The University of Winnipeg

Journal of Statistics Education v.3, n.2 (1995)

Copyright (c) 1995 by W. John Braun, all rights reserved. This text may be freely shared among individuals, but it may not be republished in any medium without express written consent from the author and advance notification of the editor.

**Key Words**: Simulation; Contingency tables;
Elementary probabilities.

The video lottery terminal dataset contains observations on the three windows of an electronic slot machine for 345 plays together with the prize paid out for each play. The prize payout distribution is so badly skewed that confidence intervals for expected payout based on the central limit theorem are not accurate. This dataset can be used at the graduate or upper undergraduate level to illustrate parametric bootstrapping. The dataset can also be used in a graduate course to illustrate tests of independence for two and three-way contingency tables involving random zeroes, or these tables may be collapsed and used as examples in an introductory course.

1 Video lottery terminals (VLTs) are electronic slot machines with several line-up style games, including one called Double Diamond. This game consists of three windows in which one of seven objects may appear (after the gambler has inserted a quarter). If the objects in the three windows match one of several possible combinations, the gambler is awarded a prize (in quarters). Otherwise, nothing is awarded.

2 The objects that can appear are blanks (0), single bars (B), double bars (BB), triple bars (BBB), double diamonds (DD), cherries (C), and sevens (7). These objects are not equally likely to appear in each window, and the object distributions may not be the same for the different windows.

3 The payouts for winning combinations of objects appearing in the three windows are given in the following table.

COMBINATION PRIZE PAYOUT DD DD DD 800 7 7 7 80 BBB BBB BBB 40 BB BB BB 25 B B B 10 C C C 10 AB AB AB 5 C C 0 5 C 0 C 5 0 C C 5 C 0 0 2 0 C 0 2 0 0 C 2

AB = ANY BAR, i.e., a single, double or triple bar. A double diamond doubles any winning combination, while two double diamonds quadruple any winning combination. For example, 7 DD DD pays 320. Cherries result in a payout regardless of what appears with them.

4 The Manitoba Lotteries Foundation provides (limited) information to the public describing the video lottery terminal games it administers. The description of the Double Diamond game claims that the "prize payout is 92%," which must be interpreted as the expected payout per play.

5 The Canadian Broadcasting Corporation (CBC) asked the author to conduct an investigation of the payout claim, since a CBC reporter suspected a payout of closer to 40%. The Manitoba Lotteries Foundation and the VLT manufacturer could not supply the required probability distributions, so these had to be estimated on the basis of a sample.

6 The author took an initial sample of 138 plays in which the actual observed payout was around 38%, appearing to confirm the reporter's claims. This result, however, could not be trusted because the payout distribution turns out to be highly skewed. A more careful analysis (which is greatly simplified by using the parametric bootstrap) gave an estimate of the expected payout that is considerably higher, together with more sensible standard error estimates.

7 Along the way, it was necessary to confirm the manufacturer's claim that the objects appear in the three windows independently of each other. The resulting three-way contingency table is riddled with zeroes so that care must be taken when computing the degrees of freedom for the chi-square test. The author has found it much quicker to use the bootstrap to test for independence. An additional sample of 207 plays was taken later to check the results of the smaller sample.

8 As stated earlier, the Manitoba Lotteries Corporation claims that its Double Diamond VLT game has a prize payout of 92%. This game provides an example of expected value that lends itself to discussion in an introductory statistics course. In fact, it is interesting to ask students how they interpret the claim. As stated, the claim is somewhat ambiguous, and not everyone realizes immediately that the claim concerns expected payout.

9 An apparent paradox can be noted: A fellow started with one thousand dollars and successfully gambled it all away within five days of intensive gambling. Does this constitute evidence that the payout is much less than 92%? (Needless to say, the unlucky fellow felt that it was very strong evidence. Of course, he was simply ignorant of the true intent of the term "payout.")

10 Because the payout distribution is highly skewed, the expected payout is not an appropriate measure of central tendency here. An interesting discussion question could center on the Manitoba Lotteries Corporation's motives in using the expected payout in its advertising instead of the median or modal payout.

11 The CBC, looking for a story, disputed the Lotteries Corporation claim of 92% and put forward their own claim of a 40% expected payout. An appropriate confidence interval is required to settle this dispute. (Alternatively, a one-tailed test, corresponding to the CBC claim or to the Lotteries Corporation claim, could be considered here.)

12 Let X denote the prize payout on any single play. The objective is to estimate \mu = E[X] with a confidence interval, using the sample of 138 plays. In this sample, the prize awarded at each play was recorded. (Information on the individual windows was collected as well, but that is not needed yet.) Since these plays were taken successively at a single machine, the question of whether they constitute a truly random sample arises. The author was assured by the manufacturer of the machines that a "good pseudorandom number generator was used to generate the sequence of objects in the three windows." As far as representing the entire population of video lottery terminals is concerned, the machines are apparently identical, so sampling from one machine should have been sufficient.

13 Using the obtained data (\bar{x} = .384, s = 1.27, s.e. = .108), an approximate 99% confidence interval for \mu is (0.11, 0.66). This would lead us to reject the Lotteries Corporation claim at the .005 level. This inference assumes that the sampling distribution of \bar{X} is close to normal because of the central limit theorem.

14 Ordinarily, a sample of 138 observations is considered to be sufficiently large that a confidence interval based on the central limit theorem should be fairly accurate. In this case, however, the standard error is severely underestimated, because events in the tail of the (highly skewed) payout distribution are not observed. Thus, the confidence interval is much narrower than it should be. Either a larger sample or a better estimate of the standard error is required.

15 A boxplot of the payout data may be useful here to convey the highly skewed character of the distribution. Students may find it surprising to see a boxplot where the "box" is a single line and all non-zero data points appear as outliers.

16 The true sampling distribution of \bar{X} can be bootstrapped using the data collected on the individual windows. For each window, the probability of each object appearing can be estimated from the sample proportion. Assuming the windows are independent (see below), a single play of the VLT game can be bootstrapped by simulation of the estimated multinomial probabilities for each window.

17 The payout table is used to determine the prize awarded for each combination generated by the simulation. Thus, the prizes for a re-sample of 138 plays can be easily obtained. A large number (say, 1000) of these re-samples is then used to obtain the bootstrap sampling distribution, from which estimates of the standard error and percentile confidence intervals (Hall 1992, Ch. I) can be obtained. This can easily be programmed. (A Fortran program that carries out these computations is available from the author. In addition, a graphics program that simulates the windows of the video lottery terminal is available for DOS machines. Instructions for obtaining this graphics program by ftp or e-mail request are given at the end of this article.)

18 The bootstrap estimate of the standard error is .39, and a 95% confidence interval (percentile-method) for \mu is (0.27, 1.29), indicating that we do not have much evidence against the claim.

19 Of course, one may argue that the confidence interval obtained above is so wide that either the CBC claim or the manufacturer's claim could be true. Is there a way of narrowing the interval enough so that at least one of the claims can be rejected? (One would ideally like to do this without taking an excessively large sample. Note that one could now obtain an estimate of the sample size required to obtain a confidence interval of specified length.)

20 We can actually get an improved estimate of the expected payout by using the payout table in conjunction with the products of the estimated probabilities (again assuming independence among the windows). The calculation is elementary but tedious; it is easily implemented in SAS or Fortran. In fact, one could use the same idea to estimate the variance of this estimator, but for this, the bootstrap is much less complicated to implement. For each of the re-samples obtained earlier, we compute the above expectation to obtain the bootstrap sampling distribution of this new estimator.

21 The expected payout is now estimated to be .65 with a standard error of .13. The 95% confidence interval (percentile-method) is (.454, .866) which fails to contain either claimed value. The 99% confidence interval is (.188, 1.021) which contains both values. Perhaps a larger sample is required after all!

22 If one pools the data from the first and second samples, one finds that the 95% and 99% confidence intervals are (.67, 1.10) and (.61, 1.21), respectively. Thus, they do not contain the CBC claim of .40, but they do contain the Lotteries Foundation claim of .92. Thus, the CBC does not have a scoop.

23 To justify the parametric bootstrap used above, we need to demonstrate that the three windows are independent. (The manufacturer said they are, but we should verify this.)

24 The \chi^2 statistic can be computed for the three-way table (a 7 x 7 x 7 table). There are many zeroes in this table including an entire layer. All of these zeroes are random, so the p-value corresponding to \chi^2 = 334.4 on 180 d.f. is meaningless. Christensen (1990, Ch. X), for example, suggests ways around this problem, but bootstrapping works as well. Using the above re-samples, compute the \chi^2 statistic each time, and compare the resulting bootstrap sampling distribution with the original observed value. (Recall that the simulation was based on the assumption that the windows are independent.) For the first sample, the p-value is about .12, indicating that independence is a safe assumption.

25 Is the distribution of objects the same in each window? A test of homogeneity among the three windows is a natural test to apply, and can be demonstrated in an introductory course. The three two-way tables can also be analyzed in an introductory setting by collapsing various rows to eliminate the random zeroes. In this case, one would presumably want to use the entire dataset. The author has also tested goodness-of-fit of various distributions of the objects in the particular windows. For example, one could test whether the seven objects (0,B,BB,BBB,C,7,DD) are distributed in the ratio 32:16:8:4:2:1:1, for any of the three windows.

26 A number of probability and stochastic processes questions arise naturally from consideration of this dataset. Some of these can be answered using the probabilities estimated from the combined dataset, or by using a bootstrap computation.

7.1 What is the probability of winning the largest prize?

7.2 What is the probability of winning nothing?

7.3 Suppose that a gambler has n quarters. Show that if the Lotteries Foundation's claim is true, then the number N of plays the gambler can expect to make is 12.5 n.

7.4 Use the bootstrap to explore the distribution of N.

7.5 Even if the mean payout is really .92, is the Lotteries Foundation supplying useful information? Use the bootstrap to estimate the median payout for various sample sizes. This indicates that the payout observed by the CBC reporter and in the first sample are not unusual, even if they are not near the true mean.

7.6 Starting with 100 quarters, what is the probability of eventually reaching 200 or more? 300 or more?

7.7 Suppose a player starts with 1000 dollars (4000 quarters), and suppose each play takes one second. Estimate the expected time until the player has nothing left.

7.8 The results for the two samples appear to be quite different. Is this difference statistically significant?

27 The file vlt.dat.txt contains the raw data. The file vlt.txt is a documentation file containing a brief description of the dataset.

The first three columns of the dataset record the objects observed in the three windows. The fourth column records the prize awarded, and the fifth column indicates the night the sample was obtained. Values are delimited by blanks; note that columns are not aligned.

Coding for variables 1, 2, 3: CODE OBJECT 0 BLANK (0) 1 SINGLE BAR (B) 2 DOUBLE BAR (BB) 3 TRIPLE BAR (BBB) 5 DOUBLE DIAMOND (DD) 6 CHERRIES (C) 7 SEVEN (7) Coding for variable 5: CODE INDICATES 1 SAMPLE TAKEN ON FIRST NIGHT 2 SAMPLE TAKEN ON SECOND NIGHT

Christensen, R. (1990), Log-Linear Models, New York: Springer-Verlag.

Hall, P. (1992), The Bootstrap and Edgeworth Expansion, New York: Springer-Verlag.

W. John Braun

Department of Mathematics and Statistics

University of Winnipeg

515 Portage Ave.

Winnipeg, Manitoba R3B 2E9

CANADA

**Download Video Lottery Program
to a Local File**

This program requires a DOS machine with a coprocessor and at least 4 MB of RAM. Start the program by typing "vlt" (without the quotes) at the DOS prompt.

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