#  Look at discrete cases

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## Important Note:  Students are not expected ##
## to be able to come up with the ideas and   ##
## programming examples demonstrated on this  ##
## script.                                    ##
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#
#  Here we will look at a situation where there
#  are a relatively small number of possible 
#  outcomes and the probability of each outcome
#  is known.  Then we will ask questions about
#  more complex probabilities.
#
#  First we will set up the problem.  Let us 
#  say that there are 7 possible outcomes, 
#  each of which is a numeric value.
#  5, 8, 11, 12, 19, 23, and 28.
#  Then we need to know the probability of 
#  getting each outcome in a random event.
#
#  We could express those probabilities as 
#  decimals or as fractions. [Remember that we 
#  can always change from one form to the 
#  other.]  For us it will be easier to give 
#  the probabilities as fractions.
#  P(X=5)=1/6   P(X=8)=1/12   P(X=11)= 1/9
#  P(X=12)=7/36 P(x=19)=1/4   P(X=23)=1/18
#  P(X=28)=5/36
#  In order to deal with this it is 
#  easier if those fractions have a common
#  denominator.
#  P(X=5)= 6/36  P(X=8)=3/36   P(X=11)= 4/36
#  P(X=12)=7/36 P(x=19)=9/36   P(X=23)= 2/36
#  P(X=28)=5/36
#  And, of course, the sum of all those 
#  probabilities is 36/36=1
#
#  We can start by creating an "urn" that has
#  our values in the proportion given by their
#  respective probabilities.

our_values <- c(5, 8, 11, 12, 19, 23,  28)
our_values
#
#  Then we want to specify the desired
#  proportion, which is just the numerator
#  of those probabilities when expressed with
#  a common denominator.
#
our_proport <- c(6, 3, 4, 7, 9, 2, 5)
our_proport
#
#  Then we can fill our urn with the values we
#  want in the proportion we need
our_urn <- rep( our_values, our_proport)
our_urn

# We could demonstrate the model by taking a 
# sample of size 1 over and over

sample(our_urn, 1)  # do this line many times

#  we could take 36000 samples by getting 1000
#  random values in the range from 1 to 36 and
#  then taking the indicated value from our urn.
   # runif() is an R function to generate random values
   #   however, those values will be will be real numbers
   #   between the second and third parameters.  
   # we want integer values, so we use as.integer()
   #   to drop off any decimal portion of each 
   #   random value

which_items <- as.integer(runif(36000,1,37))
#  look at the first 50 of those 36000 index values
head( which_items,50)
#  now get our 36000 samples, each of size 1
big_sample <- our_urn[ which_items]
#  look at the first 50 of those 36000 samples
head( big_sample,50)
#  look at the distribution of values in our
#  big_sample
table( big_sample )

# as you can see, the urn reflects or 
# desired model.
#

# What is the expected value?
mean( our_urn )
#  check that out with reference to the
#  36000 items in our big sample
mean( big_sample ) #  why the difference

#  I have had some students who worry that our
#  urn with only 36 items in it may not be representative
#  of our discrete population. 
#  Therefore, look at a big urn
big_urn <- rep( our_urn,1000)
table( big_urn )
mean( big_urn)

# what is the standard deviation 
# of values in the urn.  For this we
# need to find the population standard
# deviation.  We can do this two ways
sd( our_urn )*sqrt(35/36)
# or
source("../pop_sd.R")
pop_sd( our_urn )

# also, note
pop_sd( big_urn )

############################
##  Look at a randomly generated problem
#
our_values <- sample(-7:19,8, replace=FALSE)
our_values
#  it is easier to follow these if they
#  have been sorted
our_values <- sort( our_values )
our_values

# establish some random proportions
our_proport <- sample(3:22, 8, replace=FALSE)
our_proport
sum( our_proport )
our_urn <- rep(our_values, our_proport)
our_urn
# find the expected value
mean( our_urn )
names( our_proport ) <- our_values
our_proport
our_probabilities <- our_proport/sum(our_proport)
our_probabilities

#  find the standard deviation of this model
pop_sd( our_urn )