Here is a copy of the file :-)

L_.mw

Yes Robert Israel is a stable guy! Sometimes I dont understand
his way of thinking (I would assume too complex for my primal brain)
but he always goes the extra mile when it comes to answering queries.
This to me indicates that he has a genuine interest in helping other people.
This is a very rare and precious phenomenon especially in academia....

@Robert Israel   Ok, I think I understand where you get (k-1)/(b-1) from now.
If there are a total of b-1 values and only one is equal to the one we are after
then I understand that the probability must be 1/(b-1). Then if we have
n-1 such probabilities in a list the sum must be:

sum(1/(b-1), i = 1 .. n-1);
(1/(b-1))*(n-1);
 
                                    n - 1
                                    ----- 
                                    b - 1

                                    n - 1
                                    -----
                                    b - 1


ok, fair enough. However, the problem specification is still very confusing.

i) I thought b was the value of the best candidate ie 100. But then you write:
"On the other hand, if B = b >= k the strategy succeeds if
and only if the best of the first b-1 candidates is in a position < k."   Hummm

What I dont understand is:  If b>k how can the position of the best candidate be <k ??
Very complicated....Why is the position of the best b-1 candidates important?

ii)  Maybe it is to much to ask but could you write a ie timeline with
very clear definitions of the variables, their corresponding position and the
number of different scenarios.

1.....k....n      1.....k....b....n          1.....b....k....n       1......b-1....b....k.....n-1...n


k=threshold
n=total number
b= position best candidate I assume?!
b-1= ?
n-1=total number -1


I find the notation B, b, n, k, k-1  to be extremely confusing....he he

@Robert Israel   Ok, I think I understand where you get (k-1)/(b-1) from now.
If there are a total of b-1 values and only one is equal to the one we are after
then I understand that the probability must be 1/(b-1). Then if we have
n-1 such probabilities in a list the sum must be:

sum(1/(b-1), i = 1 .. n-1);
(1/(b-1))*(n-1);
 
                                    n - 1
                                    ----- 
                                    b - 1

                                    n - 1
                                    -----
                                    b - 1


ok, fair enough. However, the problem specification is still very confusing.

i) I thought b was the value of the best candidate ie 100. But then you write:
"On the other hand, if B = b >= k the strategy succeeds if
and only if the best of the first b-1 candidates is in a position < k."   Hummm

What I dont understand is:  If b>k how can the position of the best candidate be <k ??
Very complicated....Why is the position of the best b-1 candidates important?

ii)  Maybe it is to much to ask but could you write a ie timeline with
very clear definitions of the variables, their corresponding position and the
number of different scenarios.

1.....k....n      1.....k....b....n          1.....b....k....n       1......b-1....b....k.....n-1...n


k=threshold
n=total number
b= position best candidate I assume?!
b-1= ?
n-1=total number -1


I find the notation B, b, n, k, k-1  to be extremely confusing....he he

@Robert Israel   I still dont understand.  Why is it a ratio?
Is (k-1) the probability of success and (b-1) is the total amount?
I might be a bit slow but I found the notation to be very difficult.
Is there an easier way to explain it?

@Robert Israel   I still dont understand.  Why is it a ratio?
Is (k-1) the probability of success and (b-1) is the total amount?
I might be a bit slow but I found the notation to be very difficult.
Is there an easier way to explain it?

@Robert Israel  I still dont understand where you get the expression

(k-1)/ (b-1) from ie the probability of sucess if b>k...

is it 1/(b-1) / 1/(k-1).....   why is it a ratio?

@Robert Israel  I still dont understand where you get the expression

(k-1)/ (b-1) from ie the probability of sucess if b>k...

is it 1/(b-1) / 1/(k-1).....   why is it a ratio?

@PatrickT  Yes that sounds interesting!
The question is how do we find the probability of the second or third best etc
so we can calculate the expected value?

This was extreamly complicated stuff....
I thought I might take something away from it but it appears to be very difficult.
I am so confused that I dont even know what questions I should ask he he


Ok, I read somewhere that you can solve it recursivly.
The probability that the last person will be best is P(n)=1/n
but what is the probability that the second last person will be best P(n-1)=?
I read that it is supposed to be P(n-1)=1/n+(1-(1/(n-1)))*P(n)
but I have a hard time understanding where such an equation is coming from....

This was extreamly complicated stuff....
I thought I might take something away from it but it appears to be very difficult.
I am so confused that I dont even know what questions I should ask he he


Ok, I read somewhere that you can solve it recursivly.
The probability that the last person will be best is P(n)=1/n
but what is the probability that the second last person will be best P(n-1)=?
I read that it is supposed to be P(n-1)=1/n+(1-(1/(n-1)))*P(n)
but I have a hard time understanding where such an equation is coming from....

After I have looked at your solution I cant say that I understand that much :-)
I have number of questions:

1) What is the difference between B and b ?
2) If b>k where do you get the expression (k-1)/(b-1) from ?
3) Where do you get P(B=b)=n from ?
4) Where do you get P(A_k) = sum(P(A_k | B = b)*P(B=b), b=1..n) from ?
5) Why do you have b=k..n in the expression 1/n * sum((k-1)/(b-1), b=k .. n)
and not 1..n as in 4) ?
6) why do you use unapply in  P:= unapply(1/n*sum((k-1)/(b-1),b=k..n), k); ?
7) What does these expressions do:

map(simplify,asympt(P(n*exp(-1)+t), n, 3));
map(simplify,asympt(P(n*exp(-1) + 1 - 1/2*exp(-1)),n,3));
map(simplify,asympt(P(n*exp(-1)+1-1/2*exp(-1))-P(n*exp(-1)),n,3));

8) How is n*exp(-1) + 1 - exp(-1)/2  related to 

 0.9051258449    /1 \
                      ------------ + O|--|
                            2         | 3|
                           n          \n /

After I have looked at your solution I cant say that I understand that much :-)
I have number of questions:

1) What is the difference between B and b ?
2) If b>k where do you get the expression (k-1)/(b-1) from ?
3) Where do you get P(B=b)=n from ?
4) Where do you get P(A_k) = sum(P(A_k | B = b)*P(B=b), b=1..n) from ?
5) Why do you have b=k..n in the expression 1/n * sum((k-1)/(b-1), b=k .. n)
and not 1..n as in 4) ?
6) why do you use unapply in  P:= unapply(1/n*sum((k-1)/(b-1),b=k..n), k); ?
7) What does these expressions do:

map(simplify,asympt(P(n*exp(-1)+t), n, 3));
map(simplify,asympt(P(n*exp(-1) + 1 - 1/2*exp(-1)),n,3));
map(simplify,asympt(P(n*exp(-1)+1-1/2*exp(-1))-P(n*exp(-1)),n,3));

8) How is n*exp(-1) + 1 - exp(-1)/2  related to 

 0.9051258449    /1 \
                      ------------ + O|--|
                            2         | 3|
                           n          \n /

Have you added the ability to create a random variable within a default component in
MapleSim without having to create a custom component in Maple?

This is a very basic function which in my opinion has been overlooked in MapleSim

Thank you for that Robert!  Your explanations are very sophisticated.
I will post a reply as soon as I am done trying to understand what you
have written. I think I am starting to understand the dynamics very soon :-)