Statistical Problem

[Rev. Green]

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.)

[bobbinwales]

so for example 100 equal (apart from colour) counters in a box, 99 blue, one white, how many go's on average to pick a white if blues always returned to the pot ? Is that what you mean

[Rev. Green]

Yep, good precis, bobbinwales.

[bobbinwales]

in my example I think its 100 go's on average.... so in your case it would be n?


Or is that oversimplistic

[Rev. Green]

Sorry, bobbinwales, 2 is not the answer when there are 2.

[bobbinwales]

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

[Rev. Green]

My mistake. It is 2.

[ToraToraTora]

there is 1/100 chance every time so it could gotten on go 1 or go 100 and any go in betwees so the average number of attempts will be 50. ie if you did it an infinite number of times the average would be 50.

[ToraToraTora]

so in general terms it's n/2

[bobbinwales]

so 1/2 x1 plus 1/4 x2 plus 1/8 x 3 plus 1/16 x4 plus 1/32 x 5 ...??

[bobbinwales]

Toratoratora... I may be out of my depth here or may of misunderstood but surely it could be sometimes 101 goes or 102 or...200 or..

[Rev. Green]

Think I've solved it. Let the answer = S. There is a 1/n chance of finding it on the first go, and a 1-1/n chance of needing a second go, for which the chance is S+1, because you'll use and extra go. So S=1/n+(1-1/n)(S+1). Therefore S=n.

[ScienceNoob]

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

[ScienceNoob]

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.wikipedia.org/wiki/Summation_of_Grandi%27s_series

[Rev. Green]

The average of 1 and 100 is 50.5, not 50 Toratoratora. That's (n+1)/2 as given in the question for the case when items are kept out after being selected.

[ToraToraTora]

bob 10:47, yes ignore me, I dived in too quick.

[bobbinwales]

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?

[ScienceNoob]

https://en.wikipedia.org/wiki/Geometric_distribution

https://www.cuemath.com/geometric-distribution-formula/

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.

[Rev. Green]

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.

[Peter Pedant]

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

[Peter Pedant]

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