ChatterBank3 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

Im assuming you stop once you find the white one

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

S = 1 -1 + 1 - 1...

then S = 1 - S, giving S = 1/2 , which is at best misleading

https:/

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?

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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.

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

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