7 ms·

> 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 (i

by pif 2y ago

> 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.

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 ago
  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.
- melenaboija 2y ago
  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.
  - xnorswap 2y ago
    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.
- 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 ago
    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).
    - kgwgk 2y ago
      The main problem with your argument is that the "average lottery winner" doesn't win JackpotPool/Average winners.
- burnished 2y ago
  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.
  - xnorswap 2y ago
    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.
- jncfhnb 2y ago
  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.
  - TylerE 2y ago
    I'm not sure if this is true, as large jackpots see a higher than average number of tickets sold.
    - jncfhnb 2y ago
      It is true. The average power ball does not have a winner.
  - bluGill 2y ago
    Every 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 ago
      Winners in this case means jackpot winners. This is especially relevant as unlike partial winners, the jackpot is shared, not duplicated.
cortesoft 2y ago
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 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.
- xnorswap 2y ago
  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.
  - FabHK 2y ago
    (I think cortesoft was responding to pif, whom you also responded to, thus agreeing with you.)
mitthrowaway2 2y ago
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!

by pif 2y ago

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 ago
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.
melenaboija 2y ago
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.
- xnorswap 2y ago
  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.
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 ago
  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).
  - kgwgk 2y ago
    The main problem with your argument is that the "average lottery winner" doesn't win JackpotPool/Average winners.
burnished 2y ago
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.
- xnorswap 2y ago
  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.
jncfhnb 2y ago
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.
- TylerE 2y ago
  I'm not sure if this is true, as large jackpots see a higher than average number of tickets sold.
  - jncfhnb 2y ago
    It is true. The average power ball does not have a winner.
- bluGill 2y ago
  Every 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 ago
    Winners in this case means jackpot winners. This is especially relevant as unlike partial winners, the jackpot is shared, not duplicated.

cortesoft 2y ago

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 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.

xnorswap 2y ago
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.
- FabHK 2y ago
  (I think cortesoft was responding to pif, whom you also responded to, thus agreeing with you.)

mitthrowaway2 2y ago

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!