Hi,
Suppose people randomly pick a number between 0 and 1. What would be the expected number of people you would have to ask for numbers if you want the sum to be greater than 1 for the 1st time?
Steve
Hi,
Suppose people randomly pick a number between 0 and 1. What would be the expected number of people you would have to ask for numbers if you want the sum to be greater than 1 for the 1st time?
Steve
Last edited by Jomo; 12-05-2018 at 11:32 PM.
A mathematician is a blind man in a dark room looking for a black cat which isn’t there. - Charles R. Darwin
Would 1 to 10 and ending with > 10 be the same?
I'm a man of few words...but I use 'em often!!
Well, your post is too mysterious for li'l ole me...
going back to learning the "times 3" table; much tuffer than the "times 1"
I'm a man of few words...but I use 'em often!!
Asked 1st person to pick a number between 0 and 1. If the sum of the numbers chosen is more than 1 (which is impossible at this stage) stop. Otherwise,
Asked 2nd person to pick a number between 0 and 1. If the sum of the numbers chosen is more than 1 stop. Otherwise,
Asked 3rd person to pick a number between 0 and 1. If the sum of the numbers chosen is more than 1 stop. Otherwise,
.
.
.
What is the expected number of people you would have to ask?
A mathematician is a blind man in a dark room looking for a black cat which isn’t there. - Charles R. Darwin
Suppose that [tex]N_1~\&~N_2[/tex] choices then if [tex]\{N_1,N_2\}\subset[0.5,1][/tex] then [tex]N_1+N_2\ge1[/tex] and [tex]\mathcal{P}(N_1,N_2)=.25[/tex].
BUT if [tex]\{N_1,N_2\}\subset[0,0.5)[/tex] then [tex]N_1+N_2<1[/tex], while still [tex]\mathcal{P}(N_1,N_2)=.25[/tex].
However it gets even more messy. What if [tex]N_1=0.2~\&~N_2=0.9[/tex] with the same probability, [tex](N_1+N_2)>1[/tex].
What if [tex]N_1=0.2~\&~N_2=0.7[/tex] with the same probability, [tex](N_1+N_2)<1[/tex]
I am saying the the modeling here stumps me. I'll sleep on it.
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“A professor is someone who talks in someone else’s sleep”
W.H. Auden
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