comper binandmut atiati onsons. and in 1982 schlitz brewing company took a $4.2 million gamble that...

In 1982 Schlitz Brewing Company took a $4.2 million gamble that they couldn’t lose, because they hired

people who knew their statistics.

At halftime during the Super Bowl in front of

100 million people they did a live taste test between Schlitz

beer and their closest competitor, Michelob.

But crazier still, they pitted their brand against Michelob

drinkers!And Budweiser

drinkers. And Miller drinkers.

And they had an NFL referee oversee the

taste test

Crazy?Like a fox.

Because they couldn’t lose.

The Drinking Problem

Schlitz won their gamble because they didn’t gamble. They knew, statistically, what the outcome

would be before they started.

And it wasn’t because they produced an amazing beer because all the beers taste the same.

They knew what the outcome would be because they produced amazing statistics and manipulated

the sample!

This does not mean they cheated.It means they thought about what research design

would give them the result they wanted.

Consider…

How To Gamble and Win

First, they carefully picked their sample – 100 each of Michelob, Miller and Budweiser drinkers.

Then they calculated the probabilities of how many of them might choose Schlitz.

Think about this. Even without calculating the odds you might say that there was a 50/50 chance of

either beer being chosen.

So you can imagine what the marketing conclusion would be:

“Look! Fifty percent of Michelob (or Miller or Bud) drinkers prefer Schlitz!”


But you don’t have to rely on a guess that “50 out of 100 people will pick a Schlitz beer”.

You can actually calculate the probabilities quite easily, which the Schlitz statisticians did using what

is called a Bernoulli Trial and binomial statistics.

These type of statistics can tell you what the probability is of 1, 2, 3, … or even 100 people

choosing or not choosing a type of beer would be.

Binomial statistics is a very, very, important statistical method used widely in studies where you

want to know what the probability of ‘k’ things happening out of ‘n’ things is and this is how you do

it…

How To Gamble, Win and Know You’ll Win

Well, here's the answer anyway because we have a final lecture all about binomial statistics.

What’s the probability that 40 out of 100 Michelob drinkers will choose the Schlitz beer? Note that this would

not be a good outcome if all beer drinkers were asked!

Well, first lets assuage the fears of the marketing people by asking what are the odds of the worst possible

outcome - all 100 Michelob drinkers choosing Michelob?

1 in 1,267,650,600,228,229,401,496,703,205,376 or 1 in about 1.3 million trillion trillion.

And of all 100 choosing Schlitz?1 in 0.0000000000000000000000000007888

How To Gamble, Win, and Know You’ll Win

Now back to a more realistic question?



The answer is that the probability was 98% that at least

40 of the 100 Michelob drinkers would choose the Schlitz and an 86% chance that 45 would!

And even more strangely, if it had been Michelob doing the taste test with Schlitz drinkers, or Pepsi with Coke

the same results would occur!

It wouldn’t matter what the products taste like because they pretty much taste the same anyway!


Permutations and CombinationsImportant part of statistics because they underlie

how you figure out how many times a certain set of things can occur in a larger set of things.

For example, what are the chances of:

Two people out of 50 sharing a birthday?Finding 12, 6’-2” women in a population?

Finding a set of 6 numbers out of 49 (i.e. winning Lotto 6/49)?

Permutations and combinations underlie the Binomial and Poisson distributions that are widely

used to calculate the probabilities of ‘r’ events happening in ‘n’ trials.

Permutations and CombinationsPermutations and combinations is the name given to the

branch of mathematics that calculates how many groups of things you can get out of a larger group of things.

The difference between the two is whether you care about the order the things come in:

If you're cataloguing a group of people by their ethnicities, it doesn’t really matter what order you list them in.

If you’re cataloguing them by their ages, then it likely does.

The other factor that comes into play is whether you can have repetition in your groups.

Permutations and CombinationsHow do you choose a group ‘r’ from a larger group ‘n’?

(E.g. from the numbers 1-10 choose the first 3).

Does the order of group ‘r’ matter?

Yes – you must choose 1 then 2 then 3.

No – 1,2,3 is as good as 3,2,1 or any other.

This is a permutation. This is a combination.

Can the members of group ‘r’ be repeated?

If yes, this is a permutation or combination with

repetition.

If no, this is a permutation or combination without

repetition.

CalculatingPermutations

andCombinations

Permutations With Repetition(Remember: order matters with permutations)

These are the easiest to calculate and understand, and are what you do with a “combination” lock – which should

actually be called a permutation lock.

If you can re-use any member of the larger group ‘n’ to form your smaller group ‘r’, you can basically choose any ‘n’, ‘r’

times, thus:

n*n* … (‘r’ times) or mathematically stated as nr.

An example is a combination lock with a 3 number code - choosing any 3 numbers (r) from 10 (n) for your combination:

10*10*10 or 103 = 1,000 possible codes because any number can be used over again – the code could be 3,3,3 or any other

3 numbers such as 1,2,3 – but in that order only.

Permutations Without RepetitionNow we have to reduce the number of choices

because you cannot repeat a number.

An example would be a race with, say, 8 runners numbered 1 to 8, the question being how many ways

are there for 8 runners to place?

The reason we have to reduce the number of choices is that whoever comes in 1st place cannot be 2nd as

well, nor can 2nd be 3rd and so on for all runners.

So, for all 8 runners the arithmetic is:8*7*6*5*4*3*2*1 = 40,320

possible ways for 8 runners to place 1st through 8th.

Permutations Without RepetitionBut calculating 8*7*6*5*4*3*2*1 = 40,320 every time is tedious – what if you had the 27,000 runners in the

NY marathon?

There’s an easier mathematical way to calculate such reducing series of numbers by using factorials.

A factorial is just a mathematical way of writing “multiply this number by all whole numbers below it as

a descending series” and it is written as 8! for the example we are using.

Thus 8! means the same as 8*7*6*5*4*3*2*1

And for the NY marathon?27,000! is effectively infinite!

Permutations Without Repetition

What if you wanted just the first three placings? Now you are asking how many ways are there for 8 runners

to place 1st, 2nd, and 3rd?

That’s easy. There are 8*7*6 = 336 ways for 8 runners to place 1st, 2nd, and 3rd.

How do you write that as a factorial? That’s not so easy, but is pretty clever because it involves dividing by the 5

runners who don’t place:

(8*7*6*5*4*3*2*1) / (5*4*3*2*1) = 336or

8! / 5! = 336

Permutations Without RepetitionThis is written as:

Also seen as: P(n,r) or nPr or nPr

One more example to clarify this. How many ways can 1st and 2nd place be awarded to our 8 runners?

56

And our original 1st, 2nd and 3rd places to our 8 runners?

336

Permutations Without RepetitionThis may all seem difficult because it is math despite

my promising there would be none.

The important thing here is not that you understand why factorials (or even factorizing) works to give us

answer to permutations (and combinations) questions, but that it does work.

More importantly, as we’ll see shortly, permutations, and especially combinations, form the heart of a widely

used class of statistics that looks at the probability of getting ‘r’ results in ‘n’ trials.

These are called the Binomial and Poisson distributions.

But for now, I want to continue by looking at combinations.

Combinations Without Repetition(Remember: order doesn’t matter in combinations.)

This is how lotteries work. Numbers are drawn one at a time and if you have them – no matter what order – you win.

So the formula to do this requires that you calculate a permutation (where order matters) then reduce it by the number of extra ways

you get to order the numbers.

For example, using the numbers 1,2,3:Permutations of 1,2,3 Combinations of 1,2,3

(Order matters) (Order doesn’t matter)1,2,31,3,22,1,3 1,2,32,3,13,1,23,2,1

Combinations Without RepetitionSo how do we turn this into a combination formula?

We know that the permutations formula is:

So to remove the choices we don’t want (the various orders 1,2,3 could be put into) we multiply it by the choices:

* = =

We have now reached the reason for doing all this permutations/combinations stuff, and it is this formula, called

the Binomial Coefficient or often simply “n choose r”.

Combinations and the Binomial CoefficientThis formula, the Binomial coefficient…

or is one of the most important in statistics because it allows you to calculate the probability of ‘r’ events occurring in ‘n’ trials.

For example, what are the odds of getting 5 heads out of 9 throws? A better example is this:

You sell real estate. 70% of people like houses and 30% like condos. What is the probability that you will sell 2 houses to

the next 3 customers?

We’ll get to the answer in a little while, but hopefully you now see why combinations and permutations is important.

For now, we’ll finish up with combinations with repetition.

Combinations With RepetitionThese are the hardest to calculate and certainly to

understand, so I'm going to focus on calculation because the formula here is also used widely to solve the “combinations

with repetition problem”.

If I gave you 5 ice cream flavours (your n) and said choose 3 (your r) and don’t worry about order or repetition (you could

choose 3 of the same flavour) I would be telling you to choose r out of n+r-1.

It doesn’t matter why!

So the factorials formula for combinations with repetition is:

Permutations and Combinations SummaryThis is what you’ve learned, in case you didn’t know.

Combinations and permutations are about choosing ‘r’ things from ‘n’ things (e.g. 6 #s from 49 #s).

Combinations don’t worry about the order of the ‘r’ things but permutations do.

Both combinations and permutations can allow or not allow repeats of the ‘r’ things.

The calculation formulas for both involves factorials and factorizing.

Permutations and Combinations Summary

Withoutrepetition

Withrepetition

Permutation nrCombination

Below are the formulas for calculating permutations and combinations, with and without repetition. Simply plug in the

‘r’ (size of the group you want), and ‘n’ (size of the population from which the group is being selected).

Combinations without repetition are also called the binomial coefficient.

Note that other notational formulas exist and say the same thing. The ones above are the simplest from which to

calculate answers.

Fun With Permutations & Combinations – Lotto 6/49

Last example: how many sets of 6 numbers can you get out of out of 49 numbers if you

don’t care about the order in which they come?

Answer is 13,983,816 (@14 million) and that’s what Lotto 6/49 is all about – solving

for 49C6.

What if the order mattered – i.e solving for 49P6? That is, rather than OLGC rank ordering the six numbers that fall out of the machine, what if you had to get the order they fell in

as well?

Answer is that there are 10,068,347,520 (@10 billion) ways 6 numbers can fall out

of a machine with 49 numbers in it.

Fun With Permutations and Combinations – Lotto 6/49

So the chance of winning in any given week is:

1/13,983,816 or 0.0000000715…And therefore the chance of losing is:1-1/13,983,816 or 0.9999999285…

What if you play for 50 years (2,600 weeks)?

Chance of losing is:(1-1/13,983,816)2,600 or 0.999814…

And therefore the chance of winning is 1-0.999814 or 0.000186 or 93/500,000ths....

What if you saved the $3 a week for 2,600 weeks at 1.5% interest, compounded

annually?$11,822.94

Moral?Save your money.

BinomialAnd

PoissonDistributions

Poisson and Binomial DistributionsThe Binomial and Poisson distributions and the statistics

that go with them are used very widely in research because they can answer the question “what's the

chance that I will get ‘r’ events out of ‘n’ trials.”

Obviously, this is the same question you ask with combinations and permutations (and the math is the

same), so hence knowing how to calculate them is important.

Here are some examples of research questions that require you to know about Binomial and Poisson

distributions and calculations.

Hopefully you’ll see the usefulness of this type of stats and why they re so widely used.

Research Questions Requiring Poisson or Binomial Statistics

You throw a fair die 9 times. What is the chance of getting 5 heads?

What is the chance of getting at least half of 40 multiple choice questions correct by guessing?

A road intersection has an average of 1 collision every 4 days. What is the probability of there being 2 collisions

in 1 month?

In a given lake there are twice as many Asian carp as small mouthed bass. Four fish are caught at random.

What are the probabilities that none, 1, 2, 3, or all 4 will be small mouthed bass?

The answers to these questions are not what you think.

Normal, Poisson & Binomial DistributionsA normal distribution describes continuous data which have a

symmetric distribution with a characteristic 'bell' shape.

A binomial distribution describes the distribution of binary data (successes and failures) from a finite sample. Thus it gives the exact probability of getting ‘r’ events out of ‘n’ trials because

you know what size ‘n’ is.

A Poisson distribution describes the distribution of binary data (successes and failures) from an infinite sample. Thus it gives the

estimated probability of getting ‘r’ events in a population because you don’t know what size the population is.

Thus because binomial and the Poisson distributions are concerned with the number of ‘successes’ and ‘failures’ you get

from a series of ‘trials’, they are called BINARY.

A Bernoulli trial (or binomial trial) is an experiment whose outcome is random and can have either of two possible

outcomes, "success" or "failure”.

In practice it refers to a single experiment which can have one of two possible outcomes. These events can be

phrased into "yes or no" questions:Did the coin land heads?

Was the fish caught a small mouthed bass?Was the newborn child a girl?

Therefore success and failure are labels for outcomes, and should not be construed literally. The term "success" in

this sense consists in the result meeting specified conditions, not in any moral judgment.

Bernoulli Trial

Examples of Bernoulli trials include:

Flipping a coin, where "heads" conventionally denotes success and "tails" denotes failure. A fair coin has a

probability of success of 0.5 by definition.

Rolling a die, where a six is "success" and everything else a "failure". With a fair die the probability of “success” is

1/6th or 0.167

In conducting a political opinion poll, choosing a voter at random to ascertain whether that voter will vote "yes" in an upcoming referendum. Each voter has a 0.5 probability

of voting yes or being a “successful” outcome.

Bernoulli Trial

Consider a simple experiment. Toss a fair coin 3 times (n) and find the probability of getting 2 heads (r).

For this experiment, let heads be defined as a success and tails as a failure. Because the coin is assumed to be fair, the probability of success is 0.5 and the probability of failure is

thus 1-0.5=0.5.

All possible outcomes from a single throw?HHH, HHT, HTH, HTT, THH, THT, TTH, TTT = 8

Probability of getting any one these on any throw = 1/8

How many outcomes can we use (need 2 H)?HHT, THH, HTH = 3

Therefore, the probability of 2 heads = 3*1/8 = 3/8

Calculating Binomials

In the last slide, the answer seemed easy to get, but it isn’t as easy as it appears. Consider:

All possible outcomes from a single throw?HHH, HHT, HTH, HTT, THH, THT, TTH, TTT = 8 = P=1/8

P of getting 3 heads? = P(HHH) = 1/8 (there’s only one).P of getting 2 heads? = P(HHT)+P(HTH)+P(THH)=1/8+1/8+1/8=3/8.P of getting 1 head? = P(HTT)+P(THT)+P(TTH)=1/8+1/8+1/8=3/8.

P of getting 0 heads? = P(TTT)=1/8.

More formally, using ‘x’ as a random variable for the # of heads:P(x=3) = 1/8, P(x=2) = 3/8, P(x=1) = 3/8, P(x=0) = 1/8

Where 1/8+3/8+3/8+1/8=8/8=1.0 (unity)

Therefore the whole sample space of the 8 possible outcomes sums to P=1.0 or 100% of all possible options covered.

Calculating Binomials

Binomial FormulaAll of what we just calculated can be reduced down to a

single formula that is good for any size of ‘r’ and ‘n’ and it is one you have seen already – the combination with repetition

formula.

which is usually written as the binomial coefficient:

And using our data… = = 3That is, 3 outcomes to get the 2 heads, which we knew

already by doing it the long way in the last slide.

Binomial Formula – A Harder ProblemWhat about 5 heads in 9 throws (so n=9, k=5)?

And using our data… = = 126That is, 126 outcomes to get the 5 heads from 9 throws.

Probability of the 5 heads? For 9 tosses there are 29 = 512 outcomes so we get the probability:

P(x=5)=126/512 = 0.24609…

Or about a 25% chance of getting 5 heads in 9 throws.Easy!

Binomial Formula Using Unequal Outcome ProbabilitiesSo far we have calculated binomial probabilities using equal

probabilities for the success or failure outcomes – that is tossing a coin gives you a 50% chance of head or tail.

Even throwing a die gives you an equal chance of any number 1 to 6 and it is 0.167.

But what happens when you have unequal probabilities? Consider the following problem:

You sell real estate. 70% of your clients want a house. The other 30% want a condo.

Research question: what is the probability of selling two houses to the next three customers?

Now we are not dealing with coin tossing 50/50 chances but with dwelling purchases at 70/30.

First lets do it the hard way – use a decision tree.

C: 0.3H: 0.7

C: 0.3

H: 0.7

H: 0.7

C: 0.3

HOUSE, CONDO, HOUSE0.7*0.3*0.7=0.147 or 14.7%

HOUSE, CONDO, CONDO0.7*0.3*0.3=0.063 or 6.3%

HOUSE, HOUSE, HOUSE0.7*0.7*0.7=0.343 or 34.3%

C: 0.3

H: 0.7

H 0.7

HOUSE, HOUSE, CONDO0.7*0.7*0.3=0.147 or 14.7%

CONDO, CONDO, CONDO0.3*0.3*0.3=0.027 or 2.7%

CONDO, CONDO, HOUSE0.3*0.3*0.7=0.063 or 6.3%

CONDO, HOUSE, HOUSE0.3*0.7*0.7=0.147 or 14.7%

CONDO, HOUSE, CONDO0.3*0.7*0.3=0.063 or 6.3%

C: 0.3

C: 0.3

H: 0.7

H: 0.7

C: 0.3

First Customer

Second Customer

ThirdCustomer

Three outcomes out of 8 (the ones in red) give us the two houses and one condo we need. The individual probabilities of those outcomes is 14.7% and the total probability of 2 out

of 3 people choosing a house over a condo is 3*14.7% or 44.1%.

Binomial Formula Using Unequal Outcome ProbabilitiesNow lets do it the easy way () using a binomial

formula and a probability formula. Why?

Because we now have two different probabilities (0.7 and 0.3) we have to first calculate the probability of a

single desired outcome – the HHC or HCH, or the CHH - (which we know to be 0.147 from doing it the hard

way).

Then we have to calculate the number of outcomes we can expect – the three boxes in the decision tree.

First the probability…

Binomial Formula Using Unequal Outcome ProbabilitiesWe need two 0.7s and one 0.3.

0.7, the probability of each choice (call it ‘p’)2 is the number of H we want (call it ‘k’)

Therefore the probability of the “choices we want” (2H) is pk .

The probability of the opposite choice (1C) is 1-p.The total number of choices is n.

The number of opposite choices is n-k.And finally, the probability of “opposite choices” (1C) is

(1-p)(n-k).

So all choices together is:pk(1-p)(n-k)

Now some numbers to plug in and we’ll have the same answer as the decision tree gave us except quicker.

Binomial Formula Using Unequal Outcome ProbabilitiesThe numbers from our research question (the knowns):

P = 0.7 (chance of house).n = 3 (next three customers).

k = 2 (the 2 houses out of 3 customers we want).

The probability formula:pk(1-p)(n-k)

Plug in the numbers:0.72 (1-0.7) (3-2) = 0.72 (0.3) (1) = 0.7 * 0.7 * 0.3 = 0.147

…the probability of each outcome HHC, HCH, CHH that we got from adding up the links in the decision tree.

And the total number of outcomes (i.e. the number of boxes from the decision tree)?

Binomial Formula Using Unequal Outcome ProbabilitiesThe total number of outcomes is given by:

And using our data… = = 3And finally we get:

Number of outcomes we wanted = 3Probability of each outcome 0.147

Probability of the next 2 out of 3 choosing a house = 3*0.147=0.441

Or about a 44% chance of getting 2 houses in the next 3 customers.

Easy!

Binomial Formula Using Unequal Outcome ProbabilitiesNow an even harder question with a surprising answer.

You know that 7 out of 10 people prefer houses.

So are you correct to say that 7 out of the next 10 people will choose a house?

Put another way, what are the chances that 70% of the next 10 people choose a house?

This time, using:p = 0.7, n = 10, k = 7

We first calculate the probability of 1 H out of 10 customers, then the total number of possible H outcomes, then multiply

them to get the probability of a H 7 out of 10 times.

Binomial Formula Using Unequal Outcome ProbabilitiesThe probability formula:

pk(1-p)(n-k)

Plug in the numbers:0.77 (1-0.7) (10-7) = 0.77 (0.3) (3) = 0.0022235661

…the probability of each outcome.And the total number of outcomes is: = 120

And finally we get:120 * 0.0022235661 = 0.2668

The probability of 7 of the next 10 choosing a house is only 27%!

So, even though the average over the long term may be 70%, don’t expect 7 out of the next 10.

The General Binomial Formula in Summary

We know how to calculate getting k out of n ways:

And the probability of getting each k:

pk(1-p)(n-k)

So the full General Binomial Probability Formula for calculating the probability of getting k out of n ways is:

P(kn) = pk(1-p)(n-k)

In 1982 Schlitz Brewing Company took a $4.2 million gamble that they couldn’t lose, because they hired

people who knew their statistics.

At halftime during the Super Bowl in front of

100 million people they did a live taste test between Schlitz

beer and their closest competitor, Michelob.

But crazier still, they pitted their brand against Michelob

drinkers!And Budweiser

drinkers. And Miller drinkers.

And they had an NFL referee oversee the

taste test

Crazy? Like a fox.

Because they couldn’t lose.

Schlitz won their gamble because they didn’t gamble. They knew, statistically, what the outcome

would be before they started.

And it wasn’t because they produced an amazing beer.

They knew what the outcome would be because they produced amazing statistics and manipulated

the sample!

This does not mean they cheated.It means they thought about what research design

would give them the result they wanted.

Consider…


First, they carefully picked their sample – 100 Michelob drinkers.

Then they calculated the probabilities of how many of them might choose Schlitz.

Think about this. Even without calculating the odds you might say that there was a 50/50 chance of

either beer being chosen.

So you can imagine what the marketing conclusion would be:

“Look! Fifty percent of Michelob drinkers prefer Schlitz!”


But you don’t have to rely on a guess that “50 out of 100 people will pick a Schlitz beer”.

You can actually calculate the probabilities quite easily, which the Schlitz statisticians did using what

is called a Bernoulli Trial and binomial statistics.

These type of statistics can tell you what the probability is of 1, 2, 3, … or even 100 people

choosing or not choosing a type of beer would be.

Binomial statistics is a very, very, important statistical method used widely in studies where you

want to know what the probability of ‘k’ things happening out of ‘n’ things is and this is how you do

it…

How To Gamble, Win and Know You’ll Win

Doing The Schlitz StatisticsThe Schlitz numbers

n=100, k=40, p=0.5, 1-p=0.5

The binomial probability of exactly 40 out of 100:P(exactly k out of n) = pk(1-p)(n-k)P(exactly k out of n) = * 9.1-13*8.68-19

P(exactly k out of n) = * 7.89-31 = 0.0108 = 1.08%And of 40 or more Michelob drinkers preferring Schlitz?

The sum of the probabilities of 40 + 41 + … + 100 = 98.2%

Schlitz could be 98.2% sure that 40 or more Michelob drinkers would pick the Schlitz beer.

So there you have it.Probably the most useful stats you can know.

I have avoided examples of Poisson calculations but they are very similar – a Poisson can be

considered as a sub-class of the binomial where n is very large.

And the nice part about binomials is that there are many websites that will calculate the stats for you, this being on of them that also gives you the

formula components:


http://vassarstats.net/

textbook/ch5apx.html

An online binomial

calculator.

1-p

Now back to a more realistic question?



The answer is that the probability was 98% that at least

40 of the 100 Michelob drinkers would choose the Schlitz and an 86% chance that 45 would!

And even more strangely, if it had been Michelob doing the taste test with Schlitz drinkers or Pepsi with Coke,

the same results would apply!

It wouldn’t matter what the beverage tasted like because they pretty much taste the same anyway!


comper binandmut atiati onsons. and in 1982 schlitz brewing company took a $4.2 million gamble that...

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