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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

by kgwgk 2y ago

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.

cortesoft 2y ago
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.
- kgwgk 2y ago
  > 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 ?
- kgwgk 2y ago
  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.

by kgwgk 2y ago

cortesoft 2y ago

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.

kgwgk 2y ago
> 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 ?
kgwgk 2y ago
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.