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1 What is probability? Horse Racing. 2 Relative Frequency Probability is defined as relative frequency When tossing a coin, the probability of getting.

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Presentation on theme: "1 What is probability? Horse Racing. 2 Relative Frequency Probability is defined as relative frequency When tossing a coin, the probability of getting."— Presentation transcript:

1 1 What is probability? Horse Racing

2 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 3 But …. Some events cannot be repeated In general, how can we find a probability of an event?

4 4 Gambling The origin of modern probability theory Odds against an event A =  ( 賠率 )  = (1-P(A))/P(A)

5 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 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 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 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 9 The equivalence between P(A) and   = (1 – P(A)) / P(A) Conversely, P(A) = 1 / (1 +  )

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

11 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 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 13 P(A) =   = ( 1 -  ) /  If  >.5,  < 1 If  1

14 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 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 16 The 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)

17 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 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 19 Reality 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)

20 20 What is  ?  = B(1-t)/B(I) - 1

21 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 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 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 24 Past 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.

25 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 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 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 28 Which of the following is 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)

29 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 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 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 32 Monty Hall Problem Want to change your choice to the other unopened door. Should you change?

33 33 Improve your assessment Given the occurrence of B, what is your updated assessment of P? Answer--Bayes Theorem P(A|B)=P(B|A)P(A) / (P(B|A)P(A)+P(~B|A)P(A))


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