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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 l
by xnorswap 2y ago
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.
- xnorswap 2y agoFor 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.
- melenaboija 2y agoYou 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.
- xnorswap 2y agoI'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.
- adastra22 2y agoThis 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.
- burnished 2y agoI 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.
- xnorswap 2y agoRight, 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.
- jncfhnb 2y agoThat 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.
- TylerE 2y agoI'm not sure if this is true, as large jackpots see a higher than average number of tickets sold.
- jncfhnb 2y agoIt is true. The average power ball does not have a winner.
- bluGill 2y agoEvery lottery I know of has many winners. One big winner but many who match only one numbe, and so win a tiny amount.
- TylerE 2y agoWinners in this case means jackpot winners. This is especially relevant as unlike partial winners, the jackpot is shared, not duplicated.