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What is probability

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Horse Racing. 2. Relative Frequency. Probability is defined as relative frequency ... Usually, B(I)/B fluctuates especially near the start of the horse racing ... – PowerPoint PPT presentation

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Title: What is probability


1
What is probability?
  • Horse Racing

2
Relative Frequency
  • Probability is defined as relative frequency
  • When tossing a coin, the probability of getting a
    head is given by m/n
  • Where n number of tossings
  • m number of heads in n tossings

3
But .
  • Some events cannot be repeated
  • In general, how can we find a probability of an
    event?

4
Gambling
  • The origin of modern probability theory
  • Odds against an event A ? (??)
  • ? (1-P(A))/P(A)

5
If A Does Not Occur
  • We bet 1 on the occurrence of the event A
  • If A does not occur, we lose 1
  • In the long run, we will lose (1 P(A))
  • Notice that we just ignore N, the number of the
    repeated games

6
If A occurs
  • We will win ? in the long run for a fair
    game------ A game that is acceptable to both
    sides.
  • Why?

7
Fair Game
  • - (1 P(A)) ? P(A) 0
  • Because ? P(A) 1 P(A)
  • That is the game is fair to both sides

8
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

9
The equivalence between P(A) and ?
  • ? (1 P(A)) / P(A)
  • Conversely, P(A) 1 / (1 ?)

10
Example
  • 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)?
  • Odds4/161/4
  • P(A)1/(11/4)4/5

11
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)

12
For a fair coin
  • A---the occurrence of a head in one tossing
  • Now P(A) 0.5
  • ? (1 P(A)) / P(A) 1

13
P(A) ?
  • ? ( 1 - ? ) / ?
  • If ? .5, ?
  • If ? 1

14
A First Prize of Mark Six
  • Match 6 numbers out of 48
  • P(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

15
What is ? ?
  • ? 12,271,511
  • That is, you should win 12,271,511 for every
    dollar you bet
  • Payoff 1 (bet) 12,271,511 (gain)
  • In general, Payoff par dollar 1 ?

16
The pari-mutuel system
  • A race with N horses (5
  • The bet on the i_th horse is B(i)
  • We concern about which horse will win??
  • 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)

17
What is ? ?
  • ? B / B(I) - 1
  • Let P(I) denote the winning probability of the
    horse I
  • P(I) 1 / (? 1) B(I) / B
  • That is the proportion of the bet on the horse I
    is the winning probability of the horse i

18
Implication
  • 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?

19
Reality
  • Tracks take t (0.17
  • If horse I wins, the payoff per dollar bet on
    horse I, M(I) B(1-t)/B(I)

20
What is ? ?
  • ? B(1-t)/B(I) - 1

21
If 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
    run
  • If I win, I will win p(I) (M(I) 1) in the
    long run
  • What will happen if p(I) B(I) / B ?

22
If 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.

23
Objective 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.

24
Past data
  • In Australia and the USA, favorite (??) or
    near-favorite are underbet while longshots (??)
    are overbet.
  • But it is not so in Hong Kong.

25
Difficulty in assessing probability
  • 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

26
Contd
  • 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.

27
Accidence
  • 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

28
Which of the following is correct?
  • (1) The chance that you will be in a fatal plane
    crash is 1 in 2 million.
  • (2) In the long run, about 1 out of every 2
    million flight departures end in a fatal crash
  • (3) The probability that a randomly selected
    flight departure ends in a fatal crash is about
    1/(2,000,000)

29
Birthday
  • 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.

30
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 probability
  • Answer (1) (2)

31
Monty Hall Problem
  • Three doors with one car behind one of the doors
  • There are two goats behind the other two doors
  • You choose one door
  • Instead 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

32
Monty Hall Problem
  • Want to change your choice to the other unopened
    door.
  • Should you change?

33
Improve your assessment
  • Given the occurrence of B, what is your updated
    assessment of P?
  • Answer--Bayes Theorem
  • P(AB)P(BA)P(A) / (P(BA)P(A)P(BA)P(A))
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