Expected Value, continued

Transcription

1 Expected Value, continued

2 Data from Tuesday On Tuesday each person rolled a die until obtaining each number at least once, and counted the number of rolls it took. Each person did this twice. The data is available from the website under HowManyRolls.xls on the materials page. 2

3 To summarize the data, the average number of the 95 rolls was This is then an estimate for the theoretical expected value of how many rolls does it take. The actual expected value is 14.7, which is very close to the average of the class data. The following graph is a visual representation of the individual class data. 3

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5 With a computer program, we can simulate this experiment. The file HowManyRolls- Mapledata.pdf on the materials link of the website shows the result of doing this simulation with 10,000 trials, 100,000 rolls, and 1,000,000 rolls. 5

6 The New Mexico Lottery - Brief Review 6

7 Powerball To play you choose 5 numbers between 1 and 55 and 1 Powerball number from 1 to 42. A winning number (5 numbers and 1 Powerball number) is selected. How much you win depends on how many of your numbers match the winning numbers. 7

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9 How are the odds calculated? It comes from the ideas we have discussed. First, how many different Powerball tickets are there? First you have to choose 5 out of 55 numbers. There are 55C5 ways to do this. By using a calculator, we see that 55C5 = 3,478,761 9

10 Next, you have to choose a Powerball number. Since there are 42 choices, there are 42 ways to do this. So, the total number of ways to choose a ticket are 3,478,761 * 42 = 146,107,962 10

11 There is only one way to win the jackpot, by matching all the numbers. So, the probability of winning is 1/146,107,962. The next best way to win is to match all 5 regular numbers but not the Powerball number. To do this there is only one way to choose the 5 regular numbers, but 41 ways to choose the (wrong) Powerball number. 11

12 The probability of winning this is then which is about 41 / 146,107,962, 1 / 3,563,

13 The least amount you can win is $3 by matching the Powerball but none of the regular numbers. To win this way there is only one way to choose the Powerball number; you have to pick the winning number. For the regular numbers, you cannot choose any of the 5 winning numbers. You must then choose 5 of the 50 losing numbers. 13

14 There are 50C5 ways to choose 5 of the losing numbers. The number 50C5 is equal to The probability of winning $3 is then 50C5 / 146,107,962 = / 146,107,962, which is about.0145, or about 1/69. 14

15 If you are curious how one makes this calculation, once you use a calculator to obtain the.0145, hit the x^{-1} (or 1/x) key, and you will get something close to 69. Whatever you get, the probability is then 1 out of this number. 15

16 Pick 3 16

17 In this game you pick 3 digits (0-9). You also choose a play type. The options are straight (STR), box, or straight/box. A single play costs $1. How much you win depends on the play type you choose and how your number compares to the winning number. 17

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19 The original table can be found at 19

20 Let s consider this game. First of all, there are 10 ways to choose each of the three digits. How you choose one digit does not affect the other choices. So, to find the total number of choices, we calculate 10 * 10 * 10 = Thus, there are 1000 ways to play (e.g., ways to fill out a pick 3 ticket). 20

21 Another way to think about this is to consider your number a 3 digit number. You can choose 0 = 000, 1 = 001, 2, 3, and so on, up to 999. There are then 1000 possible numbers to choose. 21

22 As he table above indicates, the probability of winning depends on what number you choose. Let s first consider a number with 3 different digits; say you choose 123. In this case, to win playing straight, you only win if the winning number is exactly your number. This means that of the 1000 possibilities, only 1 will win. Therefore, you have a 1/1000 chance of winning. 22

23 On the other hand, if you play box, you will win as long as the three winning digits are 1, 2, 3 in any order. The winning numbers are then 123, 132, 213, 231, 312, 321 There are 6 ways to arrange the three digits. So, you have a 6/1000, or about a 1 in 167, chance to win with this game. 23

24 The expected value of playing straight is then $500 * 1/ $1 * 999/1000 = -$499 / 1000 = -$.499. The expected value of playing box is $80 * 6/ $1 * 994/1000 = -$514 / 1000 = -$

25 So, playing straight is a little better than playing box, since the expected value is less negative. Both games have an expected value of close to -$.50, or -50 on a $1 bet, which means, on average, half of the amount of money people spend playing is revenue to the state. 25

26 If you choose a number with 2 of the same digit, like 554, then the winning numbers when playing box are 554, 545, 455 only 3 possibilities. The probability of winning is then 3/1000 in this case. 26

27 Playing such a number, the expected value of playing straight is still the same, since the payout is the same, and the chance of winning is still 1/1000. If you play box, the expected value is 3/1000 * $ /1000 * $1 = -$517 / 1000 = -$.517, which is slightly worse than when playing with 3 different digits. 27

28 If you play straight/box, then you have two ways to win. Let s just consider playing 3 different digits, say 123. You win $290 if the winning number is 123, and there is a 1/1000 chance of this happening. You win $40 if the winning number is one of 132, 213, 231, 312, 321. You then have a 5/1000 chance of winning this way. You then have a 994/1000 chance of losing. 28

29 The odds listed on the ticket (1 out of 167) are the total odds of winning, meaning winning in either way. In order to find the expected value, we need to know the odds of winning in each way of the two ways. On Assignment 3 you will calculate the expected value of playing this game. 29

30 Example Here is a simple game in which there is more than one way to win. Suppose you bet $1 upon the outcome of rolling a die. You will win $2 if you roll a 6 and $1 if you roll a 1. You will lose otherwise. What is the expected value of a $1 bet? There is a 1/6 chance to win $2, a 1/6 chance to win $1, and a 4/6 chance of losing. 30

31 To find the expected value, you multiply the winning (or losing) amount by the corresponding probability, and add up these terms. For this game we get $2 * 1/6 + $1 * 1/6 - $1 * 4/6 = -$1/6, which is about 17. So, on average, you d expect to lose 1/6 of the total amount you bet. 31