Box Models - Extra Practice

Question 1

In the Normal Approximation notes, one of the examples is about dice, and sets up a box model for the process of rolling a pair of fair six-sided dice and summing the spots. In the notes we make 2 draws at random with replacement from the following box, and sum the draws:

\[ \boxed{ \fbox{1}\; \fbox{2} \;\fbox{3} \;\fbox{4}\; \fbox{5} \; \fbox{6}} \]

We might get a sum as low as 2 (1 both times), as high as 12 (6 both times), or any number in between. So… why not just make one draw from the box below? Seems easier?

\[ \boxed{ \fbox{2}\; \fbox{3}\; \fbox{4} \; \fbox{5} \; \fbox{6} \; \fbox{7} \; \fbox{8} \; \fbox{9} \; \fbox{10}\; \fbox{11} \;\fbox{12}} \]

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Because the sums are not equally likely, but the tickets in this box are. This box has 11 tickets, so one draw gives each sum the same chance: \(1/11\).

But with real dice, some sums are much more common than others. There’s only one way to roll a 2 (), but there are six ways to roll a 7 (, , , , , ). Out of the 36 equally likely ways the two dice can land:

• \(P(\text{sum} = 2) = 1/36\)
• \(P(\text{sum} = 7) = 6/36 = 1/6\)

So this box would make a 2 way too likely and a 7 not likely enough. Just listing the possible outcomes isn’t enough if the outcomes aren’t actually equally likely.

If we did indeed want to model the sum of two die rolls with a box from which we would only draw once, what would the box be? (List all the tickets in the box, which numbers, and how many of each.)

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Use 36 tickets: one for each of the 36 equally likely ways two dice can land, labeled with that roll’s sum.

Here’s the box, with matching tickets grouped together so they’re easy to count:

\[ \boxed{\begin{array}{l} \fbox{2}\;\,\fbox{3}\,\fbox{3}\;\,\fbox{4}\,\fbox{4}\,\fbox{4}\;\,\fbox{5}\,\fbox{5}\,\fbox{5}\,\fbox{5} \\ \fbox{6}\,\fbox{6}\,\fbox{6}\,\fbox{6}\,\fbox{6}\;\,\fbox{7}\,\fbox{7}\,\fbox{7}\,\fbox{7}\,\fbox{7}\,\fbox{7} \\ \fbox{8}\,\fbox{8}\,\fbox{8}\,\fbox{8}\,\fbox{8}\;\,\fbox{9}\,\fbox{9}\,\fbox{9}\,\fbox{9} \\ \fbox{10}\,\fbox{10}\,\fbox{10}\;\,\fbox{11}\,\fbox{11}\;\,\fbox{12} \end{array}} \]

Ticket	\(\fbox{2}\)	\(\fbox{3}\)	\(\fbox{4}\)	\(\fbox{5}\)	\(\fbox{6}\)	\(\fbox{7}\)	\(\fbox{8}\)	\(\fbox{9}\)	\(\fbox{10}\)	\(\fbox{11}\)	\(\fbox{12}\)
Count	1	2	3	4	5	6	5	4	3	2	1

But… where did this come from? Why two 3’s, and six 7’s?

Each ticket stands for one unique way that two dice can land. To make this more intuitive, here’s the same box again, but now each ticket also shows one unique roll of two dice that makes that sum:

2 3 3 4 4 4 5 5 5 5

6 6 6 6 6 7 7 7 7 7 7

8 8 8 8 8 9 9 9 9

10 10 10 11 11 12

From this, it’s easy to see that there are three ways to roll a 4: , , and . So the first box got three \(\fbox{4}\) tickets. There are six ways to roll a 7, so it gets six \(\fbox{7}\) tickets. And so forth.

There are exactly 36 tickets in the box, one for each unique way the dice can land. Each is equally likely to happen in a real roll of two dice. Now each possible total has the appropriate probability. For example, 6 of the 36 tickets are \(\fbox{7}\), so \(P(\text{draw a } 7) = 6/36\), just like with real dice.

Question 2

If we are setting up a box for modeling the number of heads in three tosses of a fair coin, would either of the boxes below work? If not, why not?

1. \(\boxed{\fbox{0}\; \fbox{1}}\): Draw three times at random with replacement, and sum the draws.

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Yes, this works. Each draw stands for one toss: \(\fbox{1}\) means heads and \(\fbox{0}\) means tails.

• The coin is fair (the box has one of each ticket) so each draw has a 50% chance of being heads (\(\fbox{1}\)).
• Because we draw with replacement, the box is the same every time. Same as a coin: one toss doesn’t affect the next.
• Summing all the 0s and 1s effectively just counts the 1s, so the sum of the three draws is the number of “heads.” (For example, if we draw 0, then 1, then 1, we can add 0 + 1 + 1 = 2, which is exactly how many 1s we got.)

2. \(\boxed{\fbox{0} \;\fbox{1}\; \fbox{2} \;\fbox{3}}\): Draw once, the result is the number of heads.

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No. This box lists every possible number of heads, but it gives them all the same chance (\(1/4\) each). That’s the same mistake as the 11-ticket box in Question 1.

Three tosses can come out 8 equally likely ways (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT). Count the heads in each:

• 0 heads: TTT, so \(1/8\)
• 1 head: HTT, THT, TTH, so \(3/8\)
• 2 heads: HHT, HTH, THH, so \(3/8\)
• 3 heads: HHH, so \(1/8\)

So getting exactly 1 head is three times as likely as getting 0 heads. A one-draw box that works would have 8 tickets: \(\fbox{0}\;\fbox{1}\;\fbox{1}\;\fbox{1}\;\fbox{2}\;\fbox{2}\;\fbox{2}\;\fbox{3}\).

Question 3

A standard deck of cards has 52 cards, consisting of 4 “suits” (hearts ♥︎, diamonds ♦︎, spades \(\spadesuit\), and clubs \(\clubsuit\)). Each suit has 13 cards (Ace, King, Queen, Jack, 2, 3, 4, 5, 6, 7, 8, 9, and 10). Half of the cards are red (hearts and diamonds) and half of the cards are black (spades and clubs). I would like to shuffle the cards, draw five cards with replacement, and count the number of hearts in these five cards. What would be the box model for this setup?

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Box: \(\boxed{\fbox{0}\;\fbox{0}\;\fbox{0}\;\fbox{1}}\). Draw 5 times at random with replacement, and sum the draws.

We only care about one thing for each card: is it a heart or not? So we label hearts \(\fbox{1}\) and everything else \(\fbox{0}\). (The card’s color and rank don’t matter here, so we can ignore them.)

• 13 of the 52 cards are hearts, so the chance of drawing a heart is \(13/52 = 1/4\). A box with one \(\fbox{1}\) and three \(\fbox{0}\)’s gives that same \(1/4\) chance. (A box with 13 \(\fbox{1}\)’s and 39 \(\fbox{0}\)’s would also work, since it has the same proportions.)
• We put each card back before the next draw, so we draw from the box with replacement.
• Just like the coin in Question 2, the sum of the 0s and 1s is the number of hearts.

Question 4

One ticket will be drawn at random from each of the two boxes below:

\(A: \boxed{\fbox{1}\; \fbox{2} \;\fbox{3}}\)

\(B: \boxed{ \class{tkt-b}{\fbox{1}} \;\class{tkt-b}{\fbox{2}} \;\class{tkt-b}{\fbox{3}} \;\class{tkt-b}{\fbox{4}}}\)

1. What is the chance the number drawn from \(A\) is greater than the one drawn from \(B\)?

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\(3/12 = 1/4\).

There are \(3 \times 4 = 12\) possible pairs of draws, and they’re all equally likely. A grid makes them easy to count. Each row is a ticket from \(A\), each column is a (dark) ticket from \(B\), and each cell compares them:

	\(\class{tkt-b}{\fbox{1}}\)	\(\class{tkt-b}{\fbox{2}}\)	\(\class{tkt-b}{\fbox{3}}\)	\(\class{tkt-b}{\fbox{4}}\)
\(\fbox{1}\)	1 = 1	1 < 2	1 < 3	1 < 4
\(\fbox{2}\)	2 > 1	2 = 2	2 < 3	2 < 4
\(\fbox{3}\)	3 > 1	3 > 2	3 = 3	3 < 4

\(A\) is greater than \(B\) in 3 of the 12 cells: \(\fbox{2}\,\class{tkt-b}{\fbox{1}}\), \(\fbox{3}\,\class{tkt-b}{\fbox{1}}\), and \(\fbox{3}\,\class{tkt-b}{\fbox{2}}\). So the chance is \(3/12 = 1/4\).

2. What is the chance that the number drawn from \(A\) is equal to the one drawn from \(B\)?

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\(3/12 = 1/4\). Using the same grid from part A’s solution (click above to expand), the numbers match in 3 of the 12 cells: \(\fbox{1}\,\class{tkt-b}{\fbox{1}}\), \(\fbox{2}\,\class{tkt-b}{\fbox{2}}\), and \(\fbox{3}\,\class{tkt-b}{\fbox{3}}\).

3. What is the chance the number drawn from \(A\) is smaller than the one drawn from \(B\)?

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\(6/12 = 1/2\). You can count the 6 “<” cells in the grid from part A’s solution. Or, faster: use the complement rule! In every pair drawn, A’s ticket must be either greater, equal, or smaller than B’s. You’ve already computed the first two (each was \(1/4\)), and the total probability must add up to 1, so… \(1 - (1/4 + 1/4) = 1/2\).

Intuitively, it makes sense that \(A\) is more likely to be smaller than the other way around: box \(B\) has a 4, and \(A\) can never beat that.

Question 5

I want to estimate the proportion of people in the city of Berkeley who speak at least two languages. I stand at the corner of University and Shattuck and ask each person who goes by how many languages they speak, and keep a count of how many speak at least two. Can I set this up as a box model? If so, how? If not, why not?

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Not really.

We can write down the box we’d want, but this way of collecting data doesn’t draw from it at random.

The box we’d want has one ticket per person who lives in the city of Berkeley: \(\fbox{1}\) if they speak at least two languages and \(\fbox{0}\) if not. We’d draw a bunch of tickets at random, and the proportion of \(\fbox{1}\)’s we drew would estimate the proportion for the whole city.

The problem is that a box model only fits if every ticket has the same chance of being drawn. Standing on one street corner doesn’t do that:

• Who walks by isn’t random. That corner is downtown, right next to campus, so you’ll meet lots of students and people who work downtown. Non-student residents who live in other parts of the city, or rarely come downtown, have almost no chance of being “drawn.”
• Some people who walk by don’t live in Berkeley. Commuters, visitors, and tourists end up in your count even though they aren’t tickets in the box.
• Some people won’t stop to answer.

So we’d get a proportion, but we can’t expect it be a good estimate the proportion for the whole city.

Question 6

In this increasingly gen-AI-reliant world of ours, one happy casualty is the “Homework help” company Chegg. Its business model was, for all practical purposes, selling cheating. In 2021, it had about 7.8 million subscribers. Suppose Chegg surveyed all its subscribers asking if they had ever cheated on an assignment using Chegg (the company is trying to show that they don’t enable cheating). Suppose it got back 1.5 million responses of which 87% of the respondents said that they had never used Chegg to cheat on an assignment.

1. Could this survey be set up as a box model? If so, what tickets would go in the box and how many would we draw? If not, why not?

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Not really. Like Question 5, we can describe the box, but the 1.5 million responses weren’t drawn from it at random.

The box we’d want has 7.8 million tickets, one per subscriber: \(\fbox{1}\) if they never cheated using Chegg and \(\fbox{0}\) if they did. We’d draw 1.5 million tickets at random and look at the proportion of \(\fbox{1}\)’s.

But nobody drew those 1.5 million at random. Chegg asked everyone, and the subscribers chose for themselves whether to answer. Honest people are probably more likely to respond in the first place.

There’s a second problem: even the people who do answer might not tell the truth. Chegg itself is asking whether you used Chegg to cheat. Would they really admit it? So the answers people give may not match the true “ticket” for that person.

2. Would this sample of 1.5 million be a representative sample of high school and college students?

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No. There are two main reasons:

1. Wrong group. Chegg subscribers were be surveyed, which may not be only high school and college students. Some subscribers might be in graduate programs, law school, or not enrolled in school at all. Even among subscribers that are high school or college students, the kind of students who pay for a “homework help” service are probably different in many ways from the general student population. Most of the world’s high school and college students never subscribed, so they had no chance to be in the sample.
2. Nonresponse. Same as the previous question – the 1.5 million out of 7.8 million who chose to respond may be very different from those who didn’t.

A huge sample doesn’t fix anything. 1.5 million responses sounds really impressive, but if the sampling method itself is wrong, a bigger sample just gives you a more precise estimate of the wrong thing.