4 ms·

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 chanc

by 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.
perokreco 17y ago
Uhm, that is not true. Those two events are independent, you should brush up on your probability theory.

by blhack 17y ago

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.

perokreco 17y ago

Uhm, that is not true. Those two events are independent, you should brush up on your probability theory.