1 What is probability? Horse Racing. 2 Relative Frequency Probability is defined as relative...

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What is probability?What is probability?

Horse Racing

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Relative FrequencyRelative Frequency

Probability is defined as relative frequencyWhen tossing a coin, the probability of

getting a head is given by m/nWhere n = number of tossings m = number of heads in n tossings

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But ….But ….

Some events cannot be repeated

In general, how can we find a probability of an event?

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GamblingGambling

The origin of modern probability theoryOdds against an event A = (賠率 ) = (1-P(A))/P(A)

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If A Does Not OccurIf A Does Not Occur

We bet $1 on the occurrence of the event AIf A does not occur, we lose $1In the long run, we will lose – (1 – P(A))Notice that we just ignore N, the number of

the repeated games

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If A occursIf A occurs

We will win $ in the long run for a fair game------ A game that is acceptable to both sides.

Why?

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Fair GameFair Game

- (1 – P(A)) + P(A) = 0Because P(A) = 1 – P(A)That is the game is fair to both sides

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Interpretation of Interpretation of

The amount you will win when A occurs assuming you bet $1 on the occurrence of A

Gambling--- if is found and acceptable for both sides

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The equivalence between The equivalence between P(A) and P(A) and

= (1 – P(A)) / P(A)Conversely, P(A) = 1 / (1 + )

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ExampleExample

Bet $16 on event A provided if A occurs we are paid 4 dollars (and our $16 returned) and if A does not occur we lose the $16. What is P(A)?

Odds=4/16=1/4P(A)=1/(1+1/4)=4/5

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Is it arbitrary ?Is it arbitrary ?

The axioms of probability:(1) P(A) 0(2) P(S)=1 for any certain event S(3) For mutually exclusive events A and B,P(A B)=P(A) + P(B)

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For a fair coinFor a fair coin

A---the occurrence of a head in one tossingNow P(A) = 0.5 = (1 – P(A)) / P(A) = 1

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P(A) = P(A) =

= ( 1 - ) / If > .5, < 1If < .5, > 1

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A : First Prize of Mark SixA : First Prize of Mark Six

Match 6 numbers out of 48P(A) = (6/48) (5/47) (4/46) (3/45) (2/44)

(1/43) = 1 / 12,271,512 = 8.15 x 10^{-8) = .000,000,082

In the past, when we have only 47 numbers,P(A) = (6/47) (5/46) (4/45) (3/44) (2/43)

(1/42) = 1 / 10,737,573 = .000,000,09

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What is What is ? ?

= 12,271,511That is, you should win 12,271,511 for

every dollar you betPayoff = $1 (bet) + $12,271,511 (gain)In general, Payoff par dollar = 1 +

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The pari-mutuel systemThe pari-mutuel system

A race with N horses (5 < N < 12)The bet on the i_th horse is B(i)We concern about which horse will win獨贏

The total win pool B = B(1) + … + B(N)If horse I wins, the payoff per dollar bet on

horse I M(I) = B / B(I)

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What is What is ? ?

= B / B(I) - 1Let P(I) denote the winning probability of

the horse IP(I) = 1 / ( +1) = B(I) / BThat is the proportion of the bet on the

horse I is the winning probability of the horse i

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ImplicationImplication

The probability of winning can be reflected by the number B(I)/B

Usually, B(I)/B fluctuates especially near the start of the horse racing

Does this probability reflect the reality?

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RealityReality

Track’s take t (0.17 < t < 0.185)If horse I wins, the payoff per dollar bet on

horse I, M(I) = B(1-t)/B(I)

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What is What is ? ?

= B(1-t)/B(I) - 1

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If I bet on the horse iIf I bet on the horse i

Let p(I) denote the probability of the winning of the I_th horse

If I lose, I will lose – (1-p(I)) in the long runIf I win, I will win p(I) * (M(I) – 1) in the

long run What will happen if p(I) = B(I) / B ?

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If P(I) = B(I) / BIf P(I) = B(I) / B

In the long run, I will gain p(I) (M(I) –1) = 1 – t – B(I) / B

In the long run, I will lose – (1 – p(I)).So, altogether, I will lose –t.

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Objective probabilityObjective probability

From the record, we can group the horses with similar odds into one group and compute the relative frequency of the winners of each group

We find the above objective probability is very close to the subjective probability B(I) / B.

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Past dataPast data

In Australia and the USA, favorite (大熱 ) or near-favorite are “underbet” while longshots (泠馬 ) are “overbet”.

But it is not so in Hong Kong.

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Difficulty in assessing Difficulty in assessing probabilityprobability

Example(1) Your patient has a lump in her breast(2) 1% chance that it is malignant(3) mammogram result : the lump is

malignant(4) The mammograms are 80% accurate for

detecting true malignant lumps

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ContdContd

The mammogram is 90% accurate in telling a truly benign lumps

Question 1: What is the chances that it is truly malignant?

Ans. (1) less than .1%; (2) less than 1% but larger than .1%; (3) larger than 1% but less than 50%; (4) larger than 50% but less than 80%; (5) larger than 80%.

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AccidenceAccidence

There were 76,577,000 flight departures in HK in the last two years (hypothetical)

There were 39 fatal airline accidents (again, hypothetical)

The ratio 39/76,577,000 gives around one accident per 2 million departures

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Which of the following is Which of the following is correct?correct?

(1) (1) The chance that you will be in a fatal plane crash is 1 in 2 million.

(2) (2) In the long run, about 1 out of every 2 million flight departures end in a fatal crash

(3) (3) The probability that a randomly selected flight departure ends in a fatal crash is about 1/(2,000,000)

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BirthdayBirthday

How many people would need to be gathered together to be at least 50% sure that two of them share the same birthday?

(1) 20; (2) 23; (3) 28; (4) 50; (5) 100.

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Unusual hands in card games Unusual hands in card games

(1) 4 Aces, 4 Kings, 4 Queens and one spade 2.

(2) Spade (A, K, 3) Heart (3, 4, 5) Diamond (A, 2,4) Club (7, 8, 9, 10).

Which has a higher probabilityAnswer: (1) (2)

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Monty Hall ProblemMonty Hall Problem

Three doors with one car behind one of the doors

There are two goats behind the other two doors

You choose one doorInstead of opening the selected door, the

host would open one of the other door with a goat behind it. Then he would ask if you

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Monty Hall ProblemMonty Hall Problem

Want to change your choice to the other unopened door.

Should you change?

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Improve your assessmentImprove your assessment

Given the occurrence of B, what is your updated assessment of P?

Answer--Bayes TheoremP(A|B)=P(B|A)P(A) / (P(B|A)P(A)+P(~B|

A)P(A))