# Probability Theory, Part 4: Estimating Probabilities from Finite Universes

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1 8 Resampling: The New Statistics CHAPTER 8 Probability Theory, Part 4: Estimating Probabilities from Finite Universes Introduction Some Building Block Programs Problems in Finite Universes Summary Introduction The examples in Chapter 7 dealt with infinite universes, in which the probability of a given simple event is unaffected by the outcome of the previous simple event. But now we move on to finite universes, situations in which you begin with a given set of objects whose number is not enormous say, a total of two, or two hundred, or two thousand. If we liken such a situation to an urn containing balls of different colors each with a number on it, we are interested in the probability of drawing various sets of numbered and colored balls from the urn on the condition that we do not replace balls after they are drawn. In the cases addressed in this chapter, it is important to remember that the single events no longer are independent of each other. A typical situation in which sampling without replacement occurs is when items are chosen from a finite universe for example, when children are selected randomly from a classroom. If the class has five boys and five girls, and if you were to choose three girls in a row, then the chance of selecting a fourth girl on the next choice obviously is lower than the chance that you would pick a girl on the first selection. The key to dealing with this type of problem is the same as with earlier problems: You must choose a simulation procedure that produces simple events having the same probabilities as the simple events in the actual problem involving sampling without replacement. That is, you must make sure that your simulation does not allow duplication of events that have already occurred. The easiest way to sample without replace-

2 Chapter 8 Probability Theory, Part 4: Estimating Probabilities from Finite Universes 9 ment with resampling techniques is by simply ignoring an outcome if it has already occurred. Examples 8- through 8-0 deal with some of the more important sorts of questions one may ask about drawings without replacement from such an urn. To get an overview, I suggest that you read over the titles of Examples 8- to 8-0 before beginning to work through the examples themselves. This chapter also revisits the general procedure used in solving problems in probability and statistics with simulation, here in connection with problems involving a finite universe. The steps that one follows in simulating the behavior of a universe of interest are set down in such fashion that one may, by random drawings, deduce the probability of various events. Having had by now the experience of working through the problems in Chapters 5 and 7, the reader should have a solid basis to follow the description of the general procedure which then helps in dealing with specific problems. Let us begin by describing some of the major sorts of problems with the aid of an urn with six balls. Some building-block programs Case. Each of six balls is labeled with a number between and 6. We ask: What is the probability of choosing balls, 2, and in that order if we choose three balls without replacement? Figure 8- diagrams the events we consider success pick #: 2 Figure 8-: The Event Classified as Success for Case

3 20 Resampling: The New Statistics Case 2. We begin with the same urn as in Case, but now ask the probability of choosing balls, 2, and in any order if we choose three balls without replacement. Figure 8-2 diagrams two of the events we consider success. These possibilities include that which is shown in Figure 8- above, plus other possibilities pick #: 2 Figure 8-2: An Incomplete List of the Events Classified as Success for Case 2 Case. The odd-numbered balls,, and 5, are painted red and the even-numbered balls 2, 4, and 6 are painted black. What is the probability of getting a red ball and then a black ball in that order? Some possibilities are illustrated in Figure 8-, which includes the possibility shown in Figure 8-. It also includes some but not all possibilities found in Figure 8-2; for example, Figure 8-2 includes choosing balls 2, and in that order, but Figure 8- does not pick #: 2 Figure 8-: An Incomplete List of the Events Classified as Success for Case

4 Chapter 8 Probability Theory, Part 4: Estimating Probabilities from Finite Universes 2 Case 4. What is the probability of getting two red balls and one black ball in any order? pick #: 2 Figure 8-4: An Incomplete List of the Events Classified as Success for Case 4 Case 5. Various questions about matching may be asked with respect to the six balls. For example, what is the probability of getting ball on the first draw or ball 2 on the second draw or ball on the third draw? (Figure 8-5) Or, what is the probability of getting all balls on the draws corresponding to their numbers? pick #: 2 4 or (and) pick #: 2 4 or (and) 5 6 pick #: Figure 8-5: An Incomplete List of the Events Classified as Success for Case 5

5 22 Resampling: The New Statistics Problems in finite universes Example 8-: What is the Probability of Selecting Four Girls and One Boy When Selecting Five Students From Any Twenty-five Girls and Twenty-five Boys? (Sampling Without Replacement When There are Two Outcomes and the Order Does Not Matter) The important difference between this example and the infinite-universe examples in the prior chapter is that the probability of obtaining a boy or a girl in a single simple event differs from one event to the next in this example, whereas it stays the same when the sampling is with replacement. To illustrate, the probability of a girl is.5 (25 out of 50) when the first student is chosen, but the probability of a girl is either 25/49 or 24/49 when the second student is chosen, depending on whether a boy or a girl was chosen on the first pick. Or after, say, three girls and one boy are picked, the probability of getting a girl on the next choice is (28-)/(50-4) = 22/46 which is clearly not equal to.5. As always, we must create a satisfactory analog to the process whose probability we want to learn. In this case, we can use a deck of 50 cards, half red and half black, and deal out five cards without replacing them after each card is dealt; this simulates the choice of five students from among the fifty. We can also work the example with random numbers. But if so, we cannot simply designate 0-25 as boys and as girls (ignoring 5-00 when they occur), because if we did so the probability of a girl would be the same on each pick. Instead, to simulate non-replacement, we specify that no number can be taken into the sample twice, just as no student can be chosen twice. That is, if we come across a number more than once, we simply ignore it after the first occurrence. More specifically: Step. Let 0-25 = girls, = boys Step 2. Select five non-duplicating numbers; if a number comes up more than once, ignore it. Count whether there are four numbers 0-25, and one If so, write yes, otherwise no. Step. Repeat step 2 perhaps 400 times, and count the proportion yes, which estimates the probability sought.

6 Chapter 8 Probability Theory, Part 4: Estimating Probabilities from Finite Universes 2 The results of a few experimental trials are shown in Table 8-. Table 8- Experimental Trials Experiment Numbers Chosen Success? 8, 22, [8 selected but ignored] 27, 2, 49 No 2 7, 9, 8, 7, 9 Yes, 4, 2, 29, 24 Yes It is important to notice that in this problem we do not distinguish among particular girls (or boys). That is, it does not matter which girl (or boy) is selected in a given trial. Nor did we pay attention to the order in which we selected girls or boys. This is an instance of Case 4 discussed above. Subsequent problems will deal with situations where the order of selection, and the particular individuals, do matter. A solution to this problem with RESAMPLING STATS is presented below. Let s take an overview of the steps in that program: We seek to find the probability of randomly selecting four girls out of five children in a classroom of fifty children with equal numbers of boys and girls. We simulate the class by creating an array A with numbers through 50 in order, designating through 25 as girls and 26 through 50 boys. We then REPEAT the following steps a thousand times. We first SHUFFLE the elements of Vector A into Vector B to randomize the order of the children. We next TAKE the first five children in this randomly-ordered array B and place them in C. This simulates the random selection of five children. Then we COUNT the number of girls numbers to 25 in C. Next, we SCORE the number of girls selected for each random sample (D) into z. We need not concern ourselves with whether one boy was chosen, because we know that if four (and only four) girls are chosen, one boy must have been selected as well. After the thousand samples are completed, we determine how many of the random samples of five children contained four girls by COUNTing the number of times four girls were selected. Finally, we PRINT the result.

7 24 Resampling: The New Statistics NUMBERS,50 a Constitute the set of girls (numbers -25) and boys (26-50) and put those numbers in a. REPEAT 000 Repeat the following steps 000 times. SHUFFLE a b Shuffle the numbers, call the shuffled vector b. TAKE b,5 c Take the first 5 numbers, call them c. COUNT c between 25 d Count how many girls (numbers -25) there are, put the result in d. SCORE d z Keep track of each trial result in z. END End the experiment, go back and repeat until all 000 trials are complete. COUNT z =4 k Count the number of times we got four girls, put the result in k. DIVIDE k 000 kk Convert to a proportion. PRINT kk Print the result. Note: The file fourgirl on the Resampling Stats software disk contains this set of commands. We can also find the probabilities of other outcomes from a histogram of trial results obtained with the following command: HISTOGRAM z Produce a histogram of trial results. In the following histogram we can see that in 4 percent of the trials, 4 of the 5 selected were girls.

8 Chapter 8 Probability Theory, Part 4: Estimating Probabilities from Finite Universes 25 # of girls It should be noted that for this problem as for most other problems there are several other resampling procedures that will also do the job correctly. In analytic probability theory this problem is worked with a formula for combinations. Example 8-2: Nine Spades and Four Clubs in a Bridge Hand (Multiple-Outcome Sampling Without Replacement, Order Does not Matter) This problem is similar to Example 8- except that now there are four equally-likely outcomes instead of only two. The RESAMPLING STATS solution is straightforward in the program 9spades. NUMBERS,52 a Constitute the deck of 52 cards, numbers -52 in vector a. We will let - = spades, 4-26 = clubs, etc. REPEAT 000 Repeat the trial 000 times. SHUFFLE a b Shuffle the deck, call the shuffled deck b. TAKE b, c Deal out one hand of cards. (Take the first numbers from b and call them c.) COUNT c between d Count the number of spades in c, put the result in d. IF d = 9 If we have nine spades, we ll continue on to count the clubs. If we don t have nine spades, the number of clubs is irrelevant we have

9 26 Resampling: The New Statistics not gotten the hand we are interested in. COUNT c between 4 26 e Count the clubs, put the result in e. SCORE e z Keep track of the number of clubs in each experiment in z. END End the IF condition. END End one experiment, repeat all 000 trials (experiments). COUNT z =4 k Count the number of trials where we got 4 clubs. This is the answer we want - the number of hands out of 000 with 9 spades and 4 clubs. (Recall that we only counted the clubs if the hand already had 9 spades.) DIVIDE k 000 kk Convert to a proportion. PRINT kk Note: The file 9spades on the Resampling Stats software disk contains this set of commands. Example 8-: A Total of Fifteen Points in a Bridge Hand When Ace = 4, King =, Queen = 2, and Jack =. This problem bears the same relationship to Example 8-2 that Example 4-5 bears to Example 4-4. URN 4# 4#2 4# 4#4 6#0 deck Constitute a deck with 4 jacks (point value ), 4 queens (value 2), 4 kings (value ), 4 aces (value 4), and 6 other cards with no point value REPEAT 000 Do 000 trials. SHUFFLE deck deck\$ Shuffle the deck of cards. TAKE deck\$, hand Take thirteen cards. SUM hand points Total the points. SCORE points z Keep score of the result.

10 Chapter 8 Probability Theory, Part 4: Estimating Probabilities from Finite Universes 27 END End one experiment, go back and repeat until all 000 are done. HISTOGRAM z Produce a histogram of trial results. From this histogram, we see that in about 5 percent of our trials we obtained a total of exactly 5 points. RESAMPLING STATS will calculate this for us directly if we add the following commands on to the program: COUNT z = 5 k How many times did we have a hand with fifteen points? DIVIDE k 000 kk Convert to a proportion. PRINT kk Print the result. Result: kk =.054 Note: The file bridge on the Resampling Stats software disk contains this set of commands. Example 8-4: Four Girls and Then One Boy From Twenty-five Girls and Twenty-five Boys Order Matters, Sampling Without Replacement, Two Outcomes, Several of Each Item What is the probability of getting an ordered series of four girls and then one boy, from a universe of twenty-five girls and twenty-five boys? This illustrates Case above. Clearly we can use the same sampling mechanism as in Example 8-, but now we record yes for a smaller number of composite events.

11 28 Resampling: The New Statistics We record no if only one boy is chosen but he is chosen st, 2nd, rd, or 4th, whereas in Example 8- such outcomes are recorded as yeses. Step. Let 0-25 = girls, = boys. Step 2. Select five non-duplicating numbers. Check whether the series includes four numbers 0-25 followed by a number If so, write yes, otherwise no. Step. Repeat step 2 perhaps 000 times, and count the proportion yes which estimates the probability sought. The RESAMPLING STATS program 4girlboy starts out similarly to the earlier problem Fourgirls. As in that problem, an array simulating a class of 50 children is created. But whereas in the previous problem any 4 of the first 5 children had to be girls, in this problem, the first 4 children must be girls, and then the fifth must be a boy for a trial to be considered a success. We thus TAKE the first 4 randomly-chosen children and place them in c. We must then determine the number of these children which are girls by counting those that have a number between and 25, and placing the result in d. If we find that all 4 of the children selected are girls, then we proceed to pick a fifth child and put it in e. If, however, of the first 4 children selected all were not girls, then we skip the steps through the first END statement, because we would already know that this trial will not be a success. To determine whether this fifth child selected is a boy that is, checking whether it has a number between 26 and 50 we use a COUNT statement. If it is a boy, then f will equal ; if it is not a boy, then f will equal 0. (Remember that if the first 4 children selected were not all girls, then f will not get any value.) We then SCORE the value in f to e. This ends the conditional statement and the loop. NUMBERS,50 a Constitute the set of girls (numbers -25) and boys (numbers 26-50), put them in a. REPEAT 000 Do the following experimental trial 000 times. SHUFFLE a b Shuffle the numbers, put the shuffled numbers in b. TAKE b,4 c Take the first 4 numbers, put them in c.

12 Chapter 8 Probability Theory, Part 4: Estimating Probabilities from Finite Universes 29 COUNT c between 25 d Count the number of girls, put the result in d. IF d = 4 If we have 4 girls... TAKE b 5 e Take the fifth number (not 5 numbers) from b and put it in e. COUNT e between f How many boys in e? (It would have to be either 0 or ) SCORE f z Keep track of each trial result. END End the IF condition END End the experiment, go back and repeat until all 000 are complete. COUNT z = k Count the number of trials in which we got boy (recall that we only counted the boy if we had already gotten the required 4 girls). DIVIDE k 000 kk Convert to a proportion. PRINT kk Print the result. Note: The file 4girlboy on the Resampling Stats software disk contains this set of commands. This type of problem is conventionally done with a permutation formula. Example 8-5: Four or More Couples Getting Their Own Partners When Ten Couples are Paired Randomly (Probability of Matching by Chance) (Program Couples ) Ten couples of boys and girls come to a party. In order to get them acquainted with each other, the host pairs them at random for the first dance by picking a boy s name from one hat and a girl s name from another. What is the chance that four or more couples will get the partners they came with? By now the solution seems obvious to you: Simulate the drawings from the two hats filled with boys and girls names, and see how often four or more matches occur.

13 0 Resampling: The New Statistics Step. Let ace through 0 of hearts be girls, ace through 0 of spades be boys. Step 2. Shuffle the hearts and deal them out in a row; shuffle the spades and deal in a row just below the hearts. Step. Count the pairs a pair is one card from the heart row and one card from the spade row that contain the same denomination. If 4 or more pairs match, record yes, otherwise no. Step 4. Repeat steps (2) and () perhaps 200 times. Step 5. Count the proportion yes. This estimates the probability of 4 or more pairs. Exercise for the student: Write the steps to do this example with random numbers. Now let s examine the RESAMPLING STATS solution Couples. The key step is to fill array A with 0 elements numbered to 0 in order, with each number standing for a male. We fill array B in similar fashion to represent the females. In each of the 000 samples we SHUFFLE the elements of B, which stands for the females, to create a new vector C. As long as either the males or the females are randomly ordered, the probability of getting a correct match is determined by chance. There are several ways to determine whether a match occurs. Our method is using the SUBTRACT command to compare each element of A with each element of C. A match causes a 0, this would mean that the third couple was matched correctly because they have the same numbers s. We then COUNT the number of 0 s in D to determine the number of couples correctly matched in the current sample. The result for the current sample is placed in E, then transferred to Z in order to keep SCORE for each of the 000 samples. This ends the loop. NUMBERS,0 a An array of 0 males. NUMBERS,0 b An identical array of 0 females the pair for each of the males. REPEAT 000 Do the experiment 000 times. SHUFFLE b c Shuffle the females.

14 Chapter 8 Probability Theory, Part 4: Estimating Probabilities from Finite Universes SUBTRACT a c d This operation pairs each shuffled female with a male and subtracts. If it is an original pairing (/, 2/2, etc.), the result will be a 0. The number of 0 s indicates how many of the 0 males got paired up again with their original partner. COUNT d =0 e Count the number of 0 s and put the result in e. SCORE e z Keep track of each trial result. END End the trial, go back and repeat until all 000 are complete. HISTOGRAM z Produce a histogram of the results. From this histogram, we see that in about percent of the trials did 4 or more couples end up being re-paired with their own partners. We can calculate this proportion directly with RESAMPLING STATS: COUNT z >= 4 k Determine how many trials had 4 or more males being matched with their partner after the shuffling. (Note that this is the same as the number of females being matched with their original partner.) DIVIDE k 000 kk Convert to a proportion. PRINT kk Print the result. Note: The file couples on the Resampling Stats software disk contains this set of commands.

15 2 Resampling: The New Statistics Example 8-6: Matching Hats: Another famous problem of this sort: The hat-checker at a restaurant mixes up the hats of a party of 6 men. What is the probability that at least one will get his own hat? Instead of matching men with women, as in the earlier problem, however, we are now matching men with their hats. See the program Hats for the solution. First, assign each of the 6 men a number, and place these numbers in A. Next, assign each man s hat the same number in B, but arrange them in random order by shuffling the numbers from B into C, which represents the group of mixed-up hats. The rest of the problem is the same as in Couples except that in the second COUNT statement, we now are interested in any trial where at least one (>= ) man received the right hat. NUMBERS,6 a Constitute the set of six men. NUMBERS,6 b Constitute the set of their six hats. REPEAT 000 Do 000 trials. SHUFFLE b c Mix up the hats. SUBTRACT a c d Subtract the shuffled hats from the set of men. A 0 will indicate that a man has received his own hat. COUNT d =0 e Count the number of 0 s the number of men who received their own hats back. SCORE e z Keep track of each trial result. END End one experiment, go back and repeat until all 000 are complete. HISTOGRAM z Produce a histogram of the trial results.

16 Chapter 8 Probability Theory, Part 4: Estimating Probabilities from Finite Universes From the histogram, we see that in roughly 62 percent of the trial results at least one man received his own hat back. RESAMPLING STATS will calculate this for us with the following commands: COUNT z >= k Determine how many trials resulted in at least one man getting his hat back. DIVIDE k 000 kk Convert to a proportion. PRINT kk Print the results Note: The file hats on the Resampling Stats software disk contains this set of commands. Example 8-7: Twenty executives are to be assigned to two divisions of a firm The top manager wants to spread the talent reasonably evenly, but she does not want to label particular executives with a quality rating and therefore considers distributing them with a random selection. She therefore wonders: What are probabilities of the best ten among the twenty being split among the divisions in the ratios 5 and 5, 4 and 6, and 7, etc. if their names are drawn from a hat? One might imagine much the same sort of problem in choosing two teams for a football or baseball contest. One may proceed as follows:. Put 0 balls labeled W (for worst ) and 0 balls labeled B (best) in an urn.

17 4 Resampling: The New Statistics 2. Draw 0 balls without replacement and count the W s.. Repeat (say) 400 times. 4. Count the number of times each split 5 W s and 5 B s, 4 and 6, etc. appears in the results. The problem can be done with RESAMPLING STATS as follows: URN 0# 0#2 a REPEAT 000 END SHUFFLE a b TAKE b,0 c COUNT c = d SCORE d z HISTOGRAM z

18 Chapter 8 Probability Theory, Part 4: Estimating Probabilities from Finite Universes 5 Example 8-8: Executives Moving A major retail chain moves its store managers from city to city every three years in order to calculate individuals knowledge and experience. To make the procedure seem fair, the new locations are drawn at random. Nevertheless, the movement is not popular with managers families. Therefore, to make the system a bit sporting and to give people some hope of remaining in the same location, the chain allows managers to draw in the lottery the same posts they are now in. What are the probabilities that, 2,... will get their present posts again if the number of managers is 0? The problem can be solved with the following steps:. Number a set of green balls from to 0 and put them into Urn A randomly. Number a set of red balls from to 0 and then put into Urn B. For greater concreteness one could use 0 little numbered dolls in Urn A and 0 little toy houses in Urn B. 2. Shuffle Urn A, and array all its green balls into a row (vector A). Array all the red balls from Urn B into a second row B just below row A.. Count how many green balls in row A have the same numbers as the red balls just below them, and record that number on a scoreboard. 4. Repeat steps 2 and perhaps 000 times. Then count in the scoreboard the numbers of 0,, 2,. Example 8-9: State Liquor Systems Again Let s end this chapter with the example of state liquor systems that was examined in the previous chapter and which will be discussed again later in the context of problems in statistics. Remember that as of 96, there were 26 U.S. states in whose liquor systems the retail liquor stores are privately owned, and 6 monopoly states where the state government owns the retail liquor stores. These were the 96 prices: 6 monopoly states: \$4.65, \$4.55, \$4., \$4.5, \$4.20, \$4.55, \$.80, \$4.00, \$4.9, \$4.75, \$4.74, \$4.50, \$4.0, \$4.00, \$5.05, \$4.20 Mean: \$4.5

19 6 Resampling: The New Statistics 26 private-ownership states: \$4.82, \$5.29, \$4.89, \$4.95, \$4.55, \$4.90, \$5.25, \$5.0, \$4.29, \$4.85, \$4.54, \$4.75, \$4.85, \$4.85, \$4.50, \$4.75, \$4.79, \$4.85, \$4.79, \$4.95, \$4.95, \$4.75, \$5.20, \$5.0, \$4.80, \$4.29. Mean: \$4.84 Let us now consider that all these states prices constitute one single finite universe. We ask: If these 42 states constitute a universe, and if they are all shuffled together, how likely is it that if one divides them into two samples at random (sampling without replacement), containing 6 and 26 observations respectively, the difference in mean prices turns out to be as great as \$.49 (the difference that was actually observed)? Again we write each of the forty-two observed state prices on a separate card. The shuffled deck simulates a situation in which each state has an equal chance for each price. Repeatedly deal groups of 6 and 26 cards, without replacing the cards as they are chosen, to simulate hypothetical monopoly-state and private-state samples. In each trial calculate the difference in mean prices. The steps more systematically: Step A. Write each of the 42 prices on a card and shuffle. Steps B and C (combined in this case). i) Draw cards randomly without replacement into groups of 6 and 26 cards. Then ii) calculate the mean price difference between the groups, and iii) compare the simulation-trial difference to the observed mean difference of \$4.84 \$4.5 = \$.49; if it is as great or greater than \$.49, write yes, otherwise no. Step D. Repeat step B-C a hundred or a thousand times. Calculate the proportion yes, which estimates the probability we seek. The probability that the postulated universe would produce a difference between groups as large or larger than observed in 96 is estimated by how frequently the mean of the group of randomly-chosen sixteen prices from the simulated stateownership universe is less than (or equal to) the mean of the actual sixteen state-ownership prices. Please notice how the only difference between this treatment of the problem and the treatment in Chapter 7 is that the drawing in this case is without replacement whereas in Chapter 7 the drawing is with replacement. In Chapter 7 we thought of these states as if they came from a non-finite universe, which is one possible interpretation in one

22 Chapter 8 Probability Theory, Part 4: Estimating Probabilities from Finite Universes 9 Summary This completes the discussion of problems in probability that is, problems where we assume that the structure is known. Whereas Chapter 7 dealt with samples drawn from universes considered not finite, this chapter deals with problems drawn from finite universes and therefore you sample without replacement.

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