the big picture:

The Big Picture:

• Counting events in a sample space allows us to calculate probabilities• The key to calculating the probabilities of

events is to count the occurrences of events• Our goal is to calculate probabilities of ALL

possible events• We can do this by counting the total number of

events possible in the sample space, using combinatorial mathematics to do this

Permutations Permutation is a sequence or ordering of

events. Basic Question: if I have N objects, how

many different orderings of them are there? N! Formula: N(N-1)(N-2)…(1) Example: 5(5-1)(5-2)(5-3)(5-4)

5*4*3*2*1 = 120

Permutations General formula for finding the number of

permutations of size k taken from n objects

PkN

N!

(N k)!

Example

e.g. 10 compact discs and a 6 disc carousel- What is the number 6 disc orderings that we can make from the 10 cds?:

10! = 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

(10 - 6)! 4! 4 x 3 x 2 x 1

= 10 x 9 x 8 x 7 x 6 x 5 = 151200

PkN

N!

(N k)!

Combinations

General formula for finding the number of unique combinations of k objects you can choose from a set of n objects

CkN

N!

k!(N k)!

Example

e.g. How many sets of 6 discs can we make from 10 cds (without repeating the same combinations)?

10! = 10! =10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 6!(10 - 6)! 6!(4!) (6 x 5 x 4 x 3 x 2 x 1)( 4 x 3 x 2 x 1)

= 10 x 9 x 8 x 7 = 5040 = 210(4 x 3 x 2 x 1) 24

CkN

N!

k!(N k)!

Note the distinction Before we were counting our cd groupings without

regard to the same groupings of a different order But when looking for combinations, similar

groupings are not counted as different Pavement, The New Year, Built to Spill, Pixies, Grifters,

Sonic Youth Built to Spill, Grifters, Pavement, Sonic Youth, The New

Year, Pixies 2 permutations, but 1 combination (same set of 6 are involved)

Bernoulli Trials Bernoulli trials = 2 mutually exclusive

outcomes Distribution of outcomes

e.g. the probability of various outcomes in terms of numbers of heads and tails

Order of items does not matter

N = # trials = 3

Coin toss How many possible outcomes of the 3 coin tosses

are there? List them out:

HHH HHT HTT TTT TTH THH THT HTH Now condense them ignoring order

i.e. HTT = THT = TTH Only one occurrence of heads on 3 trials

Probability of 0 heads, only 1 heads, 2 heads, all 3 heads?

Distribution of outcomes

Number of Heads

3.02.01.00.0

COIN FLIPSF

req

ue

ncy

4.0

3.0

2.0

1.0

0.0

Now how about 10 coin flips? That’d be a lot of work writing out all the possibilities. What’s another way to find the probability of coin flips?

Binomial distribution

Find a probability for an event using:

N = number of trials r = number of ‘successes’ p = probability of ‘success’ on any trial q = 1-p (probability of ‘failure’) CN

r=The number of combinations of N things taken r at a time

( ) ( )!( )

!( )!N r N r r N rr

Np r C p q p q

r N r

So if I want to know the odds of getting 9 heads out of 10 coin flips or p(H,H, H,H, H,H, H,H, H,T):

p(9) =

10(.001953)(.5)=.0098 = .01

9 10 910!( ) ( )

9!(10 9)!p H p T

9 110!(.5) (.5)

9!(10 9)!9 110*9!

(.5) (.5)9!1!

• Now if we did this for all possible hits (heads) on 10 flips:

Number Heads Probability (p value)

0 .001

1 .010

2 .044

3 .117

4 .205

5 .246

6 .205

7 .117

8 .044

9 .010

10 .001

Using these probabilities

What is the probability of getting 4 or fewer heads in 10 coin tosses?

Addition

p(4 or less) = p(4) + p(3) + p(2) + p(1) + p(0) = .205 + .117 + .044 +.010

+ .001 = p = .377

About 38% of the time

Now take it out a step Suppose you were giving some sort of treatment to

depressed individuals and assumed the treatment in general would have a 50/50 chance of doing so if it wasn’t anything special (i.e. just a placebo). Then it worked an average of 9 times out of 10 administrations.

Would you think there was something special going on or just chance?

Not just 50/50 Not every 2 outcome

situation has equal probabilities associated with each option

There are two parameters we are concerned with when considering a binomial distribution. p = the probability of a

success. n = the number of Bernoulli

trials

Number of Successes

10987654321

Pro

ba

bili

ty

.4

.3

.2

.1

0.0

p = .8 N = 10 coins

Also... More info about binomial distribution

= Np

2=Nqp

In Excel

=BINOMDIST(success,total N, prob., FALSE)

Binomial distribution shapeApproximately “normal” curve when: p is close to 0.5

If not then “skewed” distribution N large

If not then not a representative distribution

Binomial in Action Say someone claims ESP and we give them a test. The card can only

be a star or a circle on the one side and we let them guess as to what it is. We test them 20 times and they guess right on 13, which chance alone would dictate only 10 or .5 of the time.

2 Hypes: H0 = their performance is really no different from chance H1 = their performance is above chance.

The probability of guessing 13 is ~.07. Pretty unlikely!

However our question isn’t really about guessing 13 is it? It is about guessing 13 or more. Guessing at least 13 right has a chance of (or p value) ~.13 Still not that likely but not unreasonable. Think if there was a 13% chance of rain. We’d be surprised but not completely amazed.

The problem with probabilities Can be very hard to grasp

e.g. Monty Hall problem TV show “Let’s make a deal” 3 closed doors, behind 1 is a prize Select a door Monty Hall opens one of the remaining doors

that does NOT contain a prize Now allowed to keep your original door or

switch to the other one Does it make a difference if you switch?