Probability

Hong Kong

Stage 1 - Stage 3

Lesson

Suppose that you ask your grandmother for some money. Instead of just giving you the money, she offers to flip a coin.

- If the coin lands “heads” – you win – and she will give you $\$20$$20.
- If it lands “tails”, you lose, and you receive nothing.

In this example, we would say that the payout associated with the event that the coin lands “heads” is $\$20$$20, and the payout associated with “tails” is $\$0$$0.

On of the most common ways of evaluating if this is a good deal or not is to calculate the expected value.

The expected value is the average of all possible outcomes, adjusted for the probability that each outcome will occur.

If we use X to indicate the possible payouts then,

The heads scenario: $X_H=\$20$`X``H`=$20

The tails scenario: $X_T=\$0$`X``T`=$0

The **expected value** = **probability of heads** x **payout for a head** + **probability for tails** x **payout for tails**

$E(X)=$`E`(`X`)= $P(H)$`P`(`H`) × $X_H$`X``H` $+P(T)$+`P`(`T`) × $X_T$`X``T`

$E(X)=$`E`(`X`)= $\frac{1}{2}\times20+\frac{1}{2}\times0$12×20+12×0

$E(X)=$`E`(`X`)= $10$10

Later that day, whilst you are out wandering, you bump into your Uncle. You ask him for the $\$20$$20. Instead of just giving you the money he offers the following deal. I will roll a dice and give you $\$3$$3 for every spot on the side of the dice.

Using the expected value is one way of comparing the two offers. For this offer we have the following payout options:

$X_1=\$3$`X`1=$3

$X_2=\$6$`X`2=$6

$X_3=\$9$`X`3=$9

$X_4=\$12$`X`4=$12

$X_5=\$15$`X`5=$15

$X_6=\$18$`X`6=$18

Multiplying each by the respective probabilities gives

$E(X)$E(X) |
$=$= | $\frac{1}{6}\times X_1+\frac{1}{6}\times X_2+\frac{1}{6}\times X_3+\frac{1}{6}\times X_4+\frac{1}{6}\times X_5+\frac{1}{6}\times X_6$16×X1+16×X2+16×X3+16×X4+16×X5+16×X6 |

$=$= | $\frac{1}{6}\times3+\frac{1}{6}\times6+\frac{1}{6}\times9+\frac{1}{6}\times12+\frac{1}{6}\times15+\frac{1}{6}\times18$16×3+16×6+16×9+16×12+16×15+16×18 | |

$=$= | $\frac{1}{2}+1+\frac{9}{6}+2+\frac{15}{6}+3$12+1+96+2+156+3 | |

$=$= | $\$10.50$$10.50 |

So simply comparing the expected value of the offers the uncles deal appears to be a slightly better one. Of course, you might like to risk the knowledge of definitely getting at least $\$3$$3 from your uncle for the chance of the $\$20$$20 from you grandmother.

Han buys a $1500 laptop and can also purchase a 3-year extended warranty on the laptop. The manufacturer's historical records show that the probability of the laptop needing repair, which costs $403 on average, is 0.06 in the 3-year period after purchase and the probability of the laptop needing to be replaced is 0.03.

- Calculate the expected value of the extended warranty.
If the warranty costs $155, is the warranty fair value?

Yes

ANo

B

Dave and Mae are fighting over who gets to watch the TV at home. They decide to settle their differences by rolling a die. Dave rolls first. He wins control over the TV if he rolls a number greater than four. If he doesn't win, Mae gets her chance and wins if she rolls a prime number. If she doesn't roll a prime number, their parents take the remote and nobody wins.

What is Dave's chance of winning?

What is Mae's probability of winning?

Is this a fair way to settle the argument for both Dave and Mae?

Yes

ANo

B

Given the following options:

(A) a sure gain of $\$500$$500

(B) a $50$50% chance of gaining $\$1000$$1000 and a $50$50% chance of gaining nothing

Calculate the financial expectation of Option A.

Calculate the financial expectation of Option B.

Which option has greater variation?

Option A

AOption B

B