Skip to main content

Probability is Global

Today's blog post is an introduction to the Expected value and hopefully would be helpful to the kids who would be attending the upcoming Expected Value Lecture.

The handouts/ books I referred to are Evan Chen's Probability handout , AOPS introduction to Counting and probability, Calt's Expected value handout, brilliant and this IIT Delhi handout.


Probability is Global

Expected Value:   

  • The expected value is the sum of the probability of each individual event multiplied by the number of times the event happens.
  • It is denoted as E [x]
  • We have E[x]=xnP(xn)
    where xn is the value of the outcome and P(xn) is the probability that xn occurs.

Problem 1: What is the expected value of the number that shows up when you roll a fair 6 sided
dice?

Solution: Since it's a fair dice, we get each outcome to have equal probability i.e 16.
So E[x]=161+162+163+164+165+166=216=3.5

Problem 2: Find the expected value of a roll on a fair n sided dice, labelled from 1 to n.

Solution: Since it's a fair dice, we get each outcome to have equal probability i.e 1n. So E[x]=1n1+1n2++1n(n1)+1nn=n+12

Problem 3: Suppose you have a weighted coin in which heads comes up with probability 3/4 and tails 1/4 with probability . If you flip heads, you win 2 but if you flip tails, you lose 1. What is the expected win of a coin flip in dollars?

Solution: E[x]=342+14(1)=1.25

Problem 4:  At a raffle, 25 tickets are sold at 1 each for 3 prizes of 100,50, and 10. You buy 1 ticket. What is the expected value of your gain?

Solution: E[x]=12599+12549+125922251=13525=5.4

Problem 5: Linda estimates the number of questions she answered correctly on a test. She answered 10 correctly with probability 0.6,  20 correctly with probability 0.3, and 50 correctly with probability 0.1. What is the expected value of the number of questions Linda answered correctly?

Solution: E[x]=61010+31020+11050
=17

Problem 6: Mara is playing a game. There are two marbles in a bag. If she chooses the purple marble, she will win 10. If she chooses the orange marble, she will win 200. What is the expected value of Mara's winnings from the game?

Solution:  E[x]=1210+12200=105

Problem 7: In the casino game roulette, a wheel with 38 spaces (18 red, 18 black, and 2 green) is spun. In one possible bet, the player bets 1 on a single number. If that number is spun on the wheel, then they receive 36 (their original 1+35). Otherwise, they lose their 1. On average, how much money should a player expect to win or lose if they play this game repeatedly?

Solution: E[x]=1383537381=238

Problem 8: In a certain state's lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers are drawn match the numbers that a player had chosen, the player wins 1,000,000. If they match 5 numbers, then win 1,000. It costs 1 to buy a ticket. Find the expected value.

Solution: E[x]= 1(486)1000000+642(486)1000(486)253(486)1 
=1227125912271512

Linearity of Expectation:

If there exist variables a1,a2,a3,,an, independent or dependent,

E[a1+a2++an]=E[a1]++E[an]

Also E[X×Y]=E[X]×E[Y] holds when X,Y independent.

Problem 9: What is the expected value of the sum of two dice rolls?

Solution: Let the expected value of the first dice be X and the second dice be Y.
So E[X+Y]=E[X]+E[Y]=272=7.

Problem 10: Caroline is going to flip 10 fair coins one after the other. If she flips n heads, she will be paid n. What is the expected value of her payout?

Solution:  Let Xi be 1 if heads and 0. Also, denote Xi as the outcome of the i th coin flip.
So E[X]=E[X1++X10]=E[X1]++E[X10]=1012=5

Problem 11: Sammy is lost and starts to wander aimlessly. Each minute, he walks one meter forward with probability 12  ​ , stays where he is with probability 13  ​ , and walks one meter backward with probability 16. After one hour, what is the expected value for the forward distance (in meters) that Sammy has travelled?

Solution: Let Xi be the move sammy does in i minute. Note that 
E[Xi]=121+13016=13
So E[X]=E[X1++X60]=E[X1]++E[X60]=6013=20

Problem 12: 25 independent, fair coins are tossed in a row. What is the expected number of consecutive HH pairs?

Solution: So consider the consecutive pairs. Let Ci denote the ith coin in the row.. Then we consider the pairs P1=[C1,C2],P2=[C2,C3],,P24=[C24,C25].
Now, let Xi=1 if P_i HH,0 else 
Note that E[Xi]=14.  

Hence, even though they are dependent, by linearity of expectation,
 E[ no of consecutive pair]=E[X1]++E[X24]=2414=6

Problem 13: Suppose that A and B each randomly, and independently,
choose 3 of 10 objects. Find the expected number of objects chosen by both A and B.

Solution: Let X be the number of objects chosen by both A and B. Then let Xi=1 if A and B both select i,0 else . 
So E[Xi]=P[ A and B select i]=P[A selects i ]×P[ B selects i ]=9100 Alternatively, we have E[Xi]=(92)2(103)2
So E[X]=E[X1]+E[X10]=10E[Xi]=109100.

Problem 14: At a nursery, 2006 babies sit in a circle. Suddenly, each baby randomly pokes either the baby to its left or to its right. What is the expected value of the number of unpoked babies?

Solution: Let the babies be B1,B2,B2006.  
Note that any pair E[Xi]=14  ( defining 1 when unpoked)
And then we do linearity of Expectation. 
E[x]=E[X1++X2006]=E[X1]+E[X2]++E[X2006]=200614

It's the famous paradox game. :P

Problem 15: You are playing a game in which prize pool starts at 1. On every turn, you flip a fair coin. If you flip head, then the prize pool doubles. If tails, the game ends.

Solution:  Note that P(T)=12,P(HT)=14,P(HHT)=18,
E(X)=121+142+184+1168+=0.5+0.5+0.5+0.5+= A paradox. Cause expected value cant be infinite :P

Problem 16: Two random, not necessarily distinct, permutations of the digits 2017 are selected and added together. What is the expected value of this sum?

Solution:  Thanks to Pranav for the write up.
Let the permutations be P1,,P24. And the sums be S1,S2,,S288.
 Total number of permutations of 2017=4!. Total number of distinct sums =12(24)2=288
Let s be a random variable representing sum of two permutations of 2017 taken at random. Then, E[s]=sP(s)=s1288=1288s. Now, we have to calculate s. Clearly, s=3!(2000+1000+7000)+3!(200+100+700)+3!(20+10+70)+3!(2+1+7) s=611110=66660 E[s]=66660288=231.4583

The following proof is from the calt handout! The handout is very nice!!!!

Theorem: If the probability of a variable x occurring is p, then the expected number of times we must repeat the event so that we get x is 1p.

Proof:  Let X be the number of times we would have to repeat to get x.

So E[X]=1P[ x occurring in 1st turn]+2P[ x occurring in 2nd turn]+ 
p+2(p1)p+3(p1)2p+=p(1+2(p1)+3(p1)2+)
multiplying by (1p) and subtracting,
pE[X]=p(1+(1p)+(1p)2+)=p1p 
E[X]=1p


Yeah, that is it! Hope you enjoyed reading this post!
Sunaina 💚

Comments

Popular posts from this blog

LMAO Revenge

Continuing the tradition of past years, our seniors at the Indian IMO camp(an unofficial one happened this year) once again conducted LMAO, essentially ELMO but Indian. Sadly, only those who were in the unofficial IMOTC conducted by Pranav, Atul, Sunaina, Gunjan and others could participate in that. We all were super excited for the problems but I ended up not really trying the problems because of school things and stuff yet I solved problem 1 or so did I think. Problem 1:  There is a   grid of real numbers. In a move, you can pick any real number  ,  and any row or column and replace every entry   in it with  .  Is it possible to reach any grid from any other by a finite sequence of such moves? It turned out that I fakesolved and oh my god I was so disgusted, no way this proof could be false and then when I was asked Atul, it turns out that even my answer was wrong and he didn't even read the proof, this made me even more angry and guess wha...

Functional Equations 101

Let's get to the math:  Let there be two sets X and Y. A function  from X to Y denoted as f:XY is assigning a value in Y for every element in X. We say that X is the domain of the function f and Y is the range.  A function f:Xy is said to be injective if f(x)=f(x)x=x To put it in a more abstract way, if there is some aY then there is at most one bX such that f(b)=a holds true.  A function is said to be surjective when for any aY there is at least one bX such that f(b)=a holds true.  A function is bijective if for every aY there is exactly one bX such that f(b)=x. Bijective functions are basically functions which are both injective and surjective.  Bonus: A function f:XX is known as an involution if f(f(x))=xxX  As an exercise, the readers should try to prove that every function th...

Some Wandering Through Orthic Triangles

 Hi, I am Emon and I it's been long since I posted last. Today I will try to give you some ideas on how to work with a special type of triangle, known as " Orthic Triangles ". In this post, I will mainly focus on problem-solving, but still, let me first give you some ideas on what exactly it is and what properties does it have... Definition (Orthic Triangle). Let ABC be a triangle and let D,E,F be the foot of the perpendiculars from A,B,C to BC,CA and AB, respectively. Then, DEF is known as the orthic triangle of ABC. Lemma 1 (Orthic Triangle). If DEF is the orthic triangle of ABC with orthocenter H, then the following conditions are satisfied : (i) AEHF is a cyclic quadrilateral with circumdiameter AH. (ii) BCEF is a cyclic quadrilateral with circumdiameter BC. (iii) H is the incenter of DEF. Lemma 2. ABE=ADE and ACF=ADF. (We can prove this w...