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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, tha
by 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 agoI'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 agoYes, 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.