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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 $\Bbb E$ $[x]$
  • We have $$\Bbb E[x]=\sum x_n P(x_n)$$
    where $x_n$ is the value of the outcome and $P(x_n)$ is the probability that $x_n$ occurs.

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

Solution: Since it's a fair dice, we get each outcome to have equal probability i.e $\frac {1}{6}.$
So $$\Bbb E[x]=\frac 16 \cdot 1+\frac 16 \cdot 2+\frac 16 \cdot 3+\frac 16 \cdot 4+\frac 16 \cdot 5+\frac 16 \cdot 6=\frac {21}{6}=3.5$$

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 $\frac {1}{n}.$ So $$\Bbb E[x]=\frac 1n \cdot 1+\frac 1n \cdot 2+\dots+\frac 1n \cdot (n-1)+\frac 1n\cdot n=\frac {n+1}{2}$$

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: $$\Bbb E[x]=\frac{3}{4} 2+\frac{1}{4}(-1)=1.25$$

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: $$\Bbb E[x]= \frac{1}{25}\cdot 99+ \frac{1}{25}\cdot 49+ \frac{1}{25}\cdot 9-\frac{22}{25}\cdot 1=\frac{135}{25}=5.4$$

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: $$\Bbb E[x]=\frac{6}{10}\cdot 10+\frac{3}{10}\cdot 20+\frac{1}{10}\cdot 50$$

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:  $$\Bbb E[x]=\frac 12 \cdot 10+\frac 12 \cdot 200= 105$$

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: $$\Bbb E[x]=\frac{1}{38} \cdot 35-\frac{37}{38}\cdot 1=\frac{-2}{38}$$

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: $$\Bbb E[x]=\ \frac{1}{\binom{48}{6}}\cdot 1000000+ \frac{6\cdot 42}{\binom{48}{6}}\cdot 1000-\frac{\binom{48}{6}-253}{\binom{48}{6}}\cdot 1$$ 
$$ =\frac{12271259}{12271512}$$

Linearity of Expectation:

If there exist variables $a_1 , a_2 , a_3 ,\dots, a_n ,$ independent or dependent,

$$\Bbb E[a_1+a_2+\dots+a_n]=\Bbb E[a_1]+\dots+ \Bbb E[a_n]$$

Also $$\Bbb E[X\times Y]=\Bbb E[X]\times \Bbb E[Y]$$ 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 $$\Bbb E[X+Y]=\Bbb E[X]+\Bbb E[Y]= 2\cdot \frac{7}{2}=7.$$

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 $X_i$ be $1$ if heads and $0.$ Also, denote $X_i$ as the outcome of the $i$ th coin flip.
So $$\Bbb E[X]=\Bbb E[X_1+\dots + X_{10}]=\Bbb E[X_1]+\dots +\Bbb E[X_{10}]=10\cdot \frac 12=5$$

Problem 11: Sammy is lost and starts to wander aimlessly. Each minute, he walks one meter forward with probability $\frac{1}{2}$  ​ , stays where he is with probability $\frac{1}{3}$  ​ , and walks one meter backward with probability $\frac{1}{6}$. After one hour, what is the expected value for the forward distance (in meters) that Sammy has travelled?

Solution: Let $X_i$ be the move sammy does in $i$ minute. Note that 
$$\Bbb  E[X_i]=\frac{1}{2}\cdot 1+\frac{1}{3}\cdot 0-\frac{1}{6}=\frac{1}{3}$$
So $$ \Bbb E[X]=\Bbb E[X_1+\dots +X_60]=\Bbb E[X_1]+\dots + \Bbb E[X_60]=60\cdot \frac{1}{3}=20$$

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 $C_i$ denote the $i$th coin in the row.. Then we consider the pairs $$P_1=[C_1,C_2], P_2=[C_2,C_3],\dots, P_{24}=[C_{24},C_{25}].$$
Now, let $$X_i= 1 \text{ if P_i HH}, 0 \text{ else }$$
Note that $$\Bbb E[X_i]=\frac{1}{4}.$$  

Hence, even though they are dependent, by linearity of expectation,
 $$\Bbb  E[\text{ no of consecutive pair} ] =\Bbb E[X_1]+\dots +\Bbb E[X_{24}]=24\cdot \frac{1}{4}=6$$

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 $$X_i= 1\text{ if A and B both select i}, 0 \text{ else }.$$ 
So $$\Bbb E[X_i]=\Bbb P[ \text{ A and B select i}]=\Bbb P[ \text{A selects i }] \times \Bbb P[\text{ B selects i }]=\frac{9}{100}$$ Alternatively, we have $$\Bbb E[X_i]=\frac{\binom{9}{2}^2}{\binom{10}{3}^2}$$
So $$\Bbb E[X]=\Bbb E[X_1]+\dots \Bbb E[X_{10}]=10\cdot \Bbb E[X_i]= 10\cdot \frac{9}{100}.$$

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 $B_1, B_2, \dots B_{2006}.$  
Note that any pair $$\Bbb E[X_i]= \frac{1}{4}$$  ( defining $1$ when unpoked)
And then we do linearity of Expectation. 
$$\Bbb E[x]= \Bbb E[X_1+\dots +X_{2006}]=\Bbb E[X_1]+\Bbb E[X_2]+\dots+ \Bbb E[X_{2006}]= 2006\cdot \frac{1}{4}$$

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 $$P(T)=\frac{1}{2}, P(HT)=\frac{1}{4}, P(HHT)=\frac{1}{8},\dots $$
$$ \Bbb E(X) = \frac{1}{2}\cdot 1 + \frac{1}{4} \cdot 2 + \frac{1}{8}4  + \frac{1}{16}8 + \cdots  = 0.5 + 0.5 + 0.5 + 0.5 + \cdots = \infty $$ 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 $P_1,\dots, P_{24}.$ And the sums be $S_1,S_2,\dots,S_{288}.$
 Total number of permutations of $2017 = 4!$. Total number of distinct sums $= \frac{1}{2} \cdot (24)^2 = 288$. 
Let $s$ be a random variable representing sum of two permutations of $2017$ taken at random. Then, $$\Bbb{E}[s] = \sum s \cdot \Bbb{P}(s) = \sum s \cdot \frac{1}{288} = \frac{1}{288} \cdot \sum s$$. Now, we have to calculate $\sum s$. Clearly, $$s = 3! \cdot (2000 + 1000 + 7000) + 3! \cdot (200 + 100 + 700) + 3! \cdot (20 + 10 + 70) + 3! \cdot (2 + 1 + 7)$$ $$\implies s = 6 \cdot 11110 = 66660$$ $$\implies \Bbb{E}[s] = \frac{66660}{288} = 231.458\overline{3}$$

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 $\frac{1}{p}$.

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

So $$\Bbb E[X]= 1\cdot \Bbb P[\text{ x occurring in 1st turn}] + 2\cdot \Bbb P[\text{ x occurring in 2nd turn}]+\dots $$ 
$$ p+2\cdot (p-1)p+3 \cdot (p-1)^2\cdot p+\dots = p( 1+ 2(p-1)+ 3 (p-1)^2+ \dots )$$
multiplying by $(1-p)$ and subtracting,
$$ \implies p\cdot  \Bbb E[X]= p(1 +( 1-p)+ (1-p)^2+\dots ) = p \frac {1}{p}$$ 
$$ \implies \Bbb E[X]= \frac{1}{p}$$

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


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