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Jon Haugsand science forum beginner
Joined: 03 May 2005
Posts: 37

Posted: Tue Jun 14, 2005 6:31 pm Post subject:
Re: Probability question, was Re: Mensa Forgot Another Possibility!



* Scott Hemphill
Quote:  That's why I gave a reference. There are some smart and easy explanations
depending on your background. I'll give one below which uses generating
functions.

Thanks. I actually have a lot of background, but somehow generating
functions have escaped my education. Time to look into it.

Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no
http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92 

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Scott Hemphill science forum beginner
Joined: 09 Jun 2005
Posts: 21

Posted: Tue Jun 14, 2005 1:56 pm Post subject:
Re: Probability question, was Re: Mensa Forgot Another Possibility!



Scott Hemphill <hemphill@hemphills.net> writes:
Quote:  d(n) = n! sum(k=0,n) (1)^k/k!
= n! (e^1  sum(k>n) (1)^k/k!)
It's pretty easy to establish that this sum for k>n is less than 1/2
for n > 0, and since d(n) is an integer, it must be the one you get
when you round n!/e to the nearest integer. The expression has to
be fixed to get the correct answer for n = 0. Hence,

I meant to say that it's easy to establish that
n! sum(k>n) (1)^k/k! has absolute value less than 1/2 for n > 0.
Scott

Scott Hemphill hemphill@alumni.caltech.edu
"This isn't flying. This is falling, with style."  Buzz Lightyear 

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Scott Hemphill science forum beginner
Joined: 09 Jun 2005
Posts: 21

Posted: Tue Jun 14, 2005 1:41 pm Post subject:
Re: Probability question, was Re: Mensa Forgot Another Possibility!



Jon Haugsand <jonhaug@ifi.uio.no> writes:
Quote:  * Scott Hemphill
and subfactorial(n) is the number of derangements of n objects, i.e.
the number of permutations for which none of objects are in their
original position.
subfactorial(n) is further developed:
subfactorial(n) = round(n!/e) + [n==0]
where round is the function which rounds to the nearest integer
and the bracket notation yields 1 if the boolean expression inside
is true and 0 if the boolean expression is false. I've used ==
for the equality operator, and e is Euler's constant e = 2.71828....
Interesting, but hardly educational. This /is/ a mysterious formula.
Why does it work? I don't really want an answer (unless you have some
smart and easy explanation), but will try to search the answer for
myselv.

That's why I gave a reference. There are some smart and easy explanations
depending on your background. I'll give one below which uses generating
functions.
Quote:  Anyway, it might be useful to point out a recursive solution to the
whole problem:
h(n,k) = number of ways K hats land on a correct head out of N heads.
h(n,n) = 1
h(n,n1) = 0
h(n,k) = choose(n,k) * h(nk,0)
h(n,0) = n!  [ sum(i=1,n) h(n,i) ]
This formula is implementable on a computer, and follows naturally
from the problem description. It is of course the h(n,0) that is
different, i.e. your subfactorial.

Rewriting your last equation:
n! = sum(k=0,n) h(n,k)
Substituting your nexttolast equation:
n! = sum(k=0,n) choose(n,k) h(nk,0)
I'll notate h(n,0) as d(n), the number of derangements of n objects.
n! = sum(k=0,n) choose(n,k) d(nk)
1 = sum(k=0,n) 1/k! d(nk)/(nk)!
The sequence generated by the right half of this equation for n = 0, 1, ...
is the convolution of the sequences 1/n! and d(n)/n! Therefore the
generating function of the sequence for the left side of the equation,
(1, 1, ...) will be equal to the product of the generating functions
for 1/n! and d(n)/n!.
The generating function for (1, 1, ...) is sum(k>=0) 1*z^k = 1/(1z).
The generating function for 1/n! is sum(k>=0) 1/k! z^k = e^z.
Let the generating function for d(n)/n! be D(z).
Then
1/(1z) = e^z D(z), or D(z) = e^z/(1z)
D(z) = 1/(1z) (z^0/0!  z^1/1! + z^2/2!  ... )
The value of d(n)/n! will be the coefficient of z^n. Since
1/(1z) = 1 + z + z^2 + ..., the coefficient of z^n will be
sum(k=0,n) (1)^k/k!
So
d(n)/n! = sum(k=0,n) (1)^k/k!
Note that this sum goes to e^1 as n goes to infinity.
d(n) = n! sum(k=0,n) (1)^k/k!
= n! (e^1  sum(k>n) (1)^k/k!)
It's pretty easy to establish that this sum for k>n is less than 1/2
for n > 0, and since d(n) is an integer, it must be the one you get
when you round n!/e to the nearest integer. The expression has to
be fixed to get the correct answer for n = 0. Hence,
d(n) = round(n!/e) + [n==0]
Scott

Scott Hemphill hemphill@alumni.caltech.edu
"This isn't flying. This is falling, with style."  Buzz Lightyear 

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Jon Haugsand science forum beginner
Joined: 03 May 2005
Posts: 37

Posted: Tue Jun 14, 2005 12:16 pm Post subject:
Re: Probability question, was Re: Mensa Forgot Another Possibility!



* Scott Hemphill
Quote:  and subfactorial(n) is the number of derangements of n objects, i.e.
the number of permutations for which none of objects are in their
original position.
subfactorial(n) is further developed:
subfactorial(n) = round(n!/e) + [n==0]
where round is the function which rounds to the nearest integer
and the bracket notation yields 1 if the boolean expression inside
is true and 0 if the boolean expression is false. I've used ==
for the equality operator, and e is Euler's constant e = 2.71828....

Interesting, but hardly educational. This /is/ a mysterious formula.
Why does it work? I don't really want an answer (unless you have some
smart and easy explanation), but will try to search the answer for
myselv.
Anyway, it might be useful to point out a recursive solution to the
whole problem:
h(n,k) = number of ways K hats land on a correct head out of N heads.
h(n,n) = 1
h(n,n1) = 0
h(n,k) = choose(n,k) * h(nk,0)
h(n,0) = n!  [ sum(i=1,n) h(n,i) ]
This formula is implementable on a computer, and follows naturally
from the problem description. It is of course the h(n,0) that is
different, i.e. your subfactorial.

Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no
http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92 

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Martin Evans science forum beginner
Joined: 10 Jun 2005
Posts: 3

Posted: Tue Jun 14, 2005 10:07 am Post subject:
Re: Probability question



Scott Hemphill wrote:
Quote:  Scott Hemphill <hemphill@hemphills.net> writes:
Martin Evans <Martin.Evans@arm.com> writes:
Scott Hemphill wrote:
Pr = (TR1)! / (R! * (TR)!)
And what is the probability for 19 right answers out of 20? Hint: where
does the 20th answer go?
For T = 20 and R = 19 we get Pr = (20  19  1)!/(19! * (20  1)!) = 0!/19! = 0
I think this is what we would expect  the probablity of getting exactly 19 questions
right would be zero, wouldn't it? PS  I don't need the hint, thanks! :P
No. 0! is not zero. It is one.
I mean, yes the probability is zero. But, no your formula doesn't produce
that result.
Scott

Scott Hemphill hemphill@alumni.caltech.edu
"This isn't flying. This is falling, with style."  Buzz Lightyear

I had forgotten that 0! is not zero  it's all too long ago!
There is a problem with the formula I derived anyway, as you have correctly said in another
posting, but I haven't had the time to look at it yet! I need to go round the loop again,
but I think the basic approach is right. 

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Scott Hemphill science forum beginner
Joined: 09 Jun 2005
Posts: 21

Posted: Mon Jun 13, 2005 12:45 pm Post subject:
Re: Probability question



Scott Hemphill <hemphill@hemphills.net> writes:
Quote:  Martin Evans <Martin.Evans@arm.com> writes:
Scott Hemphill wrote:
Pr = (TR1)! / (R! * (TR)!)
And what is the probability for 19 right answers out of 20? Hint: where
does the 20th answer go?
For T = 20 and R = 19 we get Pr = (20  19  1)!/(19! * (20  1)!) = 0!/19! = 0
I think this is what we would expect  the probablity of getting exactly 19 questions
right would be zero, wouldn't it? PS  I don't need the hint, thanks! :P
No. 0! is not zero. It is one.

I mean, yes the probability is zero. But, no your formula doesn't produce
that result.
Scott

Scott Hemphill hemphill@alumni.caltech.edu
"This isn't flying. This is falling, with style."  Buzz Lightyear 

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Scott Hemphill science forum beginner
Joined: 09 Jun 2005
Posts: 21

Posted: Mon Jun 13, 2005 12:39 pm Post subject:
Re: Probability question



Martin Evans <Martin.Evans@arm.com> writes:
Quote:  Scott Hemphill wrote:
Pr = (TR1)! / (R! * (TR)!)
And what is the probability for 19 right answers out of 20? Hint: where
does the 20th answer go?
For T = 20 and R = 19 we get Pr = (20  19  1)!/(19! * (20  1)!) = 0!/19! = 0
I think this is what we would expect  the probablity of getting exactly 19 questions
right would be zero, wouldn't it? PS  I don't need the hint, thanks!

No. 0! is not zero. It is one.
Scott

Scott Hemphill hemphill@alumni.caltech.edu
"This isn't flying. This is falling, with style."  Buzz Lightyear 

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Martin Evans science forum beginner
Joined: 10 Jun 2005
Posts: 3

Posted: Mon Jun 13, 2005 11:09 am Post subject:
Re: Probability question



Scott Hemphill wrote:
Quote:  Pr = (TR1)! / (R! * (TR)!)
And what is the probability for 19 right answers out of 20? Hint: where
does the 20th answer go?

For T = 20 and R = 19 we get Pr = (20  19  1)!/(19! * (20  1)!) = 0!/19! = 0
I think this is what we would expect  the probablity of getting exactly 19 questions
right would be zero, wouldn't it? PS  I don't need the hint, thanks! :P
Quote: 
[1] As this is cam.misc, that is an absolute certainty, whether I am right or
wrong! :)
There are a few more newsgroups than just cam.misc.

Yes, of course there are. 

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alan truelove science forum beginner
Joined: 09 Jun 2005
Posts: 8

Posted: Sat Jun 11, 2005 5:23 am Post subject:
Re: Probability question



Piled higher and even more deeper ..
Develop iterative method for the following
and then show a VB.Net code to implement it ..
(I have a nice solution which goes on from my previous solutions, and
uses a neat combinatorial method, but this space is too small to
contain it)
OK, it's June 11th, 2005, let's just see...
..            
k groups of n objects each are presented; select one object from each
group (a 'tuple')
Given the first object, the (correct) other k1 are unique.
Repeat until all tuples have been selected.
P(k,n,x) is prob(get x correct tuples) 

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Alex Selby science forum beginner
Joined: 30 May 2005
Posts: 10

Posted: Fri Jun 10, 2005 9:53 pm Post subject:
Re: Probability question



alan truelove wrote:
Quote:  From rec.org.mensa
anyone got a slicker solution ??
'            
On Thu, 9 Jun 2005 13:56:11 +0200, "Mark"
Markiehatesspam@notelespam2.fr> wrote:
"BruceS" <bruces42@hotmail.com> wrote in message
news:1118264621.238754.254340@f14g2000cwb.googlegroups.com...
I wouldn't be at all offended if you reposted there and returned here
with any answers. I try to avoid crossposting, especially to groups I
don't follow.
Here then is the question as I remember it:
on consideration this may not be precisely on topic for sci.stat.math
although I suppose that in general people who are up on stats are up on
probability too.
A student sits a test that has 20 questions; every question has precisely 1
matching answer. There are 20 answers, no duplication. The student answers
every question with a different answer from the 20.
What is the probability, of a student answering randomly, answering
precisely 1 right? precisely 10 right?
Mark
'   
Getting 1 right: 0.3679
Getting 10 right: 0.01378 * 10^(7)
'   

The probability of getting r right out of n is D(nr)/(r!(nr)!)
where D(k) = / 1 if k=0
\ the nearest integer to k!/e if k>0
This is well approximated by 1/(r!e) if r isn't too close to n.
(Reference: Google for "hats","derangements","Poisson","inclusion")

Alex 

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Scott Hemphill science forum beginner
Joined: 09 Jun 2005
Posts: 21

Posted: Fri Jun 10, 2005 7:31 pm Post subject:
Re: Probability question



Martin Evans <Martin.Evans@arm.com> writes:
Quote:  Then we have TR columns left, each of which has the cell on the leading
diagonal forbidden to us, as well as the rows already containing 'x's for right
answers. The first column we add an 'x' to would thus have (TR1) valid cells
available. So we would have No = (TR1)! using a similar argument as that for
Nt.

This is where the error is. The first column will indeed have (TR1) valid
cells. However the second column and each additional column will will either
have (TR2) valid cells or (TR1) valid cells depending on whether the
corresponding row has already been picked.
For example, consider the case where TR = 3, and just look at the 3x3 matrix
of containing the remaining rows and columns.
X . .
X . O = chosen, X = unavailable
O . X
Working column by column, if we pick the third row for the first column,
then there is only one choice for the second column.
X . .
O X .
. . X
If we instead pick the second row for the first column, then there are
two choices for the second column.
Quote:  The chance, Pr, of getting *exactly* R right answers in T tests is therefore
Nr/Nt = (Nd * No) / Nt = (T! * (TR1)!) / (T! * (TR)! * R!)
Simplifying the above expression, we get:
Pr = (TR1)! / (R! * (TR)!)

This formula can't be right. What if T=R? You get a (1)! term (which is
infinite).
If you have a correct formula, you should expect it to get the correct
results for easily verifiable simple cases:
T = 0
R = 0, Pr = 1
T = 1
R = 0, Pr = 0
R = 1, Pr = 1
T = 2
R = 0, Pr = 1/2
R = 1, Pr = 0
R = 2, Pr = 1/2
T = 3
R = 0, Pr = 1/3
R = 1, Pr = 1/2
R = 2, Pr = 0
R = 3, Pr = 1/6
T = nonnegative integer
R = T1, Pr = 0
R = T, Pr = 1/T!
You should also expect that the sum of the probabilities for a given T
will be one!
Scott

Scott Hemphill hemphill@alumni.caltech.edu
"This isn't flying. This is falling, with style."  Buzz Lightyear 

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alan truelove science forum beginner
Joined: 09 Jun 2005
Posts: 8

Posted: Fri Jun 10, 2005 5:56 pm Post subject:
Re: Probability question



(Correctedt code!)
 but I took T=4, and got a total for all
Quote:  probs R=0,1,2,3,4 of 0.5833
Shomething wrong shurely!
(I may well have made a dumb mistake ..)
with best wishes
'   
Public Sub Button1_Click(ByVal sender As System.Object, ByVal e As 
System.EventArgs) Handles Button1.Click
Dim i, j, iii, jjj, ii, jj As Integer
ReDim P(100, 100)
ReDim Q(100, 100)
Dim T, R, Rfact, TRfact, TR1fact As Integer
Dim tot, hold As Single
T = 4
tot = 0
For R = 0 To T
TRfact = fact.fact(T  R)
TR1fact = fact.fact(T  R  1)
Rfact = fact.fact(R)
'(TR1)! / (R! * (TR)!)
hold = Rfact * TRfact
If hold > 0 Then tot = tot + TR1fact / hold
Next
MsgBox(tot)
Quote:  '   
Module fact
Function fact(ByVal j As Integer) As Integer
Dim i As Integer
fact = 1
If j = 0 Then
fact = 0
Exit Function
End If
For i = 2 To j
fact = fact * i
Next
End Function
End Module 


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alan truelove science forum beginner
Joined: 09 Jun 2005
Posts: 8

Posted: Fri Jun 10, 2005 4:29 pm Post subject:
Re: Probability question



Quote:  we get:
Pr = (TR1)! / (R! * (TR)!)
For T = 20, R = 1 I get Pr = 18! / (1! * 19!) = 1/19 = 0.0526
For T = 20, R= 10 I get Pr = 9!/(10! * 10!) = 1/(10 * 10!) = 2.76 x 10^8
I am sorry that my answers differ from the previously quoted ones! It is of
course possible that the above is in error, and maybe[1] one of the real
mathematicians here will correct me in that case. Even if the above is wrong,
the method may be of interest. Or, of course, it may be right!
   
I applaud your efforts  but I took T=4, and got a total for all
probs R=0,1,2,3,4 of 6.25
Shomething wrong shurely!
(I may well have made a dumb mistake ..)
with best wishes
'   
Public Sub Button1_Click(ByVal sender As System.Object, ByVal e As
System.EventArgs) Handles Button1.Click
Dim i, j, iii, jjj, ii, jj As Integer
ReDim P(100, 100)
ReDim Q(100, 100)
Dim T, R, Rfact, TRfact, TR1fact As Integer
Dim tot, hold As Single
T = 4
TRfact = fact.fact(T  R)
TR1fact = fact.fact(T  R  1)
tot = 0
For R = 0 To T
Rfact = fact.fact(R)
TRfact = fact.fact(T  R  1)
'(TR1)! / (R! * (TR)!)
hold = Rfact * TRfact
If hold > 0 Then tot = tot + TR1fact / hold
Next
MsgBox(tot)
end sub
'   
Module fact
Function fact(ByVal j As Integer) As Integer
Dim i As Integer
fact = 1
If j = 0 Then
fact = 0
Exit Function
End If
For i = 2 To j
fact = fact * i
Next
End Function
End Module 

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Scott Hemphill science forum beginner
Joined: 09 Jun 2005
Posts: 21

Posted: Fri Jun 10, 2005 4:21 pm Post subject:
Re: Probability question



Martin Evans <Martin.Evans@arm.com> writes:
Quote:  alan truelove wrote:
From rec.org.mensa
anyone got a slicker solution ??
'            
A student sits a test that has 20 questions; every question has precisely 1
matching answer. There are 20 answers, no duplication. The student answers
every question with a different answer from the 20.
What is the probability, of a student answering randomly, answering
precisely 1 right? precisely 10 right?
I think one possible way of visualising this problem is to think of it as a
20x20 matrix, with the rows representing the questions and the columns
representing the answers; a cell is marked (with an 'x', say) to show that a
certain question has been assigned to a certain answer. A valid test outcome
has each row and each column containing exactly one 'x'. An 'x' on the leading
diagonal represents a right answer.
Let's generalise the problem to T tests and R right answers, and let's suppose
we don't have to consider the order in which the questions are answered  we are
only interested in the outcome.
The total number of possible test outcomes is thus Nt = T! This is arrived at
by arguing that the 'x' in the first column has a choice of 20 cells, the second
a choice of 19 cells etc.
The number of ways, Nr, of generating exactly R right answers is the product of
the number of ways of scattering R 'x's down the leading diagonal, Nd, and the
number of ways of filling in the remaining columns with 'x's *off* the leading
diagonal, No.
We would have Nd = T!/((TR)! * R!) as this is the number of different ways of
arranging R things in T places, regardlessof order.
Then we have TR columns left, each of which has the cell on the leading
diagonal forbidden to us, as well as the rows already containing 'x's for right
answers. The first column we add an 'x' to would thus have (TR1) valid cells
available. So we would have No = (TR1)! using a similar argument as that for
Nt.
The chance, Pr, of getting *exactly* R right answers in T tests is therefore
Nr/Nt = (Nd * No) / Nt = (T! * (TR1)!) / (T! * (TR)! * R!)
Simplifying the above expression, we get:
Pr = (TR1)! / (R! * (TR)!)
For T = 20, R = 1 I get Pr = 18! / (1! * 19!) = 1/19 = 0.0526
For T = 20, R= 10 I get Pr = 9!/(10! * 10!) = 1/(10 * 10!) = 2.76 x 10^8
I am sorry that my answers differ from the previously quoted ones! It is of
course possible that the above is in error, and maybe[1] one of the real
mathematicians here will correct me in that case. Even if the above is wrong,
the method may be of interest. Or, of course, it may be right!

And what is the probability for 19 right answers out of 20? Hint: where
does the 20th answer go?
Quote:  [1] As this is cam.misc, that is an absolute certainty, whether I am right or
wrong!

There are a few more newsgroups than just cam.misc.
Scott

Scott Hemphill hemphill@alumni.caltech.edu
"This isn't flying. This is falling, with style."  Buzz Lightyear 

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alan truelove science forum beginner
Joined: 09 Jun 2005
Posts: 8

Posted: Fri Jun 10, 2005 4:03 pm Post subject:
Re: Probability question



Piled higher and deeper ...
3 groups of n objects each are presented; select one object from each
group (a 'triad')
Given the first object, the (correct) other two are unique.
Repeat until all triads have been selected.
(Note  this setup is modified in 'Q' and 'R' below for
computational purposes..)
P(n,x) is prob. of getting exactly x correct triads
Q(n,y) is prob. of getting y triads correct, with in each group  n
'good' objects ,plus one 'dummy' object which cannot form part of a
correct triad.
R(n,z) is prob. of getting z triads correct, with in each group  n
'good' objects + TWO dummy objects.
P(n,x)=P(n1,x1)/n^2 + 3(n1)Q(n2,x)/n^2 + (n1)(n2)R(n3,x)/n^2
Q(n,y)=P(n,y)/(n+)^2 + 3nQ(n1,y)/(n+1)^2 + n(n1)R(n2,y)/(n+1)^2
R(n,z)=4 Q(n,z)/(n+2)^2 + (n^2 + 4n)R(n1,z)/(n+2)^2
'       
'match across 3 sets of 20
' alan j truelove 6/10/05
'571 242 0153 'alan_truelove@hotmail.com
Imports System.Math
Public Class Form1
Inherits System.Windows.Forms.Form
Public Sub Button1_Click(ByVal sender As System.Object, ByVal e As
System.EventArgs) Handles Button1.Click
Dim i, j, iii, jjj, ii, jj As Integer
ReDim P(100, 100)
ReDim Q(100, 100)
ReDim R(100, 100)
'P(n,x)=P(n1,x1)/n^2 + 3(n1)Q(n2,x)/n^2 +
(n1)(n2)R(n3,x)/n^2
'Q(n,y)=P(n,y)/(n+)^2 + 3nQ(n1,y)/(n+1)^2 +
n(n1)R(n2,y)/(n+1)^2
'R(n,z)=4 Q(n,z)/(n+2)^2 + (n^2 + 4n)R(n1,z)/(n+2)^2
P(1, 1) = 1
P(1, 0) = 0
Q(1, 1) = 0.25
Q(1, 0) = 0.75
Q(0, 0) = 1
R(1, 1) = 1 / 3
R(1, 0) = 2 / 3
R(0, 0) = 1
'      
For n = 2 To 20
'  
P(n, n) = 1
For jj = 2 To n
P(n, n) = P(n, n) * jj
Next
P(n, n) = 1 / P(n, n) ^ 2
'  
P(n, n  1) = 0
pstr = n & " p " & P(n, n)
ptot = P(n, n)
For x = n  1 To 0 Step 1
' if we got a hit first
If x < n  1 Then 'not needed for P(n,n1)
P(n, x) = 0
If (x > 0) Then P(n, x) = P(n  1, x  1) / n ^ 2
If n > 1 And x < n  1 Then P(n, x) = P(n, x) + _
3 * (n  1) * Q(n  2, x) / n ^ 2
If n > 2 Then P(n, x) = P(n, x) + _
(n  1) * (n  2) * R(n  3, x) / n ^ 2
End If
' Q(n2,x) is prob  with n2 questions and 1 dummy 
' we match correctly to x out of n1 answers plus one
dummy
' the answer dummy comes because we just used up the
right question for it,
pstr = pstr & " " & Round(P(n, x), 3)
ptot = ptot + P(n, x)
Next
MsgBox(ptot & " P, n " & pstr)
'   
qstr = ""
qtot = 0
For y = n To 0 Step 1
' and the question dummy comes because we have knocked
out the right answer to it
Q(n, y) = P(n, y) / (n + 1) ^ 2
If n > 1 And y < n Then Q(n, y) = Q(n, y) + 3 * n *
Q(n  1, y) / (n + 1) ^ 2
If n > 1 And y < n  1 Then Q(n, y) = Q(n, y) + n * (n
 1) * R(n  2, y) / (n + 1) ^ 2
'if we do match 'on the first dummy question  try,
prob 1/(N+1), then we are left we
' n good questions and n good answers
'n/(n+1) is the prob of not matching dummies
' if we dont match, then we have lost the question
dummy, but another has been created
' it doesnt have a correct answer anywhere
' so we are at n1 questions + 1 dummy
'we have knocked out a good answer (for the new dummy
question) so we have n1 answers
' plus the original dummy
qstr = qstr & " " & Round(Q(n, y), 3)
qtot = qtot + Q(n, y)
Next
MsgBox(qtot & " Q, n " & qstr)
'    
rstr = ""
rtot = 0
For z = n To 0 Step 1
R(n, z) = 4 * Q(n, z) / (n + 2) ^ 2
If z < n Then R(n, z) = R(n, z) + (n ^ 2 + 4 * n) *
R(n  1, z) / (n + 2) ^ 2
rtot = rtot + R(n, z)
rstr = rstr & " " & Round(R(n, z), 3)
Next
MsgBox(rtot & " R, n " & rstr)
Next
End
End Sub
End Class
'    
Module global
Public Q(,), P(,), R(,), ptot, qtot, rtot As Single
Public n, x, y, z As Integer
Public pstr, qstr, rstr As String
End Module 

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