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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 whe
by 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 agoYou 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 agoNice. And, say the lottery jackpot is a constant 6$, then the average winning per player is 3$ (case 1) or 6$ (case 2) or 6$ (case 3), each equally likely (case 4 is not applicable), so $5. However, if A wins, A wins either $3 (case 1) or $6 (case 2), so A's expected winnings are $4.5, which is indeed < $5, as GGP asserted.
- kgwgk 2y agoThe "average lottery winner" also wins $4.5 though. (The original claim was that "if you win the lottery jackpot, then you win less than the average lottery winner".) If there are 100 draws with a $6 jackpot 25 will have no winners, 50 will have one ($6) winner and 25 will have two ($3 each) winners. 100 winners in total - half won $6 and half won $3.
- FabHK 2y agoYou are correct. So, the statement should not be "if you win the lottery jackpot, then you win less than the average lottery winner", but "if you win the lottery jackpot, then you win less than lottery winners win on average"?
- kgwgk 2y agoWhen you win the lottery you win on average what lottery winners win on average when they win the lottery. One may find ways to define things differently so something is less than something else but I’m not sure what’s the point in doing so.
- kgwgk 2y agoMaybe pif's comment "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.
- cortesoft 2y agoI 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.
- kgwgk 2y ago> 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."]
- cortesoft 2y agoRight, 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.
- kgwgk 2y agoI 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.