expectations, permutations & combinations 1 krishna.v.palem kenneth and audrey kennedy professor...

Expectations, Permutations & Combinations

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Krishna.V.PalemKenneth and Audrey Kennedy Professor of ComputingDepartment of Computer Science, Rice University

ContentsIn-class exercise on ExpectationsLaw of Large NumbersFairnessPermutations and CombinationsIn-class exercise

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In-class Exercise: Application of Expectation

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Consider the simple board without any snakes or ladders

7 8 9

6 5 4

1 2 3

Question: Calculate the expected number of rolls to reach or cross 9 from 5

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Consider the following

Denote expected number of rolls to reach 9 from square i as E(i)

Therefore E(8) = 1Because once you are at square 8 then independent of the outcome of the die the game is complete in one roll

Consider you are at square 7 and we have to calculate E(7)

7

8

9

1/6

5/61

If you are in square 7, there is a 1/6 chance that you end up in square 8 (from which it will take E(8)rolls to complete the game)

But there is a 5/6 chance that you complete the gamein one roll from square 7

Solution

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That means E(7) = probability(roll from 7 ->9) * number of rolls + [probability(roll from 7 ->8)

* probability(roll from 8->9)]*number of rolls(can also be obtained from the transition diagram)

Hence, E(7) = (5/6) * (1)+ [(1/6)* 1]*(2) = 7/6

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8

9

1/6

5/61

Transition graph for E(6)

7

8

9

1/6

5/6

1

6

1/6

1/6

4/6

Using a similar technique, we can compute that

Expected Number of Rolls

E(6) 49/36

E(5) 343/216

The one with a snake

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Question: Calculate the expected number of rolls to reach or cross 9 from 5 with a snake present

Transition Diagram for snake

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7

8

9

1

6

1/6

1/6

4/6

51/6 1/6

1/6

3/6

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Complete solution will be posted online

Calculating expected number of rolls

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So in this case, also because square 8 will lead to the completion of game no matter what

E(8) = 1

But because of the snake, when you land on square 7 you automatically move to square 5

Therefore E(7) = E(5)

Let us see for square 6. Use the previous methodology

From the transition diagram,E(6) = (1/6)*(E(7)+1) + [1/6*1](2) + (4/6)*(1)

But we know E(7) =E(5),

E(6) = (1/6)*E(5) + (7/6)

Step 1

Step 2

Step 3

Calculating expected number of rolls

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Let us write the equation for E(5)

From the transition diagram,

E(5) = (1/6)*(E(6)+1) + (1/6)*(E(7)+1) + (1/6)*(E(8)+1) + 3/6

= (1/6)*( (1/6)*E(5) + 7/6 ) + (1/6)*(E(5) + 1) + 5/6 Rearranging some terms,

E(5) * [ 1 – 1/36 – 1/6 ] = 7/36 + 1 = 43/36

Therefore, E(5) = 43/29


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Law of Large NumbersThe law of large numbers (LLN) describes the long-

term stability of the mean of a random variable.

Given a random variable with a finite expected value, if its values are repeatedly sampled, as the number of these observations increases, their mean will tend to approach and stay close to the expected valuefor example, consider the coin toss experiment. The frequency of

heads (or tails) will approach 50% over a large number of trials.

Mathematically, as an example, it can be represented as,if Mean is , then

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In-Class ExerciseLet us perform the virtual coin toss

experiment given at http://nlvm.usu.edu/en/nav/frames_asid_305_g_3_t_5.html

Let ‘x’ be the random variable denoting the number of heads in a coin toss x=1 if result is head and x=0 if result is tails

Observe the value of ‘x’ for n=10, 20, 50, 100, 500, 1000, using the virtual coin tossFor each ‘n’, compute the difference between

x that is observed and its expected value x (µ = np).

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What do you observe as the value of ‘n’ increases?

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What do you observe as the number of trials grows large ?

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Can you observe that as the number of trials grows “large” the result of the experimenttends to agree with the ideal case ?


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Fairness

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When do we say that an event is fair ?

Consider the event space

Event 1

Event 2

Event 3

Event 4

When all the events in the space are equally likely ?

By this interpretation, are all horse races unfair in horse racing?

Because usually one horse is more likely to win the race than the other.

So the question is, for a game which has unequal probabilities for differentEvents, what has to be done to make it fair ?

Fairness (Contd.)

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So to make the horse racing fair we allocate different odds for each horse.

This means that the horse which is more likely to win will give you the least amount in return.

This is because it was the option with the least risk

Therefore the method through which a game that is biased is made fair is by using “risk”.

Now no matter which horse a gambler bets, he/she will be on an equal footing with respectto others.

Fairness (Contd.)

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But what did we do here ?

We did not make the events equally likely

We made the returns or “the expectations” of all the gamblers equal.

Thus making the game fair

“A experiment is fair if the expectations of all the outcomes are equal”

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Consider the following game

Outcome of the toss

Probability of the outcome

Earnings for the player

who bets on it

HEAD ½ 2 chips

TAIL ½ 2 chips

According to the previous definition, the above game is fair to both the players

Now consider the following game

Outcome of the toss



who bets on it

HEAD 2/3

TAIL 1/3

Calculate the earnings for the outcomes that will make this game fair.

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Essentially what we want is that,

The expected earning of HEAD = The expected earning of TAIL

Prob(HEAD) * Earning (HEAD) = Prob(TAIL) * Earning(TAIL)

This implies,

Earning(HEAD) : Earning(TAIL) = Prob(TAIL):Prob(HEAD)

In this example,Earning(HEAD) : Earning(TAIL) = 1:2

Therefore to make the game fair, we have to make the earnings in the inverse ratio of the probabilities

Now consider the following game

Outcome of the die



who bets on it

1 p1 X1

2 p2 X2

3 p3 X3

4 p4 X4

5 p5 X5

6 p6 X6

Can you state the criteria to make the above game fair to all the players ?

There have been many instances when unfairness in a chance-based game was exploited to destroy kingdoms.This is the basis for the oldest epic story in the world. It is called Mahabharatha which was writtenin India. 21

The Epic of Mahabharata

Kauravas challenge the honest Pandavas to a dice game with a loaded dice in their favor.

In this unfair game played, the Pandavas lose their entire kingdom, were humiliated and banished into exile for 13 years.

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It is pivoted on a gambling game between the Kaurava and Pandava clans of the royal family for the throne of Hastinapura kingdom

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• Mahabharat is an important part of Hindu mythology

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The Epic of Mahabharata

Kauravas challenge the honest Pandavas to a dice game with a loaded dice in their favor.

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It is pivoted on a gambling game between the Kaurava and Pandava clans of the royal family for the throne of Hastinapura kingdom

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• Mahabharat is an important part of Hindu mythology

• longest epic in the world

They return after the exile for revenge and win back their lost kingdom at the epic battle of Kurushetra23


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FactorialFactorial of a non-negative integer n,

denoted by n!, is the product of all positive integers less than or equal to n.

For example,

Note that

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Permutations & Combinations

Permutation: An arrangement of elements from a set such that each element occurs atmost once, but not all elements of the given set need to be usedorder of arrangement of elements is importantIt is denoted by the formula

where n is total number of elements in the set and r is the number of elements to be arranged

Combination: An un-ordered collection of distinct elements from a given set.order of arrangements of elements is not importantIt is denoted by the formula

where n is total number of elements in the set and k is the number of elements to be selected

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Example-1

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Question: There are 6 students to be seated on a row of 6 desks. How many ways can they be arranged if we don’t care about seating orders?

Solution: Lets take one student. He/she can sit on any of the 6 desks Therefore he has 6 ways of sitting down, leaving 5 seats

available for the next person. The next student has 5 places to sit down, given first student

has already taken a seat. Therefore first and second student have 6x5 ways of sitting. Next we have third student. He/she has 4 ways of sitting down given first and second

students have already taken 2 out of 6 desk.

Example-1So the first 3 students have 6x5x4 ways of

sitting down.This can be extended till we have the last

student.If 5 students have taken 5 desks, the 6th

student has no choice but to take the last available desk. So he has 1 way of sitting given first 5 students have taken a desk each.

Total number of arrangements = 6x5x4x3x2x1 = 6!

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Example-2Question: Using the same principle, how many

different 6 letter words can you make with “potato”. Note: It does not need to make sense. For example, “otatop” is also acceptable though there

is no such word

Solution:Let us restate this problemThink of each letter as a student trying to sit at a

desk. Let the students be called as p1,o2,t3,a4,t5,o6

This is like seating the 6 students on 6 desks but its not EXACTLY the same problem.

Can you guess why?

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Example-2Let us take the following arrangement of students

p1,o2,t3,a4,t5,o6 , p1,o6,t5,a4,t3,o2 , p1,o6,t3,a4,t5,o2 ,p1,o2,t5,a4,t3,o6

Here student o2 o6 t3 t5 have taken different unique positions but the word is still the same “potato”.

So the number of unique words is not just 6!Do you know what the solution is?

Solution: 6!/(2 x 2)Explanation: Let us assume that all letters are

unique. Then, it is the same approach as finding the number of

ways of seating 6 students at 6 desksNumber of unique words would be 6!

But we know the 2 o’s and 2 t’s are not unique.

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Example-2For every unique word, we can get 2 unique student

patterns by just interchanging the the o’s.Example: p1,o2,t3,a4,t5,o6 , p1,o6,t5,a4,t3,o2

So there are 2 copies of every unique word in 6! Combinations because of the o’s.

Similarly for every word, we can get 2 unique student patterns by just interchanging the t’s.p1,o2,t3,a4,t5,o6 , p1,o6,t5,a4,t3,o2

So there are 2 copies of every unique word in 6! Combinations because of the o’s.

Therefore, the number of unique words is 6! / (No of copies due to letter o)x(No of copies due to letter t)Solution: 6! / (2x2)


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In-class ExerciseCard Terminology:

face value – same number cards (2-10, J, Q, K, A) has 4 cards of same face value

suite – set of cards with same symbol four suites – diamond, heart, spade, clubs each suite has 13 cards

Q) In a standard deck of cards, compute the number of ways you can deal each of the following five-card hands in poker.

1. Total number of different possible hands (five cards in a hand)2. Number of distinct Flush (all 5 cards have the same suite) 3. Number of distinct Four of a kind (4 same face value cards)

A) 1. C (52,5)

2. C (13,5) * C (4,1)

3. C (13,1) * C (48,1)

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Use of Combinations to Calculate Probabilities

Q) Now, compute the probability of getting a flush in a five-card poker game?

A)

Number of favorable events = C (13,5) * C (4,1)Total no. of events = C (52,5)Hence, Probability = C (13,5) * C (4, 1)/ C ( 52,5)

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Probabilityof outcome

No. of favorable events

Total no. of events

END

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Take Home - II 1. Let us go back to the Monty Hall Problem.

Let us redefine the problem.Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say Number 3, which has a goat. He then says to you, "Do you want to pick door Number 2?" We know its advantageous to switch (based on lecture 3), so we will switch to 2

Question: Can you prove the same result i.e. “Switching in the problem of Monty hall is advantageous when compared to not switching” using Bayes Theorem?

Generalizing the sum of expectations result

2. Prove that the expectation of sum of n random variables is equal to the sum of expectation of the n random variables.Let x1, x2, x3…. xn be n random variablesLet z = x1 + x2 + x3…. + xn

To prove

n

iixEzE

1

)()(

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