BerkeleyX Stat2.2x Probability Notes: Section 3 The law of averages, and expected values

[Lecture3.1 PDF]
[Lecture3.2 PDF]
[Lecture3.3 PDF]
[Lecture3.4 PDF]

Basic Formula:

• Law of Large Numbers As the number of trials increases, the chance that the proportion of successes is in the range $$p\pm\text{a fixed amount}$$ goes to $1$.
• Expected value of the Binomial If a random variable $X$ has the binomial distribution with parameters $n$ and $p$, then $$E(x)=n\cdot p$$

PRACTICE

PROBLEM 1
In each pair of events below, pick the one that has the higher chance.
1a) A: 50,000 heads in 100,000 tosses of a coin
B: 500,000 heads in 1,000,000 tosses of a coin
1b) A: In 60 rolls of a die, the number of times “six” shows up is in the range 10 plus or minus 3
B: In 600 rolls of a die, the number of times “six” shows up is in the range 100 plus or minus 3
1c) A: In 1000 draws by a random number generator, the percent of times “0” shows up is in the range 10% plus or minus 1%
B: In 10,000 draws by a random number generator, the percent of times “0” shows up is in the range 10% plus or minus 1%
1d) A: In 1000 draws by a random number generator, the percent of times “0” appears is less than 11%
B: In 10,000 draws by a random number generator, the percent of times “0” appears is less than 11%

Solution
1a) A. The chance of getting exactly half heads goes down (i.e. goes to 0) with more tosses. For example:

dbinom(x = 50000, size = 100000, prob = 0.5)
[1] 0.002523126
dbinom(x = 500000, size = 1000000, prob = 0.5)
[1] 0.0007978844 

1b) A. The chance that the number of successes falls in the range “most likely value plus or minus a fixed amount” goes down with more rolls. For example:

sum(dbinom(x = 7:13, size = 60, prob = 1/6))
[1] 0.7767122
sum(dbinom(x = 97:103, size = 600, prob = 1/6))
[1] 0.2985048

1c) B. This follows the "Law of Large Numbers".

sum(dbinom(x = 90:110, size = 1000, prob = 1/10))
[1] 0.7318197
sum(dbinom(x = 900:1100, size = 10000, prob = 1/10))
[1] 0.9991879

1d) B. “less than $11\%$” includes $10\%$, which is where the percent of 0’s is going as the number of draws increases. More formally, the chance that the percent of 0’s exceeds the actual probability (i.e. $0.1$) plus any fixed amount (e.g. $0.01$) gets smaller as the number of draws increases. So the probability of the complement gets larger.

PROBLEM 2
My friend and I gamble on a roll of a die as follows: if the die shows 1 spot or 6 spots, I give my friend \$2. If the die show 2, 3, 4, or 5 spots, my friend gives me \$1. Suppose we play this game 100 times. What is my expected net gain?

Solution
$P(\text{lose})=\frac{1}{3},\ P(\text{win})=\frac{2}{3}$. Therefore $$E=\frac{1}{3}\times(-2)+\frac{2}{3}\times1=0$$
PROBLEM 3
A die is rolled 600 times. The “multiples of 3” are the faces with 3 spots and 6 spots. Find the expected value of:
a) the proportion of times multiples of 3 appear
b) the number of times multiples of 3 appear

Solution
3a) $$P(\text{multiples of 3 appear})=\frac{2}{6}=\frac{1}{3}$$ 3b) Binomial distribution, $$E=np=600\times\frac{1}{3}=200$$
PROBLEM 4
If you bet on a “split” in roulette, your chance of winning is 2/38. The bets pays 17 to 1; this means that if you bet \$1 on a split and win the bet, your net gain is \$17; if you lose the bet, you lose your dollar and so your net gain is -\$1. Suppose you place 200 bets on a split; assume your bets are independent of each other. Find your expected net gain.

Solution
The expected value of the proportion of the net gain is $$E_p=\frac{2}{38}\times 17+\frac{36}{38}\times(-1)=-\frac{1}{19}$$ Hence the expected value of the net gain is $$E=200\times(-\frac{1}{19})\doteq-10.52632$$ In other words you expect to come out losing about \$10.53.


EXERCISE SET 3

PROBLEM 1
In a roll of a die, let a "six" be the face with six spots. Consider the two events below.
Event A: more than 16,000 sixes in 100,000 rolls
Event B: more than 160,000 sixes in 1,000,000 rolls.
Which probability is larger?

Solution
According to the Law of Large Numbers, $P(A) < P(B)$. Alternatively, we can test it using smaller examples: $$P(\text{more than 160 sixes in 1000 rolls})$$ $$=\sum_{k=161}^{1000}C_{1000}^{k}\cdot(\frac{1}{6})^k\cdot(\frac{5}{6})^{1000-k}\doteq0.6971976$$ And $$P(\text{more than 1600 sixes in 10000 rolls})$$ $$=\sum_{k=1601}^{10000}C_{10000}^{k}\cdot(\frac{1}{6})^k\cdot(\frac{5}{6})^{10000-k}\doteq0.9626252$$ R code:

sum(dbinom(161:1000, 1000, 1/6))
[1] 0.6971976
sum(dbinom(1601:10000, 10000, 1/6))
[1] 0.9626252

PROBLEM 2
In the bet on a “single number” at roulette, there is chance 1/38 of winning. The bet pays 35 to 1; you can take this to mean that if you win the bet, your net gain is \$35, and if you lose the bet then you lose \$1 (that is, your net gain is -\$1).
You don’t need to know any more about roulette to solve this problem, but the description given above is what would happen if you were to bet $1 on a single number at a Las Vegas roulette table.
Suppose you bet repeatedly on a single number; assume that the bets are independent of each other. Find the expected value of your net gain from 38 bets.

Solution
$$E=38\times(\frac{1}{38}\times35+\frac{37}{38}\times(-1))=38\times(-\frac{2}{38})=-2$$
PROBLEM 3
400 draws are made at random with replacement from 5 tickets that are marked -2, -1, 0, 1, and 2 respectively. Find the expected value of:
3A
the number of times positive numbers appear
3B
the sum of all the numbers drawn
3C
the sum of the positive numbers drawn

Solution
3A) $$E(\text{number of times positive numbers appear})=400\times\frac{2}{5}=160$$ 3B) $$E(\text{sum of all the numbers drawn})=0$$ 3C) $$E(\text{sum of the positive numbers drawn})$$ $$=400\times(1\times\frac{1}{5}+2\times\frac{1}{5})=400\times\frac{3}{5}=240$$


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