Quizzes & Puzzles5 mins ago
Statistical Problem
21 Answers
If one of a group of n items is distinguishable, then it can be found in (n+1)/2 attempts, on average, by examining a random item from the group, provided that the same item is not chosen twice. How many attempts are needed, on average, if items are examined at random without checking whether they have been examined previously? (This is only for interest - I've never studied statistics. Don't put great effort into answering.)
Answers
If there are N items and you are looking for precisely one of them, with replacement, then you want to work out the sum to infinity of 1/N + 2*( 1/ N)*( N- 1)/ N + 3*( 1/ N)*[( N- 1)/ N]^ 2 + 4*( 1/ N)*[( N- 1)/ N]^ 3 ... etc. Write as ( 1/ N)* Sum( i= 1, infinity) i [( N- 1)/ N]^( i- 1) This is the same as N* d/ dN{ Sum( i= 1, infinity) [( N- 1)/ N]^( i)} which is the same as N* d/dN...
11:06 Mon 25th Jul 2022
why not-suppose its a white and a blue.... sometimes you'll get white in one go (half the time), sometimes not until second go goes (25% of the time?) , sometimes not until 3rd go(1/8thof the time?) , sometimes 4th time if your unlucky.... sometimes 10 or moreth times if your very unlucky.
Im assuming you stop once you find the white one
Im assuming you stop once you find the white one
If there are N items and you are looking for precisely one of them, with replacement, then you want to work out the sum to infinity of
1/N + 2*(1/N)*(N-1)/N + 3*(1/N)*[(N-1)/N]^2 + 4*(1/N)*[(N-1)/N]^3 ...
etc. Write as (1/N)*Sum(i=1,infinity) i [(N-1)/N]^(i-1)
This is the same as
N* d/dN{Sum(i=1,infinity) [(N-1)/N]^(i)}
which is the same as
N* d/dN {((N-1)/N)/(1-((N-1)/N))
from geometric sum. This just simplifies to
N* d/dN{N-1} = N* 1 = N
1/N + 2*(1/N)*(N-1)/N + 3*(1/N)*[(N-1)/N]^2 + 4*(1/N)*[(N-1)/N]^3 ...
etc. Write as (1/N)*Sum(i=1,infinity) i [(N-1)/N]^(i-1)
This is the same as
N* d/dN{Sum(i=1,infinity) [(N-1)/N]^(i)}
which is the same as
N* d/dN {((N-1)/N)/(1-((N-1)/N))
from geometric sum. This just simplifies to
N* d/dN{N-1} = N* 1 = N
I think my approach is better because you don't need to assume that the sum exists and is finite. If you did then for example you can "prove" that if
S = 1 -1 + 1 - 1...
then S = 1 - S, giving S = 1/2 , which is at best misleading
https:/ /en.wik ipedia. org/wik i/Summa tion_of _Grandi %27s_se ries
S = 1 -1 + 1 - 1...
then S = 1 - S, giving S = 1/2 , which is at best misleading
https:/
another way of looking at this is too look at the average number of attempts figure for all possible items chosen.
If theres 1 item then its clearly 1.
if theres 2 items then 2 makes sense. If one is red and one is blue then it takes 2 go's on average to get a red (just as its 2 for a blue.). Could'nt be 1 for a red as it would have to be the same for blue too as there equally likely, but clearly one go can't give red AND blue, only red or a blue.
If 3 items (one red a blue &1 yellow) then 3 makes sense- 3 go's for a red, 3 for a blue, 3 for a yellow as there all equally likely .
Well makes sense to me in my mind but i realise I'm not explaining it well.
Maybe sciencenoob can explain it better but without all them complicated formulas which lost me?
If theres 1 item then its clearly 1.
if theres 2 items then 2 makes sense. If one is red and one is blue then it takes 2 go's on average to get a red (just as its 2 for a blue.). Could'nt be 1 for a red as it would have to be the same for blue too as there equally likely, but clearly one go can't give red AND blue, only red or a blue.
If 3 items (one red a blue &1 yellow) then 3 makes sense- 3 go's for a red, 3 for a blue, 3 for a yellow as there all equally likely .
Well makes sense to me in my mind but i realise I'm not explaining it well.
Maybe sciencenoob can explain it better but without all them complicated formulas which lost me?
https:/ /en.wik ipedia. org/wik i/Geome tric_di stribut ion
https:/ /www.cu emath.c om/geom etric-d istribu tion-fo rmula/
Maybe thinking in terms of dice rolls helps, if you roll a fair dice six times then you expect to hit each number exactly once in the first six rolls.
https:/
Maybe thinking in terms of dice rolls helps, if you roll a fair dice six times then you expect to hit each number exactly once in the first six rolls.
Your Answer.. If the number of goes is n for some value of n, then it is n for the next higher value of n because, if there are n+1 items you'll select the correct item on the first go 1 in n+1 times, otherwise (i.e. n out of n+1 times) you'll have to find it among the remaining n items, which will take n further goes on average, i.e. n+1 goes on total, including the go you've already had. Repeating this argument shows it'll be n for all higher values of n. But the number of goes is n when n is 1, so it is n for all positive integers.
so much groping around that it reminds me of - - the house of commons
ter saah ! TTT will confirm this is a paraprosdokian....
There is a probability function - the expectation
the chance of drawing three barls if you try 25 times
and here the chance of drawing any blu barl if the chancce is .01 and you try one hundred times is - you are likely to draw ONE blue barl
If you have 1 in 100 blues, and you draw fifty times
the chance of drawing one is 1/2
ter saah ! TTT will confirm this is a paraprosdokian....
There is a probability function - the expectation
the chance of drawing three barls if you try 25 times
and here the chance of drawing any blu barl if the chancce is .01 and you try one hundred times is - you are likely to draw ONE blue barl
If you have 1 in 100 blues, and you draw fifty times
the chance of drawing one is 1/2
on these figures
1/100 and 100 goes - you are very likely to get a POISSON distribution where the chances are really "no blue barl "and "one blue barl" - and there are no other relevant possibilities
This means if one has a chance of P(x) the other is 1-P(x)
yes I know I am courting screams of "who said blue barl what den?" from knowledgeable and articulate ABers
1/100 and 100 goes - you are very likely to get a POISSON distribution where the chances are really "no blue barl "and "one blue barl" - and there are no other relevant possibilities
This means if one has a chance of P(x) the other is 1-P(x)
yes I know I am courting screams of "who said blue barl what den?" from knowledgeable and articulate ABers