a coincidence!