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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.
by adastra22 2y ago
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.
- xnorswap 2y agoI 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).
- kgwgk 2y agoThe main problem with your argument is that the "average lottery winner" doesn't win JackpotPool/Average winners.