6 ms·

If lottery tickets are generated randomly how does the person in front of you winning decrease your chance of winning?

by brent 17y ago

run4yourlives 17y ago
Sounds like the whole 3 doors game show problem: Does an additional data point decrease your odds, or does it do nothing to alter the situation? IMO my answer to your question would be, it doesn't.
swolchok 17y ago
If there are N participants, k winners total, and y'all really do choose tickets randomly (proven false, but let's go with your assumption), your chances of winning are k/N. If you know that one of the winners is not you, your chances of winning are (k-1)/N. EDIT: above is assuming that you can win more than once if k > 1. If not, the math is more complicated than I can be bothered to try to format in an HN post, but the principle should hold.
- smanek 17y ago
  I don't think there are a fixed number of winners in advance ... My expected return may go down if the person in front of my wins (I have to split with more people, so each winner gets less) - but my chances of winning don't change.
- brent 17y ago
  That is an interesting form of lottery. Do you have any references to lotteries run in this manner (and proof that the tickets are not randomly generated)? I think the form of lottery I (and others) are referring to is the one where there are N unique tickets and you have a 1/N chance of matching those N numbers. Therefore, there is no dependence between different tickets and the likelihood of matching the winning numbers.
  - swolchok 17y ago
    Well, the tickets where y'all fill in the bubbles with pencil to choose numbers and the form can be read by an optical scanner are surely not randomly generated. People will draw pictures, play birthdays, and attempt to choose layouts that "look random", none of which result in a uniform ticket distribution, which is what most people seem to mean when they say that something is "random". The takeaway is that "distributed uniformly at random" is not a great assumption at all when inputs are generated by a process. It's one of the relatively few things I did learn in my graduate algorithms class.
ricree 17y ago
It shouldn't, but since lotto winnings are shared, it would decrease the already low expected payout.
ahoyhere 17y ago
My assumption was operating on the "one big win" model. But - forest, trees, guys? Seriously?
- brent 17y ago
  I think the most common model is that tickets are randomly generated and winners share the pot. Two winning tickets do not decrease one's odds from having a winning ticket since they are independent. Clearly it isn't a forest/trees issue, I just thought it was a bizarre addition since under any reasonable assumptions it is incorrect.
blhack 17y ago
Lets say that the lotter involves guessing the correct 1 digit number (0-9). There is a one in ten chance of winning. The guy in front of you has a 1/10 chance of hitting the winner. You ALSO have a 1/10 chance of hitting the winner. The likelihood of you BOTH hitting the winner in a row is 1/10*1/10 = 1/100. So, yes, the person in front of you winning DOES decrease your chances of winning. You could also think about it like this:
- likpok 17y ago
  No. You do not understand conditional probability. You have made a few simplifying assumptions. One, that the pmf of picking an number is uniform, and two, that the two choices are independent. WLOG the first person is male (ease of pronoun usage) Neither of these are particularly odd assumptions. Therefore. You have a 1/10 chance of winning. He has a 1/10 chance of winning. The probability of you BOTH winning is 1/100. The probability of you winning GIVEN that he won is 1/10. This is because P(A|B) = P(AB)/P(B) => 1/10 = (1/10 x 1/10)/(1/10), which is in fact a consequence of independence. To explain: Conditional probability is special. Unlike the intersection, you have to correct for all the stuff that you know wasn't possible (i.e., guy 1 choosing the wrong number). In this case, that exactly compensates for the low probability of the intersection. EDIT: Asterisks for times and italics don't mix
  - blhack 17y ago
    I'm sorry, I'm sure my lack of a maths degree is showing, but wouldn't the probability of hitting a random digit (0-9) be 1/10, but the probability of the hitting the SAME random digit twice IN A ROW be 1/100?
    - swolchok 17y ago
      Yes, that's the prior probability, before you observe the first outcome. This is a common gamblers' fallacy; past events do not affect future events from which they are independent! I'm assuming that you and/or the GP are assuming that everyone can win and everyone's ticket has a random digit drawn separately. However, if we assume that you know what number you picked, you don't know the result of the lottery, you know that the person in front of you won, and he chose his digit uniformly at random, the probability is still 1/10, because it has already been decided that he and the lottery chose the same digit.

If lottery tickets are generated randomly how does the person in front of you winning decrease your chance of winning?

by brent 17y ago

If lottery tickets are generated randomly how does the person in front of you winning decrease your chance of winning?

Sounds like the whole 3 doors game show problem: Does an additional data point decrease your odds, or does it do nothing to alter the situation? IMO my answer to your question would be, it doesn't.

swolchok 17y ago

If there are N participants, k winners total, and y'all really do choose tickets randomly (proven false, but let's go with your assumption), your chances of winning are k/N. If you know that one of the winners is not you, your chances of winning are (k-1)/N. EDIT: above is assuming that you can win more than once if k > 1. If not, the math is more complicated than I can be bothered to try to format in an HN post, but the principle should hold.

smanek 17y ago
I don't think there are a fixed number of winners in advance ... My expected return may go down if the person in front of my wins (I have to split with more people, so each winner gets less) - but my chances of winning don't change.
brent 17y ago
That is an interesting form of lottery. Do you have any references to lotteries run in this manner (and proof that the tickets are not randomly generated)? I think the form of lottery I (and others) are referring to is the one where there are N unique tickets and you have a 1/N chance of matching those N numbers. Therefore, there is no dependence between different tickets and the likelihood of matching the winning numbers.
- swolchok 17y ago
  Well, the tickets where y'all fill in the bubbles with pencil to choose numbers and the form can be read by an optical scanner are surely not randomly generated. People will draw pictures, play birthdays, and attempt to choose layouts that "look random", none of which result in a uniform ticket distribution, which is what most people seem to mean when they say that something is "random". The takeaway is that "distributed uniformly at random" is not a great assumption at all when inputs are generated by a process. It's one of the relatively few things I did learn in my graduate algorithms class.

ricree 17y ago

It shouldn't, but since lotto winnings are shared, it would decrease the already low expected payout.

ahoyhere 17y ago

My assumption was operating on the "one big win" model. But - forest, trees, guys? Seriously?

brent 17y ago
I think the most common model is that tickets are randomly generated and winners share the pot. Two winning tickets do not decrease one's odds from having a winning ticket since they are independent. Clearly it isn't a forest/trees issue, I just thought it was a bizarre addition since under any reasonable assumptions it is incorrect.

blhack 17y ago

Lets say that the lotter involves guessing the correct 1 digit number (0-9). There is a one in ten chance of winning. The guy in front of you has a 1/10 chance of hitting the winner. You ALSO have a 1/10 chance of hitting the winner. The likelihood of you BOTH hitting the winner in a row is 1/10*1/10 = 1/100. So, yes, the person in front of you winning DOES decrease your chances of winning. You could also think about it like this:

likpok 17y ago
No. You do not understand conditional probability. You have made a few simplifying assumptions. One, that the pmf of picking an number is uniform, and two, that the two choices are independent. WLOG the first person is male (ease of pronoun usage) Neither of these are particularly odd assumptions. Therefore. You have a 1/10 chance of winning. He has a 1/10 chance of winning. The probability of you BOTH winning is 1/100. The probability of you winning GIVEN that he won is 1/10. This is because P(A|B) = P(AB)/P(B) => 1/10 = (1/10 x 1/10)/(1/10), which is in fact a consequence of independence. To explain: Conditional probability is special. Unlike the intersection, you have to correct for all the stuff that you know wasn't possible (i.e., guy 1 choosing the wrong number). In this case, that exactly compensates for the low probability of the intersection. EDIT: Asterisks for times and italics don't mix
- blhack 17y ago
  I'm sorry, I'm sure my lack of a maths degree is showing, but wouldn't the probability of hitting a random digit (0-9) be 1/10, but the probability of the hitting the SAME random digit twice IN A ROW be 1/100?
  - swolchok 17y ago
    Yes, that's the prior probability, before you observe the first outcome. This is a common gamblers' fallacy; past events do not affect future events from which they are independent! I'm assuming that you and/or the GP are assuming that everyone can win and everyone's ticket has a random digit drawn separately. However, if we assume that you know what number you picked, you don't know the result of the lottery, you know that the person in front of you won, and he chose his digit uniformly at random, the probability is still 1/10, because it has already been decided that he and the lottery chose the same digit.