29 Sep
2003
29 Sep
'03
11:38 p.m.
greetings. some questions from some of us on the problem set....
we have some questions about the Pepsi problem, part b. Are people assuming that the
question is asking what the probability that any one person, anybody, or any of the ten
people selected has picked the same number?
we were looking at the conditional probability that you have the right number, given that
you were one of the ten. using bayes, this seemed to be 1 in 10,000, though if you find
the probability that ANY of the finalists have the number it is 1 in 1000.
the whole thing seems problematic because you have to account for the people who have 3,
4, 5 and 6 matching numbers in the final ten (though the probability of any of these
people existing is pretty trivial), and thus the different mix of possibilities in the
group of 10.
any help is appreciated...
best,
jason
30 Sep
30 Sep
10:42 a.m.
New subject: ps 2
On Mon, 29 Sep 2003, Jason Lakin wrote:
greetings. some questions from some of us on the
problem set....
we have some questions about the Pepsi problem, part b. Are people assuming that the
question is asking what the probability that any one person, anybody, or any of the ten
people selected has picked the same number?
any one person of the ten people in the final group. So if I is the set of
the 10 closest people, you want to know if there is an i in I so that P(i
wins a billion) = 1/1000.
we were looking at the conditional probability that
you have the right number, given that you were one of the ten. using bayes, this seemed
to be 1 in 10,000, though if you find the probability that ANY of the finalists have the
number it is 1 in 1000.
everyone in the group of 10 closet people does not have the same
probability of winning. in particular, someone must be closest, even if
we can't identify them.
the whole thing seems problematic because you have to account for the people who have 3,
4, 5 and 6 matching numbers in the final ten (though the probability of any of these
people existing is pretty trivial), and thus the different mix of possibilities in the
group of 10.
I don't understand this sentence.
any help is appreciated...
best,
jason
7:19 p.m.
New subject: ps 2
"everyone in the group of 10 closet people does not have the same
probability of winning. in particular, someone must
be closest, even if
we can't identify them."
So why mention the ten people anyway? Can't we just calculate the
probability of winning the million dollars given that person x is the
closest (1/1000)?
I'm approaching the problem from two different angles which give me
different answers. If I calculate the probability that you have the right
number given you're in the ten closest, and then use THAT probability to
see the odds of AT LEAST one person out of the ten getting the number
right, then it's less than 1/1000.
But if I start with calculating the odds you get the number right given
that you are THE closest, then your odds are 1/1000 ...
Anyone else experiencing something similar?
--Andrew
On Tue, 30 Sep 2003, Jacob M. Kline wrote:
On Mon, 29 Sep 2003, Jason Lakin wrote:
greetings. some questions from some of us on the
problem set....
we have some questions about the Pepsi problem, part b. Are people assuming that the
question is asking what the probability that any one person, anybody, or any of the ten
people selected has picked the same number?
any one person of the ten people in the final group. So if I is the set of
the 10 closest people, you want to know if there is an i in I so that P(i
wins a billion) = 1/1000.
we were looking at the conditional probability
that you have the right number, given that you were one of the ten. using bayes, this
seemed to be 1 in 10,000, though if you find the probability that ANY of the finalists
have the number it is 1 in 1000.
everyone in the group of 10 closet people does not have the same
probability of winning. in particular, someone must be closest, even if
we can't identify them.
the whole thing seems problematic because you have to account for the people who have 3,
4, 5 and 6 matching numbers in the final ten (though the probability of any of these
people existing is pretty trivial), and thus the different mix of possibilities in the
group of 10.
I don't understand this sentence.
any help is appreciated...
best,
jason
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7:30 p.m.
New subject: ps 2
On Tue, 30 Sep 2003, Andrew Calvin Fong wrote:
"everyone in the group of 10 closet people does not have the same
it ought to be the same or very very close, depending on whether you
sample with or without replacement.
probability of winning. in particular, someone
must be closest, even if
we can't identify them."
So why mention the ten people anyway? Can't we just calculate the
probability of winning the million dollars given that person x is the
closest (1/1000)?
I'm approaching the problem from two different angles which give me
different answers. If I calculate the probability that you have the right
number given you're in the ten closest, and then use THAT probability to
see the odds of AT LEAST one person out of the ten getting the number
right, then it's less than 1/1000.
But if I start with calculating the odds you get the number right given
that you are THE closest, then your odds are 1/1000 ...
good.
Anyone else experiencing something similar?
--Andrew
On Tue, 30 Sep 2003, Jacob M. Kline wrote:
On Mon, 29 Sep 2003, Jason Lakin wrote:
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gov1000-list(a)fas.harvard.edu
http://www.fas.harvard.edu/mailman/listinfo/gov1000-list
greetings. some questions from some of us on the
problem set....
we have some questions about the Pepsi problem, part b. Are people assuming that the
question is asking what the probability that any one person, anybody, or any of the ten
people selected has picked the same number?
any one person of the ten people in the final group. So if I is the set of
the 10 closest people, you want to know if there is an i in I so that P(i
wins a billion) = 1/1000.
we were looking at the conditional probability
that you have the right number, given that you were one of the ten. using bayes, this
seemed to be 1 in 10,000, though if you find the probability that ANY of the finalists
have the number it is 1 in 1000.
everyone in the group of 10 closet people does not have the same
probability of winning. in particular, someone must be closest, even if
we can't identify them.
the whole thing seems problematic because you have to account for the people who have 3,
4, 5 and 6 matching numbers in the final ten (though the probability of any of these
people existing is pretty trivial), and thus the different mix of possibilities in the
group of 10.
I don't understand this sentence.
any help is appreciated...
best,
jason
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gov1000-list(a)fas.harvard.edu
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3 comments
3 participants
participants (3)
-
Andrew Calvin Fong -
Jacob M. Kline -
Jason Lakin