Carnival Booth Lab: teacher answer key
Public sample teacher key. Includes model answers; open the student worksheet when assigning the activity.
(+)-S.MD.2: Calculate and interpret the expected value of a random variable. (+)-S.MD.6: Use probabilities to make fair decisions.
Using the worksheet
- Worksheet opens the questions. New questions follow each run; misses add a reflection prompt.
- Recorded automatically: the prediction, the setting the student chose and its expected value E, how many players played, the profit per player and cash box, and pass or miss. On Level 4, each submitted estimate is one run: it lists the afternoon predictions and runs since the last estimate, the estimate, and the student's data so far.
- My work offers Copy my work and Turn in a PDF. Nothing is submitted automatically. Review written reasoning separately from game success; blank responses remain visible.
- The game's completion code (CARN-levels-stars) still works as before. My work adds the answers and runs behind it. The game forgets its progress when the page reloads, but the worksheet keeps answers and runs on that device.
- On shared devices, use Clear my work and tap again to confirm. The game's Start over does not clear worksheet answers. If browser storage is blocked, export before leaving.
Numbers like "#4" match the printed worksheet. Questions marked "After a miss" appear in the game only after a missed run; on paper they are combined into one question near the end.
Level 1: Fair game
- To pass: prize $8. E = 2 − 0.25 × 8 = $0.00 per player (break even). Only $8 passes: each $1 of prize moves E by $0.25, and E must be less than $0.01 from the target.
- Common misses: $6 (took the $2 cost off the prize first) gives E = +$0.50, and the debrief names that mistake. $10 gives −$0.50. $5, the starting prize, gives +$0.75.
- Stars: ★★★ if the first new design passes, ★★ on the second, ★ after that. Re-running the same prize (10 players, then 10,000) is free. Runs after passing are practice for stars, but they still show in the run log.
- Each play earns the booth +$2 (the player loses, 75%) or −$6 (the player wins $8, 25%). Runs of the $8 booth: 10 players land on +$2.00, +$1.20, +$0.40, −$0.40, −$1.20, −$2.00 or lower per player, never exactly $0.00, and about half show a loss. 100 players: about −$0.72 to +$0.64. 1,000 players: about −$0.22 to +$0.22. 10,000 players: about −$0.07 to +$0.07 per player (cash box about −$680 to +$680). These ranges hold 95% of runs.
#1. Record what happened: the profit per player in your shortest run and your longest run of the $8 booth. Give the number of players for each. (+)-S.MD.2
Example: "10 players: +$0.40 per player. 10,000 players: −$0.02 per player." A 10-player run can land anywhere from about −$2.00 to +$2.00 and is never exactly $0.00. A 10,000-player run usually lands between −$0.07 and +$0.07. Check it against the run log on the My work page.
#2. Your $8 booth is a fair game. What does fair mean here? In the long run, who comes out ahead: the booth or the players? (+)-S.MD.6
A fair game has an expected value of $0: E = 2 − 0.25 × 8 = $0.00. Each player pays $2 and wins back 0.25 × $8 = $2 on average, so in the long run neither the booth nor the players come out ahead. One player still either loses $2 or comes out $6 ahead. A fair booth raises $0 for the carnival on average, so a fundraiser booth needs E above $0.
| Common wrong answer | What it shows |
|---|---|
| "Fair means everyone has the same chance to win." | Mixes up fair (E = $0) with equal chances. Here only 25% of players win. |
| "The booth comes out ahead because 75% of players lose." | Counts how often players lose but not how much the winners get: each winner takes $8. |
#3. Why can a short run land far from E = $0.00, even though the game is fair? What happens as more players play? (+)-S.MD.2
In a short run, one or two extra winners change the average a lot: with 10 players, each extra winner moves profit per player by $0.80, so 10-player runs land anywhere from about −$2.00 to +$2.00. As more players play, the lucky and unlucky plays balance out and profit per player settles close to E. This is the law of large numbers: 10,000 players usually land within $0.07 of $0.00.
| Common wrong answer | What it shows |
|---|---|
| "The booth is broken if a short run isn't $0." | Expects every short run to match E. |
| "With more players the cash box gets closer to $0." | It is profit per player that settles near E. The cash box can still drift by hundreds of dollars over 10,000 players. |
After a miss: One of your booths was not on target. What went wrong? Use E = cost − chance × prize with your numbers. (+)-S.MD.2
$6: "We took the $2 cost off the prize first. The booth pays the whole prize, so E = 2 − 0.25 × 6 = +$0.50, not $0. Solving 2 − 0.25 × prize = 0 gives prize = $8." $10: "E = 2 − 0.25 × 10 = −$0.50, so players win too much on average. A smaller prize raises E."
Level 2: Profit target
- To pass: chance to win 25%. E = 3 − 0.25 × 10 = +$0.50 per player. Only 25.0% passes: each 0.1% moves E by $0.01, and E must be less than $0.01 from the target.
- Common misses the debrief names: 30% is the break-even chance (E = $0.00). 35% adds the $0.50 instead of subtracting it (E = −$0.50). 5% divides the target by the prize (E = +$2.50).
- Each play earns the booth +$3 (the player loses, 75%) or −$7 (the player wins $10, 25%). Runs of the 25% booth: 10 players land on whole dollars from about −$2 to +$3 per player (about 1 in 5 show a loss, and 1 in 4 show exactly $0.00). 100 players: about −$0.40 to +$1.30. 1,000 players: about +$0.23 to +$0.77. 10,000 players: about +$0.42 to +$0.58 per player (cash box about $4,150 to $5,850; expected +$5,000). These ranges hold 95% of runs.
#4. Record your longest run of the 25% booth: how many players, the profit per player, and the cash box. How close was the cash box to E × players? (+)-S.MD.2
Example: "10,000 players: +$0.53 per player, cash box +$5,300. E × players = $0.50 × 10,000 = $5,000, so it was $300 off." A 10,000-player run usually lands between $4,150 and $5,850. Check it against the run log on the My work page.
#5. Look at the histogram of single plays. What can the booth earn on one play? Why does no single play ever earn exactly $0.50? (+)-S.MD.2
One play earns the booth +$3 (the player loses) or −$7 (the player wins: $3 − $10). Those are the only two results. $0.50 is the average over many plays: 0.75 × $3 + 0.25 × (−$7) = $2.25 − $1.75 = $0.50. The expected value is the mean of the results, not a result one play can have.
| Common wrong answer | What it shows |
|---|---|
| "Each player gives the booth $0.50." | Thinks E is what happens on every play. |
| "$0.50 is the most common result." | Mixes up the mean with the most likely result. The most common play earns +$3. |
After a miss: One of your booths was not on target. What went wrong? Use E = cost − chance × prize with your numbers. (+)-S.MD.2
30%: "E = 3 − 0.30 × 10 = $0, the break-even chance. The treasurer wanted $0.50 left over." 35%: "We added the $0.50: 3 − 0.35 × 10 = −$0.50, a loss." 5%: "We divided 0.50 by 10. Solving 3 − 10 × chance = 0.50 gives chance = 0.25 = 25%."
Level 3: Two prizes
- To pass: big prize chance 5.6%. E = 5 − 0.30 × 4 − 0.056 × 50 = 5 − 1.20 − 2.80 = +$1.00 per player. Only 5.6% passes: each 0.1% moves E by $0.05.
- Common misses the debrief names: 8% leaves out the small prize (5 − 0.08 × 50 = 1, but the real E = −$0.20 per player, so 10,000 players sink the cash box by about $2,000 on average). 7.6% is the break-even chance (E = $0.00).
- Each play earns the booth +$5 (no prize, 64.4%), +$1 (small prize, 30%), or −$45 (big prize, 5.6%). The rare −$45 makes short runs swing hard: more than half of 10-player runs have no big winner and show about +$3 to +$5 per player. Runs of the 5.6% booth: 100 players about −$1.40 to +$3.00 per player. 1,000 players about +$0.28 to +$1.68. 10,000 players about +$0.78 to +$1.22 (cash box about $7,800 to $12,200; expected +$10,000). These ranges hold 95% of runs.
- The 8% booth (small prize left out) over 10,000 players usually lands between about −$4,600 and +$600. About 1 run in 15 still ends above $0, which is why the expected value, not one run, shows the mistake.
#6. Record your longest run of the 5.6% booth: how many players, the profit per player, and the cash box. How close was it to E? (+)-S.MD.2
Example: "10,000 players: +$1.09 per player, cash box +$10,900, $0.09 from E = +$1.00." A 10,000-player run usually lands between +$0.78 and +$1.22 per player. This booth swings more than the Level 1 and 2 booths because of the rare $50 prize. Check it against the run log on the My work page.
#7. A booth that forgets the $4 small prize sets the big prize chance to 8%. What is that booth's real E? What would 10,000 players do to its cash box? (+)-S.MD.2
E = 5 − 0.30 × 4 − 0.08 × 50 = 5 − 1.20 − 4.00 = −$0.20 per player. Over 10,000 players the cash box would sink by about $2,000 (−$0.20 × 10,000). Every prize counts: 30% of players win $4, which costs the booth 0.30 × $4 = $1.20 per player on average.
| Common wrong answer | What it shows |
|---|---|
| "+$1.00, because 5 − 0.08 × 50 = 1." | Left out the small prize. Every prize subtracts chance × prize. |
| "The small prize is too small to matter." | 30% of players win it, so it costs $1.20 per player, more than the whole $1.00 target. |
After a miss: One of your booths was not on target. What went wrong? Use both prizes in E = cost − chance × prize. (+)-S.MD.2
8%: "We left out the small prize. 30% of players win $4, which costs $1.20 per player, so E = 5 − 1.20 − 4.00 = −$0.20." 7.6%: "E = 5 − 1.20 − 0.076 × 50 = $0, break even." The fix: 5 − 1.20 − 50 × chance = 1, so 50 × chance = 2.80 and chance = 0.056 = 5.6%.
Level 4: Bad luck or bad booth?
- The booth is the correct Level 2 booth: cost $3, prize $10, 25% chance, E = +$0.50 per player. An afternoon is 20 players: it collects $60 and pays $10 to each winner, so afternoon profit = 60 − 10 × winners. On average 5 players win, for +$10 an afternoon.
- An afternoon loses money with 7 or more winners. 6 winners is exactly $0, which is not a loss. The exact chance of a losing afternoon is 21.4% (P(7 or more winners out of 20) = 0.214), about 1 afternoon in 5. The game shows 21.4% after a good estimate.
- To pass: run at least 100 afternoons (Submit estimate unlocks at 100), then submit an estimate from 16% to 26%. Stars: ★★★ on the first estimate, ★★ on the second, ★ after that. On the worksheet, each submitted estimate is one run.
- What the data shows (95% of runs): 10 afternoons can land anywhere from 0% to 50%. 100 afternoons: about 14% to 30% (about 4 runs in 5 land inside 16% to 26%). 1,000 afternoons: about 19% to 24%.
- The game names these wrong estimates: about 50% (a booth that earns +$0.50 per player does not lose half its afternoons), 75% (the chance that one player loses), 0%, and 100%. Watch for 25%: it is the chance that one player wins, but it falls inside 16% to 26%, so the game accepts it. Check the reasoning in the winners question.
#8. Before you run: an afternoon collects 20 × $3 = $60 and pays $10 to each winner. How many winners out of 20 make the afternoon lose money? Show the math. (+)-S.MD.2
7 or more. Afternoon profit = 60 − 10 × winners. With 6 winners, 60 − 60 = $0 (break even, not a loss). With 7 winners, 60 − 70 = −$10, a loss.
| Common wrong answer | What it shows |
|---|---|
| "6 or more" | 6 winners is exactly $0, break even. A loss needs profit below $0. |
| "5 or more" | 5 winners is the average afternoon: 60 − 50 = +$10. |
#9. Record your data: how many afternoons you ran in all, how many lost money, and the estimate you submitted. (+)-S.MD.2
Example: "1,110 afternoons, 236 lost money (21.3%). We estimated 21%." With 1,000 or more afternoons the percent almost always lands between 19% and 24%. The exact chance is 21.4%. Check it against the run log on the My work page.
#10. The treasurer saw one losing afternoon and wants to shut the booth down. Is that a fair decision? Use your data and the expected value. (+)-S.MD.6
No. The booth is designed right: E = +$0.50 per player, about +$10 an afternoon on average, and over all our afternoons it still made close to $0.50 per player. But about 21% of afternoons (1 in 5) still lose money by chance, when 7 or more of the 20 players win. One losing afternoon is what a correct booth does now and then, so it is weak evidence. A fair decision looks at many afternoons or at E, not one afternoon.
| Common wrong answer | What it shows |
|---|---|
| "Yes. It lost money, so it's broken." | Judges the booth on one short run. |
| "No, because a correct booth never loses." | Right answer, wrong reason: correct booths lose about 1 afternoon in 5. |
After a miss: One of your estimates was not accepted. What went wrong? Use your data. (+)-S.MD.2
50%: "We thought a booth that makes $0.50 loses half the time, but only 236 of 1,110 afternoons lost money (21.3%)." 75%: "75% is the chance that one player loses, not the chance that an afternoon loses money." 0% to 15%: "We thought a correct booth never loses, or we trusted a short run of 10 afternoons."
When a booth or an estimate missed (#11)
Any of the After a miss answers above. Look for E = cost − chance × prize with the student's own numbers (on Level 4, their data) and the specific thing that was off.
Wrap-up: the long run
#12. Can a few plays tell you whether a booth is designed right? Use evidence from at least two levels. (+)-S.MD.2, (+)-S.MD.6
Claim: No. A few plays can land far from the expected value, so judge a booth by its expected value or by a very large number of plays. Evidence: In Level 1, the fair $8 booth (E = $0.00) can show anywhere from −$2.00 to +$2.00 per player over 10 players, but 10,000 players land within about $0.07 of $0. In Level 2, every single play earns +$3 or −$7, never $0.50, yet 10,000 players average close to +$0.50 (cash box near $5,000). In Level 4, that same correct booth still lost money on about 21% of 20-player afternoons. Reasoning: The expected value is the long-run average of all the possible results, weighted by their chances. By the law of large numbers, the average settles near E only as the number of plays grows. So one losing afternoon is not proof that a booth is broken, and a fair decision uses E or lots of data.