B.2 Discrete Distribution (Dice Experiment Using Three Regular Dice)

Learning Objectives

By the end of this section, you will be able to:

In this section, you will learn to:

compare the data your class actually collects against the theoretical discrete distribution the dice game should follow, and say whether the two agree;
build the theoretical probability distribution for the number of matches and for the profit, and compute the mean and standard deviation of each;
explain what a long-run average profit tells you about a betting game, and why one round of class data is not the final word on it.

B.2.1 Stats Lab: Discrete Distribution (Dice Experiment Using Three Regular Dice)

This section is a lab, not a reading. In §4.1 through §4.3 you built probability distributions on paper and computed what they predict. Here you will play a small betting game, collect what your class actually gets, and lay the two side by side. The point of the lab is the comparison: the theory tells you what should happen in the long run, and the class data tells you what happened in the short run. They will not match exactly, and learning to judge whether they match closely enough is the skill.

Class Time:

Names:

Round answers to relative frequency and probability problems to four decimal places throughout. That precision is not fussiness — several of the probabilities in this lab are small enough that two decimal places would round two different answers to the same number.

B.2.2 Supplies

Three regular six-sided dice

B.2.3 Procedure

The experimental procedure is to bet on a specific number appearing on the dice. Then roll three dice and count the number of matches. The number of matches decides your profit:

This is the oldest carnival game there is

The dice game below is Chuck-a-Luck, run at fairs and on midways for well over a century. It survives because it looks fair: your number can show on any of three dice, and you get paid more when it shows more often. This lab is how you find out whether "looks fair" and "is fair" are the same thing.

no match on any die: you lose $1, so your profit is $-\$1$;
a match on exactly one die: you win $1;
a match on exactly two dice: you win $2;
a match on all three dice: you win $3.

Bet on a specific number appearing on the dice, roll three dice, and count the matches.
What is the theoretical probability of one die matching the specific number? _____
Choose one number to place a bet on. Roll the three dice. Count the number of matches.
Let $X$ = number of matches. Theoretically, $X \sim$ _____(_____,_____)
Let $Y$ = profit per game.

Three dice, not one die rolled three times

Rolling three dice at once and rolling one die three times give exactly the same probabilities, because the dice do not talk to each other. Whichever picture is easier to hold, hold that one — the three rolls are independent, they all have the same chance of matching, and there are a fixed number of them. Those are the conditions that make $X$ a binomial count.

Before you roll anything, notice what kind of quantity $X$ is. Each die either matches your number or it does not, which is two outcomes. The chance of a match is the same on every die, and one die's result tells you nothing about another's. You have a fixed number of trials — three — and you are counting how many of them succeed. Every one of those phrases is a condition from §4.3, which means the theoretical distribution for this game is one you already know how to write down and one you already know how to take the mean and standard deviation of.

Try It Now B.2.1

Riley is filling in items 2 and 4 of the Procedure before their group rolls anything. What is the theoretical probability that one die matches the number they bet on, and what is the full theoretical distribution of $X$, the number of matches on three dice?

Solution

Step 1 — one die. A regular die has six equally likely faces and exactly one of them is the number Riley bet on, so

$$ P(\text{one die matches}) = \frac{1}{6} \approx 0.1667 $$

Step 2 — check the binomial conditions. There are $n = 3$ trials (the three dice); each trial is a match or not a match; the probability of a match is $p = \frac{1}{6}$ on every die; and the dice are independent.

Step 3 — name the distribution.

$$ X \sim B\left(3, \frac{1}{6}\right) $$

Answer: the theoretical probability of one die matching is $\frac{1}{6} \approx 0.1667$, and $X \sim B\left(3, \frac{1}{6}\right)$. Riley should write $p$ as the exact fraction $\frac{1}{6}$ rather than 0.1667 wherever they can — that value gets cubed later, and a rounded $p$ compounds into a visibly wrong probability by the time you reach $P(X = 3)$.

B.2.4 Organize the Data

Definition B.2.1: Relative Frequency

The relative frequency of an outcome is the number of times that outcome occurred divided by the total number of trials:

$$ \text{relative frequency} = \frac{\text{frequency}}{\text{total number of games played}} $$

It is a proportion between 0 and 1, and the relative frequencies of all four match counts must sum to 1. Relative frequency is an empirical quantity — it is measured from data that actually happened — as opposed to a probability, which is a theoretical quantity computed from the rules of the game.

In Table B.2.1, fill in the $y$ value that corresponds to each $x$ value using the payoff schedule above. Next, record the number of matches picked for your class. Then calculate the relative frequency.

Complete the table.

**Table B.2.1** — Your class's empirical results: matches, the profit each match count pays, how often it happened, and how often it happened as a proportion.
$x$	$y$	Frequency	Relative Frequency
0
1
2
3

Try it in rāSHio

Your class will finish with one long list of match counts — a 0 here, a 1 there, one game in a hundred with a 3. Rather than tally them by hand, open rāSHio, choose File → Delimited List… and paste the whole list in (comma, semicolon, tab or newline all work), then run Graph → Frequency Table with Discrete values ticked. That box is the one that matters here: it gives each of 0, 1, 2 and 3 its own row instead of binning them, which is exactly the four-row shape Table B.2.1 wants, and it prints the relative frequency column beside the count.

Figure B.2.1 — Building the class’s frequency and relative-frequency columns in rāSHio: Graph → Frequency Table, Discrete values.

Calculate the following:

a. $\overline{x} =$ _____

b. $s_x =$ _____

c. $\overline{y} =$ _____

d. $s_y =$ _____

Explain what $\overline{x}$ represents.
Explain what $\overline{y}$ represents.
Based upon the experiment:

a. What was the average profit per game?

b. Did this represent an average win or loss per game?

c. How do you know? Answer in complete sentences.

Construct a histogram of the empirical data. Put the number of matches on the horizontal axis and the relative frequency on the vertical axis, and draw one bar for each of $x = 0, 1, 2, 3$.

The bar and the average are the same number twice

$\overline{x}$ is the balance point of the histogram you just drew — the place a ruler would sit if the bars were weights on a plank. So a histogram leaning hard toward $x = 0$ and an $\overline{x}$ well under 1 are not two findings. They are one finding, drawn and then computed.

Use the sample statistics keys on your calculator rather than adding the values by hand, but be careful about which mean you are asking for. The four rows of Table B.2.1 are not four games — they are four categories, each of which happened some number of times. If your calculator asks for a frequency list alongside the data list, that frequency column is what goes in it. Entering 0, 1, 2, 3 with no frequencies gives you the average of the four labels, which is 1.5 and means nothing.

Try it in rāSHio

The same idea covers item 2. With your match counts loaded, rāSHio’s Stats → Summary Statistics returns $\overline{x}$ and $s_x$ in one pass. For $\overline{y}$ and $s_y$, add a second column holding each game’s profit — $-1$, $1$, $2$ or $3$, matching that game’s match count — and run the same panel on it. Two columns, one tool, all four numbers.

Figure B.2.2 — Reading $\overline{x}$, $s_x$, $\overline{y}$ and $s_y$ in rāSHio: Stats → Summary Statistics.

Try it in rāSHio

For item 6, Graph → Histogram in rāSHio will draw the empirical picture from the same loaded column, and you can draw the theoretical one beside it later by entering the four probabilities. Set both to the same vertical range before you compare them — a chart auto-scaled to its own tallest bar will make two very different distributions look alike.

Try It Now B.2.2

Ms. Ocampo's class plays 120 games, and she records 68 games with no matches, 43 with exactly one match, 8 with exactly two, and 1 with all three. Complete the relative frequency column, then find $\overline{x}$, the average number of matches per game.

Solution

Step 1 — check the total.

$$ 68 + 43 + 8 + 1 = 120 $$

Step 2 — divide each frequency by 120.

$$ \frac{68}{120} = 0.5667 \qquad \frac{43}{120} = 0.3583 \qquad \frac{8}{120} = 0.0667 \qquad \frac{1}{120} = 0.0083 $$

Those four proportions sum to 1.0000, which is the arithmetic check that you divided by the right total.

Step 3 — find the mean number of matches. Weight each match count by how often it happened:

$$ \overline{x} = \frac{(0)(68) + (1)(43) + (2)(8) + (3)(1)}{120} = \frac{62}{120} \approx 0.5167 $$

Answer: the relative frequencies are 0.5667, 0.3583, 0.0667, and 0.0083, and $\overline{x} \approx 0.5167$ matches per game across the games her class played. Hold on to that number — you are about to compute what the theory predicts the average number of matches should be, and how close 0.5167 lands to it is the whole question this lab is asking.

Try It Now B.2.3

Using the same 120 games Ms. Ocampo's class played in Try It Now B.2.2, find $\overline{y}$, the average profit per game. Then answer items 3, 4, and 5: what do $\overline{x}$ and $\overline{y}$ represent, and did her class win or lose?

Solution

Step 1 — attach a profit to each match count. From the payoff schedule, 0 matches pays $-1$ dollar, 1 match pays 1, 2 matches pay 2, and 3 matches pay 3.

Step 2 — weight each profit by how often it happened.

$$ \overline{y} = \frac{(-1)(68) + (1)(43) + (2)(8) + (3)(1)}{120} = \frac{-68 + 43 + 16 + 3}{120} = \frac{-6}{120} = -0.05 $$

Step 3 — say what the two averages mean. $\overline{x} \approx 0.5167$ is the average number of dice that matched, per game, across the 120 games her class actually played. $\overline{y} = -\$0.05$ is the average profit per game across those same 120 games.

Answer: $\overline{y} = -\$0.05$. Ms. Ocampo's class lost money — five cents per game on average, or about $6 over the 120 games. You know because $\overline{y}$ is negative, and a negative average profit means that across all the games played, more was paid out in lost dollars than was collected in winnings. Both $\overline{x}$ and $\overline{y}$ are sample statistics: they describe these 120 games and nothing else. The Greek-letter versions, $\mu_x$ and $\mu_y$, describe the game itself.

B.2.5 Theoretical Distribution

Build the theoretical PDF chart for $x$ and $y$ based on the distribution from the Procedure section.

**Table B.2.2** — The theoretical distribution: what the rules of the game predict, before any dice are rolled.
$x$	$y$	$P(x) = P(y)$
0
1
2
3

Try it in rāSHio

You can fill this whole table from rāSHio’s Calculators → Binomial panel: enter $n = 3$ and $p = 0.1667$ (or $1/6$), then read $P(X = x)$ off the Exact tab for each of $x = 0, 1, 2, 3$. The other three tabs answer the Use the Data questions further down without any extra setup — At Least gives $P(x \geq 2)$ directly, and Range gives $P(0 < x < 3)$ once you ask for 1 through 2. The panel also prints the complement beside whatever it shades, so you rarely have to subtract from 1 yourself.

Figure B.2.3 — Filling the theoretical PDF chart in rāSHio: Calculators → Binomial with $n = 3$, $p = 1/6$. The walkthrough is shown with other numbers as an illustration, not with this lab’s.

Calculate the following:

a. $\mu_x =$ _____

b. $\sigma_x =$ _____

c. $\mu_y =$ _____

Explain what $\mu_x$ represents.
Explain what $\mu_y$ represents.
Based upon theory:

a. What was the expected profit per game?

b. Did the expected profit represent an average win or loss per game?

c. How do you know? Answer in complete sentences.

Construct a histogram of the theoretical distribution, on the same axes and the same scale you used for the empirical one.

Notice that $x$ and $y$ share a single probability column. That is not a shortcut in the table's design — it is a fact about the game. Your profit is decided entirely by how many dice matched, so "two matches" and "a profit of $2" are the same event described twice. Any probability you compute for a match count is automatically the probability of the profit that goes with it, which is why the two-decimal question "what is my expected profit" can be answered from a distribution built for counting matches.

Try it in rāSHio

Once Table B.2.2 is filled in, rāSHio’s Calculators → Expected Value takes it from there: paste the outcome column and the probability column, and it returns the mean and standard deviation of that distribution. Run it once on $x$ with its probabilities to get $\mu_x$ and $\sigma_x$, then again on $y$ with the same probability column to get $\mu_y$ and $\sigma_y$ — which is the shared-probability fact above, put to work.

Figure B.2.4 — Turning the theoretical PDF chart into $\mu$ and $\sigma$ in rāSHio: Calculators → Expected Value.

The expected profit is nobody's actual profit

$\mu_y$ will come out to a number of cents you cannot possibly win or lose on a single game — the payouts are whole dollars. That is normal. An expected value is the long-run average of many games, not a prediction about the next one, and a game with a small negative expected profit is exactly how a midway makes a living off players who never lose much on any one roll.

Try It Now B.2.4

Complete Table B.2.2. Find $P(x)$ for each of the four match counts, fill in the profit column, and confirm the probabilities sum to 1.

Solution

Step 1 — use the binomial formula with $n = 3$ and $p = \frac{1}{6}$. Working over a common denominator of $6^3 = 216$ keeps every value exact:

$$ P(X = 0) = \binom{3}{0}\left(\frac{1}{6}\right)^{0}\left(\frac{5}{6}\right)^{3} = \frac{125}{216} \approx 0.5787 $$ $$ P(X = 1) = \binom{3}{1}\left(\frac{1}{6}\right)^{1}\left(\frac{5}{6}\right)^{2} = \frac{75}{216} \approx 0.3472 $$ $$ P(X = 2) = \binom{3}{2}\left(\frac{1}{6}\right)^{2}\left(\frac{5}{6}\right)^{1} = \frac{15}{216} \approx 0.0694 $$ $$ P(X = 3) = \binom{3}{3}\left(\frac{1}{6}\right)^{3}\left(\frac{5}{6}\right)^{0} = \frac{1}{216} \approx 0.0046 $$

Step 2 — read the profits off the payoff schedule. Zero matches pays $-\$1$, one match pays $1, two pays $2, three pays $3.

Step 3 — check the total.

$$ \frac{125 + 75 + 15 + 1}{216} = \frac{216}{216} = 1 $$

Answer:

**Table B.2.3** — The completed theoretical distribution for the three-dice game.
$x$ (matches)	$y$ (profit, dollars)	$P(x) = P(y)$
0	-1	0.5787
1	1	0.3472
2	2	0.0694
3	3	0.0046

More than half the time you lose your dollar, and the three-match jackpot arrives about once in every 216 games. Both facts are already visible in the table, before any mean is computed.

Try It Now B.2.5

Find $\mu_x$, $\sigma_x$, and $\mu_y$, then answer question 5: is this game a win or a loss in the long run, and how do you know?

Solution

Step 1 — the mean and standard deviation of the match count. For a binomial these come straight from $n$ and $p$:

$$ \mu_x = np = 3\left(\frac{1}{6}\right) = 0.5 $$ $$ \sigma_x = \sqrt{npq} = \sqrt{3\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)} = \sqrt{\frac{15}{36}} \approx 0.6455 $$

Step 2 — the expected profit. Profit is not binomial, so use the general definition $\mu = \sum y \cdot P(y)$:

$$ \mu_y = (-1)\left(\frac{125}{216}\right) + (1)\left(\frac{75}{216}\right) + (2)\left(\frac{15}{216}\right) + (3)\left(\frac{1}{216}\right) $$ $$ \mu_y = \frac{-125 + 75 + 30 + 3}{216} = \frac{-17}{216} \approx -0.0787 $$

Step 3 — read the sign. The expected profit is negative, so in the long run this is a loss — about 7.9 cents per game.

Answer: $\mu_x = 0.5$, $\sigma_x \approx 0.6455$, and $\mu_y \approx -\$0.0787$. The game loses money on average. You know because the expected profit is negative, and an expected value is precisely the average profit per game over many, many games. Play 216 games and you should expect to be down about $17. Play ten games and almost anything can happen — which is why the class data you collected will not reproduce $-0.0787$ and is not supposed to.

Try It Now B.2.6

Find $\sigma_y$, the standard deviation of the profit, and explain why it is so much larger than $|\mu_y|$.

Solution

Step 1 — find $\sum y^2 P(y)$. Square each profit, weight it by its probability, and add:

$$ \sum y^2 P(y) = \frac{(-1)^2(125) + (1)^2(75) + (2)^2(15) + (3)^2(1)}{216} = \frac{125 + 75 + 60 + 9}{216} = \frac{269}{216} \approx 1.2454 $$

Step 2 — subtract the square of the mean. With $\mu_y = -\frac{17}{216} \approx -0.0787$:

$$ \sigma_y^2 = 1.2454 - (-0.0787)^2 \approx 1.2454 - 0.0062 = 1.2392 $$

Step 3 — take the square root.

$$ \sigma_y = \sqrt{1.2392} \approx 1.1132 $$

Answer: $\sigma_y \approx \$1.1132$, roughly fourteen times the size of $|\mu_y| \approx \$0.0787$. That gap is the whole business model of a betting game. A typical game swings your bankroll by about a dollar in one direction or the other, so the five- or ten-cent drift against you is invisible underneath the noise for as long as you care to watch. It only becomes visible in thousands of games — and the house is the only one at the table who plays thousands of games.

Try It Now B.2.7

Hannah has to sketch the theoretical histogram called for in question 6. Draw it the way she should, and describe its shape in one sentence.

Solution

She puts the four match counts on the horizontal axis and the theoretical probabilities on the vertical axis, then draws a bar of height $P(x)$ above each one.

The four heights come straight from Table B.2.3: 0.5787 above $x = 0$, 0.3472 above $x = 1$, 0.0694 above $x = 2$, and 0.0046 above $x = 3$ — a bar so short it barely lifts off the axis, which is the honest picture of a one-in-216 event.

Shape: the distribution is strongly skewed right — the tallest bar sits at $x = 0$, each bar after it is much shorter than the last, and the balance point $\mu_x = 0.5$ sits between the first two bars rather than under the tallest one.

Hannah should draw her empirical histogram on this same vertical scale. If you scale each histogram to its own tallest bar, two very differently shaped distributions can be made to look identical, and the comparison the lab is asking for becomes impossible to make by eye.

B.2.6 Use the Data

Definition B.2.2: The Law of Large Numbers

As the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency probability of that event approaches zero.

The same six questions, asked of two different things

The two lists below look like duplicates and are not. The first asks what the rules of the game say should happen; the second asks what your class actually got. Every gap between a matched pair is a gap between theory and one afternoon of dice — which is the finding this lab exists to produce.

In the notation of this lab: the more games your class plays, the closer each $RF$ value gets to its matching $P$ value. The law promises nothing about any particular game and nothing about any particular class period — it is a statement about what happens as the number of trials grows without bound.

Throughout this section, RF stands for relative frequency. Use the data from the Theoretical Distribution section to calculate the following answers. Round your answers to four decimal places.

$P(x = 3) =$ _____
$P(0 < x < 3) =$ _____
$P(x \geq 2) =$ _____

Use the data from the Organize the Data section to calculate the following answers. Round your answers to four decimal places.

$RF(x = 3) =$ _____
$RF(0 < x < 3) =$ _____
$RF(x \geq 2) =$ _____

The inequalities are where most of the lost points in this lab live, so read them slowly. The phrase $0 < x < 3$ is strictly between: it means $x = 1$ or $x = 2$, and it excludes both 0 and 3. The phrase $x \geq 2$ includes 2 itself, so it means $x = 2$ or $x = 3$. Because $X$ only takes the whole-number values 0, 1, 2, and 3, you can always settle an ambiguous inequality by writing out which of those four values it actually admits, then adding up just those probabilities.

Try It Now B.2.8

Andre is working through the first list in Use the Data. Using the theoretical distribution, compute $P(x = 3)$, $P(0 < x < 3)$, and $P(x \geq 2)$ for him.

Solution

Step 1 — $P(x = 3)$. This is a single value, read straight off Table B.2.3:

$$ P(x = 3) = \frac{1}{216} \approx 0.0046 $$

Step 2 — $P(0 < x < 3)$. Strictly between 0 and 3 means $x = 1$ or $x = 2$:

$$ P(0 < x < 3) = \frac{75}{216} + \frac{15}{216} = \frac{90}{216} \approx 0.4167 $$

Step 3 — $P(x \geq 2)$. At least 2 means $x = 2$ or $x = 3$:

$$ P(x \geq 2) = \frac{15}{216} + \frac{1}{216} = \frac{16}{216} \approx 0.0741 $$

Answer: 0.0046, 0.4167, and 0.0741. Now he puts his class's three $RF$ values beside them. The middle pair will usually be close, because $P(0 < x < 3)$ is a large probability and large probabilities settle down quickly. The first pair will usually be embarrassing, because $P(x = 3) \approx 0.0046$ and a class like his that plays a hundred games will most often see zero triples and report $RF = 0.0000$. That is not evidence against the theory. It is what a probability of one in 216 looks like in a hundred trials.

B.2.7 Discussion Question

For questions 1 and 2, consider the graphs, the probabilities, the relative frequencies, the means, and the standard deviations.

Knowing that data vary, describe three similarities between the graphs and distributions of the theoretical and empirical distributions. Use complete sentences.
Describe the three most significant differences between the graphs or distributions of the theoretical and empirical distributions.
Thinking about your answers to questions 1 and 2, does it appear that the data fit the theoretical distribution? In complete sentences, explain why or why not.
Suppose that the experiment had been repeated 500 times. Would you expect Table B.2.1 or Table B.2.2 to change, and how would it change? Why would the other table not change?

Question 4 is the one to think hardest about, and it is the one the Law of Large Numbers answers. One of these two tables is built from data and one is built from the rules of the game, and only one of them can be affected by playing more games. Say which, say why, and say which direction the changing one would move.

Try It Now B.2.9

Answer Discussion Question 4. If the experiment were repeated 500 times instead, which of Table B.2.1 and Table B.2.2 would you expect to change, how would it change, and why would the other one hold still?

Solution

Step 1 — ask where each table's numbers come from. Table B.2.1 is filled in from games the class played, so every number in it is measured. Table B.2.2 is computed from $X \sim B\left(3, \frac{1}{6}\right)$, which is fixed by the rules of the game: three dice, six faces, one winning number.

Step 2 — decide which one 500 games can touch. Playing more games produces new counts, so the frequency and relative frequency columns of Table B.2.1 would change. Nothing about playing more games alters the number of faces on a die, so Table B.2.2 would not change at all.

Step 3 — say which direction. By the Law of Large Numbers, the relative frequencies in Table B.2.1 would move toward the probabilities in Table B.2.2 — 0 matches settling near 0.5787, 1 match near 0.3472, 2 near 0.0694, and 3 near 0.0046. The $x = 3$ row is the one to watch: at 120 games you would often record 0.0000 there, but at 500 games you would expect two or three triples to have shown up.

Answer: Table B.2.1 changes and Table B.2.2 does not. The empirical relative frequencies drift toward the theoretical probabilities as the number of games grows, because the theoretical table describes the game itself while the empirical table describes one sample of plays. More data gives you a better look at a distribution that was never moving.

Key Terms

empirical distribution — the distribution of outcomes actually observed in an experiment, summarized by the relative frequency of each outcome.

expected value — the long-run arithmetic average of a random variable over many repetitions of an experiment; also called the mean, written $\mu = \sum x P(x)$.

mean of a probability distribution — the long-term average of many trials of a statistical experiment.

relative frequency — the number of times an outcome occurred divided by the total number of trials.

standard deviation of a probability distribution — a number that measures how far the outcomes of a statistical experiment fall from the mean of the distribution, $\sigma = \sqrt{\sum \left[(x - \mu)^2 \cdot P(x)\right]}$.

the Law of Large Numbers — as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency probability approaches zero.

theoretical distribution — the distribution predicted by the rules of the experiment, computed before any data are collected.

**Table B.2.2** — The theoretical distribution: what the rules of the game predict, before any dice are rolled.
\(x\)	\(y\)	\(P(x) = P(y)\)
0
1
2
3