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

> 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.

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.

For anyone unconvinced, let's simulate this.

Instead of 1 in 14 million, we'll just do 1 in 2, and 8 players.

So we'll check how many bits are set in the average random byte:

    void Main()
     {
      byte[] buffer = new byte[256*1024];
      Random.Shared.NextBytes(buffer);
      var avg = buffer.Average(b =>     System.Runtime.Intrinsics.X86.Popcnt.PopCount(b));
      Console.WriteLine(avg);
     }

Okay, bounces around 3.998 to 4.001, seems normal.

Now let's check how many bits are set in the average random byte given that the low bit is 1 (i.e. player 1 has won!)

    void Main()
     {
      byte[] buffer = new byte[256*1024];
      Random.Shared.NextBytes(buffer);
      var avg = buffer
       .Where(b => (b & ((byte)0x01)) == 0x01)
       .Average(b =>     System.Runtime.Intrinsics.X86.Popcnt.PopCount(b));
      Console.WriteLine(avg); 
     }

Now ~=4.500

Which is 1+3.5

In this case, we're 1+ average from the 7 other players, so being an average of 7 others not 8 others is significant.

If we simulate with millions of players, you'll see that removing 1 person from the pool makes essentially no difference.

You are making an assumption you did not explain before and makes it confusing, you don’t know the number of winning tickets or the number of winning tickets affects the prize quantity, and not all lotteries work like that.

To understand why your totally understandable conclusion is wrong, it helps me to think about what it means to determine the average number of other winners when I win.

The reason is similar to the Monty Hall Problem (https://en.wikipedia.org/wiki/Monty_Hall_problem)

To understand, lets think about the simplest representation of this problem... 2 people playing the lottery, and a 50/50 chance to win.

So, we can map out all the possible combinations:

A wins (50%) and B wins (50%) - 25% of the time

A wins (50%) and B loses (50%) - 25% of the time

A loses (50%) and B wins (50%) - 25% of the time

A loses (50%) and B loses (50%) - 25% of the time

So we have 4 even outcomes, so to figure out the average number of winners, we just add up the total number of winners in all the situations and divide by 4... so two winners in the first scenario, plus one winner in scenario 2, plus one winner in scenario 3, and zero winners in scenario 4, for 4 total winners in all situations... divide that by 4, and we see we have an average of 1 winner per scenario.

This makes sense... with 50/50 chance of winning with 2 people leads to an average of 1 winner per draw.

Now lets see what happens if we check for situations where player A wins; in our example, that is the first two scenarios. We throw out scenario 3 and 4, since player A loses in those two scenarios.

So scenario one has 2 winners (A + B) while scenario two has 1 winner (just A)... so in two (even probability) outcomes where A is a winner, we have a total of 3 winners... divide that 3 by the two scenarios, and we get an average of 1.5 winners per scenario where A is a winner.

Why does this happen? In this simple example it is easy to see why... we removed the 1/4 chance where we have ZERO winners, which was bringing down the average.

This same thing happens no matter how many players and what the odds are... by selecting only the scenarios where a specific player wins, we are removing all the possible outcomes where zero people win.

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.

I think you forgot to mention the condition 'given that you already have a ticket', and whatever justifications are required to assume that two more winning tickets will be present (if each ticket has independent odds of being a winner then you end up with a distribution of other tickets yeah?). Otherwise your premise doesn't quite lead to your conclusion.

That doesn’t work very well when the average number of winners is much less than 1. The math might work out that the “expected value” is more than one winner but in a realistic lottery you should expect to be the only winner.

You're both right! This is where the subjective Bayesian framework helps clarify things. The passive-voice term "expected winners" leaves ambiguous a key idea: Expected by whom?

The number of winners you expect depends on what information you have, namely, whether or not you know that you are holding a winning lottery ticket or not!

I'm not sure if this is true, as large jackpots see a higher than average number of tickets sold.

Every lottery I know of has many winners. One big winner but many who match only one numbe, and so win a tiny amount.

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).

You said my conclusion was wrong (edit: Apologies, I confused the nesting level here), then proved it correct by calculating the expected number of winners given you win as 3/2.

Winners in this case means jackpot winners. This is especially relevant as unlike partial winners, the jackpot is shared, not duplicated.

Right, I didn't spell out the format of the lottery. I'm assuming a "pick X numbers from Y" format, rather than a raffle style lottery.

This allows for multiple independent winners.

Nice. And, say the lottery jackpot is a constant 6$, then the average winning per player is 3$ (case 1) or 6$ (case 2) or 6$ (case 3), each equally likely (case 4 is not applicable), so $5.

However, if A wins, A wins either $3 (case 1) or $6 (case 2), so A's expected winnings are $4.5, which is indeed < $5, as GGP asserted.

(I think cortesoft was responding to pif, whom you also responded to, thus agreeing with you.)

I'm essentially assuming the UK lottery (from 96-200x) rules, where players:

Pick 6 numbers from 49, so odds are independent, and roughly 1 in 14m.

The prize jackpot is determined from a set percentage from the ticket sales, and is shared between jackpot winners.

How much the jackpot prize is therefore determined by total sales and how many winners there are.

I'm also assuming your individual ticket contribution doesn't materially affect either the prize pool or the number of people playing. For a large N, small p, this holds true.

Ah, thank you, navigating the nesting on here is difficult sometimes and this has proven a very contentious topic!

It is true. The average power ball does not have a winner.

The main problem with your argument is that the "average lottery winner" doesn't win JackpotPool/Average winners.

Maybe pif's comment

"The total number of expected winners (including you) is the same as the average number of winners"

means

"The total number of expected winners (including you) is the same as the average number of winners when there is at least one winner"

All the possible outcomes where zero people win are irrelevant when it comes to the calculation of how much "the average lottery winner" wins.

The "average lottery winner" also wins $4.5 though. (The original claim was that "if you win the lottery jackpot, then you win less than the average lottery winner".)

If there are 100 draws with a $6 jackpot 25 will have no winners, 50 will have one ($6) winner and 25 will have two ($3 each) winners.

100 winners in total - half won $6 and half won $3.

I mispoke a bit when saying it is ALL because of the case where zero people win.

It still holds for non-zero cases, too.

Since whether any individual wins is independent of other people winning, selecting only the situations where you win doesn't change the odds of other people winning, it simply adds a 100% chance of you winning. So it has all the same combination of winners, plus you.

I don't have time right now to type out a more full explanation, but I hope this somewhat makes sense given my previous comment.

> So it has all the same combination of winners, plus you.

And the same is true when you condition on having at least one winner. One winner doesn't change the odds of other people winning.

[edit: this may not be correct, never mind "In your example it doesn't matter whether you condition on A winning, on B winning or on at least one of A and B winning."]

Right, one winner doesn't change the odds... but we are choosing to throw out all the scenarios where that winner doesn't win, which DOES change the overall odds distribution. We are changing our selection criteria.

I think my previous comment was wrong. Anyway, the point is that the original claim

"if you win the lottery jackpot, then you win less than the average lottery winner"

seems wrong unless the winnings of "the average lottery winner" are defined in a quite unnatural way.

In your example the average lottery winner wins 3/4 of the jackpot. Half the winners take it all, the other half have to split it with someone else.

No, it it still the case that “if you win the lottery jackpot, then you will win less than the average lottery winner”. Let me see if I can explain in another way that might make this more clear… the example of only 2 people actually confuses the issue.

So in our example with 50% chance of winning, we know the average number of winners will be n/2, where n is the number of players. This means that the average lottery winner will win prize_pool / (n/2).

Now, let’s say we know I won. That means the average number of other winners is going to be (n-1) / 2. If you add in the known winner (me), we would have an average of 1 + (n-1)/2 winners… meaning the prize per person when I win is going to be prize_pool / (1 + (n-1)/2).

You can clearly see that the prize pool will be smaller when you know I am a winner. If it isn’t clear, just sub in 10 for N and solve it… the average winner will get prize_pool / (10/2) or prize_pool / 5. When I win, the average winner will get prize_pool / (1 + (10-1)/2), or prize_pool / 5.5. You can see that when I win, the average is lower.

This of course works whenever you start with the assumption that a particular person wins… you are turning the 1/2 chance for that person into a 100% chance, which increases the overall average number of winners.

> This means that the average lottery winner will win prize_pool / (n/2).

Does it?

Take the case n=2. Run the lottery a few times and take all the winners.

Half the winners win prize_pool.

Half the winners win prize_pool/2.

How do you define “average lottery winner” so the average lottery winner will win prize_pool / (n/2) = prize_pool ?

Taking the n=10 case, because you think n=2 is confusing.

> the average winner will get prize_pool / (10/2) or prize_pool / 5.

No, the average winner will get expected_prize_pool / expected_number_of_winners.

If 5 is the number of winners averaged over all draws - including those without winners - the (average) pot they share has also to take into account draws without winners.

The average prize shared in this case is not prize_pool, it’s 1023/1024 times prize_pool.

You are correct. So, the statement should not be "if you win the lottery jackpot, then you win less than the average lottery winner", but "if you win the lottery jackpot, then you win less than lottery winners win on average"?

When you win the lottery you win on average what lottery winners win on average when they win the lottery.

One may find ways to define things differently so something is less than something else but I’m not sure what’s the point in doing so.