Math 147 Lecture Notes: Lecture 21

Math 147 Lecture Notes: Lecture 21 Walter Carlip March, 2018 The Probability of an Event is greater or less, according to the number of Chances by which it may happen, compared with the whole number of Chances by which it may either happen or fail. Wherefore, if we constitute a Fraction whereof the Numerator be the number of Chances whereby an Event may happen, and the Denominator the number of all the Chances whereby it may either happen or fail, that Fraction will be a proper designation of it happening. Abraham de Moivre (1667-1754) Copyright c 2018 by Walter Carlip 1 Lecture 21

Today s Topics Announcements Box Models Sum of Draws Expected Value Standard Error Copyright c 2018 by Walter Carlip 2 Lecture 21

Announcements Don t forget that Exam II will be held in the evening next Wednesday. Everything we have covered or is covered in the text, through next Wednesday s lecture is fair game for the exam. That will include Chapters 1 through Chapter 17 of the text, and whatever portion of Chapter 18 we complete. While the emphasis will be on the newer material (Chapters 10 17), these chapters made use of the earlier material, so you will be responsible for the earlier material as well. The exam will certainly be closed book. You will be permitted to bring one 3 5 inch index card with notes. Last time people brought in other sizes of notes. Do not repeat this error. Index cards are cheap and available at the campus bookstore. Do not bring your text or other notes or study materials. You will need a simple calculator capable of performing basic arithmetic operations, including square roots. Do NOT bring a calculator with advanced functions! You may not borrow a calculator, and proctors will not give out loaners. Penalties will be higher this time! You may not use a laptop or pad computer, cell phone, pda, or other electronic device. If you bring any such device to class, it must be powered off and stowed away out of sight. Photo IDs may be checked during the exam, so have your ID available and in sight. Copyright c 2018 by Walter Carlip 3 Lecture 21

I should not need to say this, but do not converse or exchange anything with another student during the exam. Copyright c 2018 by Walter Carlip 4 Lecture 21

Announcements The deadline for requesting accommodations or rescheduling of the exam because of a conflict is today. Don t forget to email me. Unfortunately, I must cancel office hours today. I will hold some extra office hours tomorrow (Thursday) from 2:00 4:00pm. Copyright c 2018 by Walter Carlip 5 Lecture 21

Box Models In the next few sections we will be studying chance events, in particular, learning to identify the expected result and then to predict how far from the expected result a typical case should be. In other words, we will try to quantify the chance error. Here are some typical examples of the kind of questions we will be studying. How many Heads appear in some number of coin tosses? The expected result is half Heads, but chance comes into the outcome of each throw, resulting in chance error in the total number of Heads. How much money is is won or lost playing roulette? We ll need to know more about roulette to determine the expected result. Chance comes in on each spin of the wheel. What is the percentage of Democrats in a random sample of voters? The expected result depends upon the population we are selecting from. A chance process is used to choose the sample. Different samples will have different proportions of Democrats. Copyright c 2018 by Walter Carlip 6 Lecture 21

Box Models We will have a general strategy for studying such questions. First find an analogy between the process being studied and drawing numbers at random from a box. Connect the expected value and the chance variability in the process being studied with the expected value and chance variability in the sum of numbers drawn from the box. The box of numbers corresponding to a process we wish to study is called a Box Model for the process. The expected value and chance variability for draws from the box will be easy to study mathematically, and the results will be used to draw conclusions about the original process. This is a common tool in all of mathematics: Find an abstract model for some problem we wish to study, and use our mathematical analysis of the abstract model to draw conclusions about the original problem. Copyright c 2018 by Walter Carlip 7 Lecture 21

Sum of Draws Imagine the following process. A box is filled with tickets numbered 1 through 6: 1 2 3 4 5 6 A card is drawn at random from the box and returned to the box. A second card is drawn from the box and returned. Question: What is the sum of the draws? This is a chance process, since the cards are drawn at random. The smallest sum you can attain is 2 (when 1 is drawn both times) and the largest is 12 (when 6 is drawn both times). The draws are independent, so the chance of the result being 2 is 1 6 1 6 = 1 36 and the chance of the result being 12 is also 1 6 1 6 = 1 36. What about the other possible sums (3 through 11)? We already know the answer! This box model describes the process of rolling a pair of dice and adding the spots. We analyzed that problem in Lecture 18 by counting the possible outcomes (36 different ways the dice can fall) and identifying which of these equally likely outcomes have a given sum. Copyright c 2018 by Walter Carlip 8 Lecture 21

Possible Rolls of Two Dice Copyright c 2018 by Walter Carlip 9 Lecture 21

Sum of Draws Now imagine the same box 1 2 3 4 5 6 and independent draws with replacement, but now suppose we draw 25 times. Question: Now what is the sum of the draws? The analysis is now much more complicated. We can see easily that the smallest possible sum is 25 (when you select 1 each time) and the largest is 6 25 = 150 (when you select 6 each time). You should be able to see that every value in between could arise. There is chance variability in the sums obtained, because the draws are at random. The next slides show some data: The sums of 25 picks (or rolls) for three trials. The results of 1000 trials, with mean, maximum, and minimum. The results of 1000 trials displayed as a histogram. The results of 20, 000 trials displayed as a histogram. Copyright c 2018 by Walter Carlip 10 Lecture 21

Sum of Twenty-five Dice Rolls Copyright c 2018 by Walter Carlip 11 Lecture 21

Thousand Sums of Twenty-five Dice Copyright c 2018 by Walter Carlip 12 Lecture 21

Thousand Sums Histogram Copyright c 2018 by Walter Carlip 13 Lecture 21

Twenty Thousand Sums Histogram Copyright c 2018 by Walter Carlip 14 Lecture 21

Sum of Draws For the remainder of the course, when we refer to the sum of draws from a box, we will always be referring to the process in which we Draw tickets at random from a box. Add up the number on the tickets to obtain the outcome. Copyright c 2018 by Walter Carlip 15 Lecture 21

Making a Box Model To make a box model you must analyze the problem at hand and decide: What numbers should go into the box? How many of each kind? How many draws are to be made? As an example, we will examine the gambling game of roulette. Copyright c 2018 by Walter Carlip 16 Lecture 21

Roulette A Nevada roulette wheel has 38 pockets. One is numbered 0 and another 00. The remaining pockets are numbered 1 through 36. The croupier spins the wheel and throws a ball on the wheel. The ball is equally likely to land in any of the 38 pockets. Before the ball lands bets are placed on the table (illustrated on the next slide). Copyright c 2018 by Walter Carlip 17 Lecture 21

The Roulette Table Copyright c 2018 by Walter Carlip 18 Lecture 21

The Roulette Wheel Copyright c 2018 by Walter Carlip 19 Lecture 21

Red or Black With the exception of the 0 and 00 pockets, which are colored Green, the other pockets alternate Red and Black. One possible bet is to place a dollar on Red. If the ball lands in a Red pocket, you get your dollar back and win an additional dollar. If the ball lands on Black or Green, the croupier smiles and rakes in your dollar. What would a Box Model for this bet look like? On each bet you either win $1.00 or lose $1.00, so the box should contain tickets labelled $1 and $-1. Since 18 of the pockets are colored Red, the box should contain 18 tickets labelled $1. The remaining 20 pockets are Black or Green, so 20 of the tickets should be labelled $-1. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1 Copyright c 2018 by Walter Carlip 20 Lecture 21

Using the Box Model As far as the chances go, betting on Red is just like drawing from the box. The advantage of the box is its abstraction: unnecessary details, like the colors, the wheel, the croupier s smile, have been eliminated. Copyright c 2018 by Walter Carlip 21 Lecture 21

Using the Box Model Imagine now that you have bet on Red ten times. The table below shows ten games, with the matching box draws and cumulative winnings. Note that the net winnings (or losses) are equal to the sum of the draws from the box. Copyright c 2018 by Walter Carlip 22 Lecture 21

The Box Model The box model is built to correspond to the bet being made. The numbers on the tickets correspond to the amounts that can be won or lost on a single play. The chance of drawing a given number from the box should be equal to the chance of winning that amount in a single play. (Of course, losing corresponds to winning a negative amount.) The number of draws is equal to the number of plays. Copyright c 2018 by Walter Carlip 23 Lecture 21

Another Roulette Example A $1.00 bet is placed in the four numbers intersection of 23,24,26,27. According to the rules, if one of these numbers comes up, you win $8.00 and otherwise you lose your dollar. Suppose that you play this game 100 times. Your net gain is like the sum of draws made at random with replacement from the box. Solution: What numbers go into the box? On one bet you either win $8.00 or lose $1.00, so the tickets should be numbered either 8 or -1. How many of each kind? Since every pocket is equally likely to come up and four of them are winners, there should be four tickets labelled 8. The remaining 34 pockets result in a loss of $1, so there should be 34 tickets labelled -1. How many draws? Of course the number of draws should be 100, because you play the game 100 times. Copyright c 2018 by Walter Carlip 24 Lecture 21

Expected Value When we try to make sense of chance processes, we will usually be interested in two things. The first of these is the expected value of the process. Thus, for example, when flipping a coin 1000 times and counting Heads, the expected value is 500. A box model for coin flipping and counting Heads would look like this: 0 1 If we imagine choosing 1 to correspond to flipping Heads and 0 to Tails, it is clear that the sum of 1000 draws corresponds to the number of Heads chosen. Our first task will be to determine the expected value of a given box model. Copyright c 2018 by Walter Carlip 25 Lecture 21

Chance Error The second thing we will be interested in is the chance error. When we flip a coin 1000 times and count Heads, the expected value is 500, but, because chance is involved, we will sometimes get 510, or 520, or 490, or 475. As usual, the difference between the value actually attained and the expected value is the chance error. If we get 510 Heads, the chance error is 510 500 = 10; if we get 520 Heads, the chance error is 520 500 = 20; if we get 490 Heads, the chance error is 490 500 = 10; and if we get 475 Heads, the chance error is 475 500 = 25. The chance error may be positive or negative. We will be interested in trying to quantify the chance error, and, in particular, to identify a standard error, which will be the typical size of the chance error. (The standard error will play a role similar to the standard deviation in our study of data distributions.) Copyright c 2018 by Walter Carlip 26 Lecture 21

The Expected Value Let s start by trying to identify the expected value for the process of rolling a die 600 times and adding the results. What should the expected value be? In 600 rolls, how many times should we expect to come up? Since we expect each number to come up about 1 6 of the time, we should expect to come up on about 1 6 of the 600 rolls, that is, 100 times. Similarly, we should expect each of,,,, and to come up 100 times. Putting this together, we should expect the sum of the 600 rolls to be 100 1+100 2+100 3+100 4+100 5+100 6 = 100 (1+2+3+4+5+6) = 100 21 = 2100. Copyright c 2018 by Walter Carlip 27 Lecture 21

The Expected Value Now what should we expect with 25 rolls? We can repeat the analysis we made for 600 rolls. We still expect about 1 6 of the rolls to come up with each of the six possible values. That is, we expect about... 25 6 25 6 25 6 25 6 25 6 25 6 Therefore, the expected value should be 25 6 1+ 25 6 2+ 25 6 3+ 25 6 = 25 6 (1+2+3+4+5+6) = 25 (1+2+3+4+5+6) 6 = 25 21 6 = 87.5 4+ 25 6 5+ 25 6 6 Copyright c 2018 by Walter Carlip 28 Lecture 21

Clever Trick! We did something tricky in the computation. Notice that by shifting the 6 under the sum (1+2+3+4+5+6), we obtained the average value for the box. We can summarize our analysis as follows: Theexpected value of the sumof somenumberof draws from the box 1 2 3 4 5 6 is (number of draws) (average of the box). Copyright c 2018 by Walter Carlip 29 Lecture 21

Look at the Data! Notice that in the simulated data, the average of the sums over 1000 repetitions was 87.795, and for 20, 000 repetitions, the average of the sums was 87.5434. This is about what we would expect! Copyright c 2018 by Walter Carlip 30 Lecture 21

Thousand Sums Histogram Copyright c 2018 by Walter Carlip 31 Lecture 21

Twenty Thousand Sums Histogram Copyright c 2018 by Walter Carlip 32 Lecture 21

Any Box Our analysis works for any box model. Imagine another box, say 1 1 2 2 2 Imagine 500 draws from the box. How many of each number would you expect? We would expect about 1 5 of the draws to be each of the numbers in the box, that is 500 5 of 1 (first one) 500 5 of 1 (second one) 500 5 of 2 (first two) 500 5 of 2 (second two), and 500 of 2 (third two). 5 Therefore, the expected value should be 500 5 1+ 500 5 1+ 500 5 2+ 500 5 = 500 5 (1+1+2+2+2) = 500 (1+1+2+2+2) 5 = 500 8 5 = 100 8 = 800. 2+ 500 5 2 Copyright c 2018 by Walter Carlip 33 Lecture 21

The General Rule Notice that as before, the expected value was equal to the number of draws times the average value of the box. This is always true. Theexpected value of the sumof somenumberof draws from the any box is (number of draws) (average of the box). Copyright c 2018 by Walter Carlip 34 Lecture 21

Red or Black Let s go back and apply what we ve learned about expected value to roulette, beginning with the basic dollar bet on Red. We place a dollar on Red. If the ball lands in a Red pocket, you get your dollar back and win an additional dollar. If the ball lands on Black or Green, the croupier smiles and rakes in your dollar. Recall the Box Model for this bet. On each bet you either win $1.00 or lose $1.00, so the box should contain tickets labelled $1 and $-1. Since 18 of the pockets are colored Red, the box should contain 18 tickets labelled $1. The remaining 20 pockets are Black or Green, so 20 of the tickets should be labelled $-1. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1 Copyright c 2018 by Walter Carlip 35 Lecture 21

Using the Box Model Now let s apply what we just learned about expected value. What is the expected value for 100 bets on Red? The box contains 18 cards labeled 1 and 20 cards labeled 1, so the sum of the values in the box is 2 and the average of the box is 2 38 = 1 19. Therefore, the expected value of 100 bets on Red is 100 1 19 = 100 19 5.26. While chance error might produce a positive value, or a negative value less than $ 5.26, the expected outcome of betting on Red is to lose (on the average) about $ 5.26 for every $100.00 bet. Copyright c 2018 by Walter Carlip 36 Lecture 21

Another Roulette Example: Intersections A $1.00 bet is placed in the four numbers intersection of 23,24,26,27. According to the rules, if one of these numbers comes up, you win $8.00 and otherwise you lose your dollar. Suppose that you play this game 100 times. What is your expected winnings? Solution: As we determined earlier, your winnings are like the sum of 100 draws from the box 8 8 8 8-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1-1 -1 Since every pocket is equally likely to come up and four of them are winners, there should be four tickets labelled 8. The remaining 34 pockets result in a loss of $1, so there should be 34 tickets labelled -1. Stated in terms of probability, the chances of winning are 4 34 38 and the chances of losing are 38. To compute the expected value of the sum of 100 draws, we compute the average of the box and multiply by the number of draws. Copyright c 2018 by Walter Carlip 37 Lecture 21

Intersection Bets The box contains 4 cards labeled 8 and 34 cards labeled 1, so the sum of the values in the box is 4 8+34 ( 1) = 32 34 = 2 and the average of the box is 2 38 = 1 19. Therefore, the expected value of 100 bets on an intersection is therefore 100 1 19 = 100 19 5.26. While chance error might produce a positive value, or a negative value less than $ 5.26, the expected outcome of betting on an intersection is to lose (on the average) about $ 5.26 for every $100.00 bet. (Sound familiar?) Copyright c 2018 by Walter Carlip 38 Lecture 21

Quick Summary Sum of Draws The sum of draws refers to the process in which we Draw tickets at random from a box. Add up the number on the tickets to obtain the outcome. Expected Value The expected value of the sum of some number of draws from a box of numbered tickets is expected value = (number of draws) (average of box). Chance Error The chance error is the difference between an actually observed value and the expected value: chance error = observed value expected value. The chance error is positive if the observed value is greater than the expected value and negative if the observed value is less than the expected value. Copyright c 2018 by Walter Carlip 39 Lecture 21

Standard Error We will be interested in trying to quantify the chance error, and, in particular, to identify a standard error, which will be the typical size of the chance error. (The standard error will play a role similar to the standard deviation in our study of data distributions.) We first examine this question with a particular box and number of draws in mind, say, 25 draws from the box 0 3 4 8 10 The smallest possible sum of 25 draws from this box is 0 and the largest is 250. Other sums land somewhere between, and the expected value is 25 0+3+4+8+10 5 = 25 25 5 = 25 5 = 125. Copyright c 2018 by Walter Carlip 40 Lecture 21

Chance Error If we draw 25 times from the box 0 3 4 8 10 and compute the sum, the expected value is 125, but because chance is involved, the observed result may be larger or smaller than 125. If, for example, the sum of draws is 130, then the chance error is 130 125 = 5. If, on the other hand, the sum of draws is 119, then the chance error is 119 125 = 6. How large is the chance error likely to be? It turns out that there is a value, the standard error (abbreviated SE) which expresses how large we should expect the chance error to be. A sum of draws is likely to be close to its expected value, but off by a chance error roughly similar in size to the standard error. Copyright c 2018 by Walter Carlip 41 Lecture 21

The Standard Error The standard error for the sum of draws (with replacement) from a box is computed using the square root law: The standard error for the sum of random draws (with replacement) from a box of numbered tickets is number of draws (standard deviation of the box). Some observations about the standard error are in order: If the SD of the box is large, the numbers in the box are more spread out, and the sum of draws is less predictable. The larger the SD of the box, the larger the standard error. As the number of draws increases, the standard error also increases. The more draws, the larger the sum and more variable the results. The sum is likely to be somewhat farther from the expected value when there are more draws. Although the standard error increases when the number of draws increases, it does not increase as rapidly as the number of draws. The sum of 100 draws is only 100 = 10 times more variable than a single draw. Copyright c 2018 by Walter Carlip 42 Lecture 21

Example We have already computed that the expected value for 25 draws from the box 0 3 4 8 10 is 125. We now compute the standard error to estimate how far from the expected value a typical observed value will be. First we compute the SD of the box. The average of the box is 0+3+4+8+10 5 The deviations from average are = 25 5 = 5. 5 2 1 3 5 The SD of the box is then ( 5)2 +( 2) 2 +( 1) 2 +(3) 2 +(5) 2 5 = 64 5 3.58 Finally SE for the sum of 25 draws (with replacement) from the box is 64 25 5 3.58 = 17.9. 5 This tells us is that a typical draw of 25 will miss the expected value of 125 by about 18. (We will make this observation more precise soon.) Copyright c 2018 by Walter Carlip 43 Lecture 21

Homework Please read Chapters 17, 18, and 19 carefully!. Homework Week Date Assignment Chapter Page Set Problems 9 March 14, 2018 Reading 17 Set #17 290 A 1, 2, 3, 6 293 B 1, 2, 3 296 C 1, 2, 3, 4, 8 299 D 1, 2, 3 303 E 1, 2, 3 304 Review 1, 2, 3, 6, 10 10 March 19, 2018 Reading 18 Set #18 312 A 1, 2, 3 318 B 1, 2, 3, 5 324 C 1, 2, 5, 6, 7 327 Review 1, 2, 4, 5, 7 10 March 23, 2018 Reading 19 Set #19 349 A 3, 5, 6, 11 351 Review 2, 5, 7, 9 Copyright c 2018 by Walter Carlip 44 Lecture 21