#This is a script to demonstrate some aspects
#   of topic 11, probability.

################################################
## Important Note:  Students are not expected ##
## to be able to come up with the ideas and   ##
## programming examples demonstrated on this  ##
## script.                                    ##
################################################
#
#
#   First, we want to install gnrnd4(), just in case
#   we need it.
source("../gnrnd4.R")
#
# to demonstrate a deterministic  task just look at
#    your algorithm (the set of steps you follow) for doing 
#    long division.  Try doing 507605 / 13
#          39046 R 7
#       +-------- 
#    13 | 507605
#         39
#         --
#         117
#         117
#         ---
#           06
#            0
#           --
#            60
#            52
#            --
#             85
#             78
#             --
#              7
# You follow those prescribed steps and each time
# you get the same answer
#  
# of course we could do this in R
as.integer( 507605/13)
507605 - 39046*13
# or better for finding the remainder
507605 %% 13
#
#  Here is another deterministic task
#     ( remember that we already loaded gnrnd4)
gnrnd4(873461401,  400001)
L1
#   You can run those two last lines any number of
#   times and you always get the same 15 values:
#   three 1's, six 2's, two 3's, and four 4's,
#   and you get them in the same order:
#   4 4 4 2 2 2 4 2 3 1 3 1 2 2 1

#  here is a  non-deterministic  task
L2 <- as.integer( runif(15,1,5)  )
L2
#  Every time you run those last two lines you
#  get a different listing of 15 values, each
#  of which is between 1 and 4, inclusive.


    ###################################
    ## The following is beyond the   ##
    ## expectations for what you     ##
    ## should be able to do for this ##
    ## class.                        ##
    ###################################

#  As described in the web page,
#  here is a  Risk "battle" between 
#  3 attacking red armies and 2 defending
#  blue armies.  Run the next  line
#  each time there is a battle
{
blue_die<- as.integer(runif(2,1,7))
red_die <- as.integer(runif(3,1,7))
blue_die <-sort( blue_die, decreasing=TRUE)
red_die <- sort( red_die, decreasing=TRUE)
print( c(" red=",red_die))
print( c("blue=",blue_die))
R<-0
B<-0
if( red_die[1] > blue_die[1] )
{ B<- -1} else { R <- -1}
if( red_die[2] > blue_die[2] )
{ B<- B - 1} else { R <- R -1}
print( c("red=",R,"blue=",B))
}  # end of block


###################################
## The following is beyond the   ##
## expectations for what you     ##
## should be able to do for this ##
## class.                        ##
###################################

#  Demo of a sample space
#
#  make a table of outcomes from two
#  lists of possible values
{
m_list <- 4:19
t_list <- sort(sample( m_list, 6))
s_list <- sort(sample( m_list, 6))
t_list
s_list
demo_table <- matrix(0,nrow=6, ncol=6)
rownames( demo_table ) <- s_list
colnames( demo_table ) <- t_list
demo_table
for( i in 1:6 )
{ k <- as.integer( t_list[i]/3 )
  for( j in 1:6)
    { m <- as.integer(s_list[j]/2)
      demo_table[j,i] = k*m
  
    }
}
print("top/3 * s/2 with integer division")
demo_table
}
# now find the frequency of each different
#  table value 
table( demo_table )
#  using that we can express the
#  probability of getting any particular
#  result.  

##  Look ways of counting
#
#   straight multiplication
#
#   example, we have 4 things in group 1
#            we have 3 things in group 2
#            we have 5 things in group 3
#   how many arrangements are there of 
#   a thing from group 1, a thing from 
#   group 2, and a thing from group 3?
#
# conceptually, that will be
4*3*5

###################################
## The following is beyond the   ##
## expectations for what you     ##
## should be able to do for this ##
## class.                        ##
###################################

#
#  we could construct such a situation
#  and then create a listing of all such 
#  threesomes
group_1 <- c("brown", "black", "gray",
             "tan")
group_2 <- c("big","little", "tiny")
group_3 <- c("mouse","cat","dog",
             "hamster","gerbil")
#
{ L4 <- NULL 
  for( i in 1:4)
  {  g_1 <- group_1[i]
     for ( j in 1:3)
     { g_2 <- group_2[j]
        for ( k in 1:5 )
        { new_item <- paste(g_1,g_2,
                            group_3[k])
          L4 <- c(L4,new_item)
        }
     }
  }
L4
}


#################
#  permutations
#
#  Here we start with a fixed number of 
#  different things and we ask for the
#  number of ways to arrange some 
#  selection of those things.
#  
#  for example, we had group 1 above
group_1
#  how many different ways are there 
#  to arrange those four things

# to do this we want to use the R command
# permutations, but that is not a built-in
# function.  We will need to load it 
# from the web.  This may take a moment or
# two ...
install.packages("gtools")
library(gtools)
# now we can use the function.
# Note that in this approach we do not
# see the function listed in our environment.
permutations( 4, 4, group_1)

# a slightly different question
# could be raised if we do not want to
# use all of the items in our arrangement:
#
# Show all of the arrangements of the
# values in group 1 if we only take 2
# at a time
permutations(4, 2, group_1)
#  What if we look at group_3.  
#
# Show all of the arrangements of the
# values in group 3 if we only take 2
# at a time
permutations(5, 2, group_3)

# it was interesting to use permutations()
# to produce all of the possible 
# arrangements.  From the list we can see
# the actual number of such arrangements.
# However, we want to get that number
# of arrangements without getting the list.
#
# Look at the web pages for the math behind
# this, but we will just use the function
# nPr() or num_perm() to do this.
# These are functions that I have provided.
source("../permutations.R")
#  we could find the number of permutations
#  of 5 things taken two at a time via
nPr(5,2)
# or by using the alternative function
num_perm(5,2)
     ######################################
     ##  the two functions merely arrive ##
     ##  at the answer using different   ##
     ##  approaches (i.e., algorithms)   ##
     ######################################

# How many permutations are there for 
#     15 things taken 7 at a time?
nPr(15,7)
#  As you can see we really would not want 
#  to produce the list of all those
#  32,432,400 things just to find out
#  how may permutations there are.

   ########################################
   ##  You are expected, in this course, ##
   ##  to be able to find the number of  ##
   ##  permutations of n things taken r  ##
   ##  at a time.  I suggest using nPr() ##
   ##  or num_perm() to do this.         ##
   ########################################

#  find the number of permutations of 24
#  things take 5 at a time
num_perm( 24, 5 )

############################
#    combinations
#
###################################
## The following is beyond the   ##
## expectations for what you     ##
## should be able to do for this ##
## class.                        ##
###################################

#  When we look at combinations we 
#  disregard the order of the items.
#  so in group 1 we had 4 things:
group_1
#  if we ask for the permutations of those 
#  4 things taken 2 at a time we can get
permutations(4,2,group_1)
#  but notice that we have both 
# [1,] "black" "brown"
# and
# [4,] "brown" "black"
#  These have the same items, just in 
#  a different order
# For combinations we do not want to 
# see these as different results.  
# We can use the function combinations()
# to get the list of such combinations.
#
# Note that combinations was loaded at 
# the same time we got permutations.

combinations(4,2,group_1)
#
# remember finding permutations of
# the 5 things in group 3 taken 3 at a time
permutations( 5, 3, group_3)
# but the combination of cat dog gerbil
# appears 6 times as
# [1,] "cat"     "dog"     "gerbil"
# [4,] "cat"     "gerbil"  "dog"
#[13,] "dog"     "cat"     "gerbil"
#[16,] "dog"     "gerbil"  "cat"
#[25,] "gerbil"  "cat"     "dog"
#[28,] "gerbil"  "dog"     "cat"
#
# In fact each item in the permutation
# list appears 6 times in the list if
# we disregard order.
# Look at the combinations of 5 things
# taken 3 at a time from group 3
combinations( 5, 3, group_3)
#
#  Again, we rarely want to see all of 
#  the combinations.  Usually, we just
#  want to find the number of combinations
#  of n things taken r at a time.
#
#  See the web pages for the math behind 
#  this, but we have two functions, nCr()
#  and num_comb() that just find the 
#  desired number, although they arrive
#  at the answer in slightly different
#  ways.
#
#  We need to load these functions
#
source("../combinations.R")
# Then, the number  of combinations of 
#   5 things taken 3 at a time is 
nCr(5,3)
# or we could do
num_comb( 5, 3 )

# How many combinations are there of 27
#  things taken 5 at a time?
nCr( 27,5 )
# How many combinations are there of 27
#  things taken 22 at a time?
nCr( 27,22 )
#  For discussion: why did we get the
#  same answer for the two cases?
#
# How many combinations are there of 31
#  things taken 6 at a time?
nCr( 31, 6 )
# How many combinations are there of 31
#  things taken 25 at a time?
nCr( 31, 25 )

########################################
##  You are expected, in this course, ##
##  to be able to find the number of  ##
##  combinations of n things taken r  ##
##  at a time.  I suggest using nCr() ##
##  or num_comb() to do this.         ##
########################################


########################################
##  Just for completeness, you might  ##
##  note that doing the               ##
##  source("../combinations.R")       ##
##  actually loads all 4 functions    ##
##  nPr(), nCr(), num_perm() and      ##
##  num_comb().                       ##
########################################


#########################################
#  Contingency tables
#
###################################
## The following is beyond the   ##
## expectations for what you     ##
## should be able to do for this ##
## class.                        ##
###################################
#
#  Let us make up one
gnrnd4(187341901, 63500300)
L1
c_table <- matrix( L1, nrow=4)
c_table
rs <- rowSums( c_table )
c_table <- cbind( c_table, rs)
c_table
cs <- colSums( c_table )
c_table <- rbind( c_table, cs)
c_table
rownames( c_table) <- c("AA","AAA","C","D","Total")
colnames( c_table) <- c("Ever","Rayo","Maxw",
                        "ProC","Amaz","total")
c_table
##########################################
##  Now that we have the table you are  ##
##  expected to be able to use the      ##
##  values in the table to answer some  ##
##  probability questions.              ##
##########################################

#  if an item is selected at random,
#     what is the probability that 
#     it as a AA package?  
#                 P(AA)=2670/12587
#     it is a ProC package? 
#                 P(Proc)=2242/12587
#     it is a Rayo C package?
#                 P( Rayo and C)=546/12587
#     it is a Rayo or a C Package?
#                 P( Rayo or C)=(2643+3331-546)/12587
#     it is a Rayo or ProC package?
#                 P( Rayo or ProC)=(2643+2242)/12587
#     it is Not a D package?
#                 P( D' )=(12587-3438)/12587
#     it is Not a Amaz package?
#                 P( Amaz' ) = 1 - 2704/12587
#     it is neither a 
#          C nor Ever package?
#                 P( C' or Ever')=(12587-2695-3331+662)/12587
#     given it is a D it is a Maxw 
#          package?
#                 P( Maxw | D ) =671/3438
#     given that it is a ProC it is a AAA
#          package?
#                 P( AAA | Proc ) = 468/2242