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.
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.
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.
"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.