Simulation practice problems with random digit tables

(a) Assign 0, 1, and 2 to the ginger blend and 3, 4, 5, 6, 7, 8, and 9 to every other order. All ten digits are used, each is equally likely, and 3 of the 10 go to ginger, so the scheme gives a ginger probability of $\frac{3}{10} = 0.30$, exactly the advertised rate.

(a) Which three digits you pick does not matter. Assigning 4, 7, and 9 to ginger works just as well, because the digits are equally likely and it is the count, not the identity, that sets the probability.

(b) Ignoring 0 means every 0 is thrown out and a new digit read, so only 9 digits stay in play: 1 through 9. Three of those nine go to ginger, giving $\frac{3}{9} = \frac{1}{3} \approx 0.3333$, not $0.30$.

(b) That is the error to watch for. Discarding is legitimate only when the target proportion cannot be built from the digits you have. Here $0.30 = \frac{3}{10}$ divides the ten digits exactly, so there is nothing to discard, and throwing a digit away inflates the ginger rate by about $0.033$.

(c) $0.25$ is not a multiple of $\frac{1}{10}$, so single digits cannot carry it without waste. One correct scheme: 0 and 1 are ginger, 2 through 7 are other orders, and 8 and 9 are discarded and redrawn. That leaves 8 usable digits with 2 assigned to ginger, so the probability is $\frac{2}{8} = 0.25$.

(c) A discard-free alternative is to read the digits in pairs: 00 through 24 is ginger and 25 through 99 is not, which is $\frac{25}{100} = 0.25$ with every pair usable. It burns 2 digits per customer instead of an average of $\frac{1}{0.8} = 1.25$, so choose whichever you would rather spend, digits or bookkeeping.

(d) One trial is 4 digits, one for each of the next 4 customers, read straight off the table with repeats kept, because separate customers place separate orders and the same digit can legitimately come up twice. For each trial you record a single yes or no: yes if at least one of the 4 digits is 0, 1, or 2. One digit is one customer, not one trial, and confusing those two is the most common way this question goes wrong.

(a) Read the digits in pairs. A pair from 01 to 20 selects the member with that number. The pairs 00 and 21 through 99 match no member, so discard each one and read the next pair. A pair that repeats a member already chosen is also discarded, because the trip needs 4 different people. Stop as soon as 4 distinct members have been selected.

(b) Two-digit pairs run from 00 to 99, which is 100 equally likely pairs, and only 20 of them, 01 through 20, select a member. So $\frac{20}{100} = 0.20$ are usable and $\frac{80}{100} = 0.80$ are discarded on the range check. Expect to read about 5 pairs for every member you manage to select.

(c) Split the 20 digits into pairs: 73, 14, 08, 20, 51, 96, 14, 03, 82, 17.

(c) Work through them in order. 73 is above 20, so discard it. 14 is in range, so member 14 is the first selection. 08 is in range, so member 8 is the second. 20 is in range, so member 20 is the third.

(c) Keep going. 51 is above 20, so discard. 96 is above 20, so discard. 14 is in range but member 14 is already on the list, so discard it as a repeat. 03 is in range and new, so member 3 is the fourth selection and the trial ends.

(c) The trial selected members 14, 8, 20, and 3, using the first 16 digits. The pairs 82 and 17 were never needed and stay in the table for the next trial.

(d) The two situations model different processes. Here the 4 selections come from one fixed group of 20 people and nobody can go twice, so this is sampling without replacement and a repeated number is impossible in the real process. Four separate customers share no pool at all, so two of them ordering the same drink is not only possible but expected, and throwing out those repeats would distort the model.

(a) Assign 0 and 1 to a pack with a holographic card and 2 through 9 to a pack without one. That is 2 of the 10 digits, so $\frac{2}{10} = 0.20$, matching the advertised rate with no discards needed.

(a) One trial is 4 digits, one per pack. Repeated digits inside a trial are kept, since 4 packs are 4 separate openings. For each trial you record yes if at least one of the 4 digits is 0 or 1, and no otherwise.

(b) Scan each line for a 0 or a 1.

(b) Count the yes rows: trials 1, 3, 5, 7, 9, and 10, which is 6 of the 10 trials. The estimate is $\frac{6}{10} = 0.60$.

(c) Take the complement. A single pack has no holographic card with probability $1 - 0.20 = 0.80$, and 4 independent packs all miss with probability $0.80^4$.

(c) Evaluate: $0.8^2 = 0.64$ and $0.64^2 = 0.4096$, so $0.80^4 = 0.4096$ and $1 - 0.4096 = 0.5904$.

(c) The simulation gave $0.60$ against the exact $0.5904$, a difference of about $0.01$. With only 10 trials the estimate can land on nothing but $0, 0.1, 0.2, \ldots, 1$, so coming this close is partly luck. A few hundred trials would pin the value down reliably.

(d) That student is estimating the wrong quantity. Counting holographic digits out of all 40 digits treats each single digit as its own trial, so $\frac{7}{40} = 0.175$ estimates the per-pack rate, which the app already told you is $0.20$. The question asked about a group of 4 packs, and the unit of the trial has to match the unit of the question.

(a) Number the 8 tutors 1 through 8 and let Rosa be tutor 1. Assign digit 1 to Rosa and digits 2 through 8 to the other seven tutors, one digit each, then discard 0 and 9 whenever they appear, since there is no tutor 0 and no tutor 9.

(a) The discard is what makes the scheme fair. With 0 and 9 thrown out, 8 usable digits remain and each tutor holds exactly one, so every tutor has probability $\frac{1}{8} = 0.125$ on any Saturday. Handing 0 and 9 to two of the tutors instead would give those two $\frac{2}{10}$ and everyone else $\frac{1}{10}$.

(a) One trial is 20 usable digits, one per Saturday, reading past any 0 or 9 without counting it. Repeats are kept, because a tutor can be picked in consecutive weeks. For each trial you record the number of 1s.

(b) The mean of the 100 simulated counts is the total number of Rosa shifts divided by 100. Multiply each count by its frequency: $0 \times 7 = 0$, $1 \times 20 = 20$, $2 \times 27 = 54$, $3 \times 22 = 66$, $4 \times 14 = 56$, $5 \times 7 = 35$, $6 \times 2 = 12$, $7 \times 1 = 7$, and the last column contributes $0$.

(b) Add and divide: $0 + 20 + 54 + 66 + 56 + 35 + 12 + 7 = 250$, so the simulated mean is $\frac{250}{100} = 2.50$ shifts. Random assignment predicts $20 \times \frac{1}{8} = 2.5$ shifts, so the simulation is behaving the way it should.

(c) Count the trials that matched or beat Rosa's result. Trials with 6 shifts: 2. With 7 shifts: 1. With 8 or more: 0. That is $2 + 1 + 0 = 3$ trials out of 100.

(c) The estimate is $\frac{3}{100} = 0.03$. Under random assignment, about 3 percent of 20-Saturday stretches would hand one particular tutor 6 or more shifts.

(d) An estimate of $0.03$ sits below the 5 percent mark people usually treat as surprising, so Rosa's 6 shifts are unusual under the random-assignment claim, and that is some evidence the software is not assigning shifts the way the center describes.

(d) Now state the limits. A probability of $0.03$ is rare, not impossible, so the simulation proves nothing: roughly 3 stretches in 100 would look like this even with a perfectly fair scheduler. The estimate also carries its own uncertainty from using only 100 trials. And Rosa was singled out after her total was seen, which makes an unusual result much easier to find than it would be if a tutor had been named in advance.

(a) Digit assignment: 0, 1, 2, 3, 4, and 5 mean the loaf passes the crust check, and 6, 7, 8, and 9 mean it does not. That is 6 of the 10 digits, so the pass probability is $\frac{6}{10} = 0.60$, and no digit needs to be discarded.

(a) One trial is 10 consecutive digits, one for each loaf in a batch of 10. Repeated digits are kept, since 10 loaves are 10 separate checks.

(a) What you record: count the digits from 0 to 5 in the trial, which is the number of loaves that passed, then record yes if that count is 8, 9, or 10, and no otherwise.

(a) Repeat for a large number of trials, at least a few hundred. The estimate is the number of yes trials divided by the number of trials.

(b) Count the pass digits, 0 through 5, in each line.

(b) Check two lines digit by digit. Trial 1: 3 passes, 7 fails, 1 passes, 5 passes, 9 fails, 0 passes, 4 passes, 8 fails, 2 passes, 6 fails, for 6 passes, short of 8. Trial 3: 1, 4, 0, 2, 5, and 3 all pass, 9 and 7 both fail, then 0 and 4 pass, for 8 passes, which reaches the mark.

(c) One trial in five said yes, so these trials give $\frac{1}{5} = 0.20$.

(c) The baker should not report it, because 5 trials cannot resolve a probability. The estimate can only come out as $0$, $0.2$, $0.4$, $0.6$, $0.8$, or $1$, so it is coarse before it is even inaccurate. The law of large numbers is what fixes this: as the number of trials grows, the proportion of yes trials settles toward the true probability, and a few hundred trials would give the baker a number worth acting on.

(d) The assignment reads each loaf as an independent draw with the same pass probability of $0.60$. If loaves in one batch share an oven position, a dough, or a bake time, a bad batch could fail together, and then the digits would understate how often a batch comes out very good or very bad.

(a) Single digits cannot carry 45 percent or 35 percent, since neither is a multiple of $\frac{1}{10}$, so read the table in pairs. Assign 00 through 44 to commuters, which is 45 of the 100 pairs, or $0.45$; 45 through 79 to recreational riders, which is 35 pairs, or $0.35$; and 80 through 99 to delivery riders, which is 20 pairs, or $0.20$.

(a) Check that the assignment covers everything exactly once: $45 + 35 + 20 = 100$ pairs, so every pair from 00 to 99 lands in exactly one category and nothing has to be discarded. The three categories are mutually exclusive and cover every rider, which is what a valid assignment requires.

(a) One trial is 6 pairs, so 12 digits, one pair per rider. Repeats are kept, since 6 riders are 6 separate people. Record the number of pairs falling in 80 through 99, then note whether that count is 2 or more.

(b) Count the digits in each block. Digits 0 to 4 is 5 of the 10 digits, so $\frac{5}{10} = 0.50$ for commuters. Digits 5 to 7 is 3 digits, so $\frac{3}{10} = 0.30$ for recreational riders. Digits 8 and 9 is 2 digits, so $\frac{2}{10} = 0.20$ for delivery riders.

(b) The delivery share happens to be right, but commuters are overstated by $0.05$ and recreational riders understated by $0.05$. For this one question, which only counts delivery riders, the scheme would still give the right answer; for any question touching the other two categories it would not, and a scheme that is only accidentally right is not worth using.

(c) Split line 1 into pairs: 07, 62, 91, 38, 45, 83. Classify each one. 07 is in 00 to 44, a commuter. 62 is in 45 to 79, recreational. 91 is in 80 to 99, delivery. 38 is a commuter. 45 is the first pair of the recreational block, so recreational. 83 is delivery. That is 2 delivery riders, so trial 1 records yes.

(c) Split line 2 into pairs: 54, 20, 76, 09, 33, 68. Classify each one. 54 is recreational, 20 is a commuter, 76 is recreational, 09 is a commuter, 33 is a commuter, and 68 is recreational. That is 0 delivery riders, so trial 2 records no.

(d) Two trials give an estimate of $0$, $0.5$, or $1$ and nothing else, which carries no real information about a probability. Run at least a few hundred trials, then divide the number of yes trials by the number of trials. For reference, the exact answer here is $1 - 0.8^6 - 6(0.2)(0.8^5) = 1 - 0.262144 - 0.393216 = 0.34464$, and the 2-trial estimate of $0.5$ is nowhere near it.

(a) Any assignment that uses all ten digits gives each outcome a probability of the form $\frac{k}{10}$ for some whole number $k$. Since $\frac{1}{3} = 0.3333\ldots$ is not $\frac{k}{10}$ for any whole $k$, no such assignment is exact: $\frac{3}{10} = 0.30$ is too low and $\frac{4}{10} = 0.40$ is too high.

(a) Discarding fixes it. Assign 1, 2, and 3 to a jam, assign 4, 5, 6, 7, 8, and 9 to a clean startup, and discard every 0. That leaves 9 usable digits with 3 of them meaning a jam, so the probability is exactly $\frac{3}{9} = \frac{1}{3}$.

(b) One digit in ten is a 0, so $\frac{1}{10} = 0.10$ of the digits are discarded and $0.90$ are usable. To collect 5 usable digits you need about $\frac{5}{0.90} = 5.6$ digits on average, so plan on roughly 6 digits per trial and accept that trials vary in length.

(b) One trial is 5 usable digits, one per startup, ignoring any 0 you read past. Repeats are kept, since 5 startups are 5 separate events. Record the number of digits from 1 to 3, then note whether that count is 2 or more.

(c) Trial 1 reads 6, 4, then a 0 that is skipped, then 9, 5, 2. The five usable digits are 6, 4, 9, 5, and 2, and only the 2 falls in 1 to 3, so trial 1 has 1 jam and records no.

(c) Trial 2 picks up where trial 1 stopped: 7, 1, 3, then a skipped 0, then 8, 6. The usable digits are 7, 1, 3, 8, and 6, and both the 1 and the 3 are jams, so trial 2 has 2 jams and records yes.

(c) Trial 3 reads 4, 9, 2, 5, then a skipped 0, then 1. The usable digits are 4, 9, 2, 5, and 1, and the 2 and the 1 are jams, so trial 3 has 2 jams and records yes.

(c) Trial 4 reads 7, 3, 8, then a skipped 0, then 6, 4. The usable digits are 7, 3, 8, 6, and 4, and only the 3 is a jam, so trial 4 has 1 jam and records no. Every trial here happened to use exactly 6 digits because each block of 6 contained exactly one 0, which will not usually be the case.

(d) Two of the 4 trials recorded yes, so the estimate is $\frac{2}{4} = 0.50$.

(d) Compute the exact value. $\left(\frac{2}{3}\right)^5 = \frac{32}{243} = 0.1317$ to four decimal places, and $\left(\frac{2}{3}\right)^4 = \frac{16}{81} = 0.1975$, so $5 \times \frac{1}{3} \times 0.1975 = 1.6667 \times 0.1975 = 0.3292$. Then $1 - 0.1317 - 0.3292 = 0.5391$. The 4-trial estimate of $0.50$ lands in the right neighborhood, which at this trial count is luck rather than evidence.

(a) Find the renter share of the applicant pool: $\frac{200}{500} = 0.40$. Assign digits 0, 1, 2, and 3 to a renter and digits 4 through 9 to a non-renter. That is 4 of the 10 digits, so the scheme gives $\frac{4}{10} = 0.40$ with no discards.

(b) One trial is 25 digits, one for each of the 25 subsidies awarded. Record the number of digits from 0 to 3, which is the simulated number of renters among the winners, and note whether that count is 5 or fewer, since 5 is what actually happened.

(c) Walk the 25 digits in order and count the ones from 0 to 3. The digits are 8, 2, 5, 0, 9, 3, 6, 1, 7, 4, 0, 8, 2, 9, 5, 3, 6, 0, 4, 7, 1, 9, 8, 2, 5. The ones that count are the 2 in position 2, the 0 in position 4, the 3 in position 6, the 1 in position 8, the 0 in position 11, the 2 in position 13, the 3 in position 16, the 0 in position 18, the 1 in position 21, and the 2 in position 24.

(c) That is 10 renters in this trial. Since 10 is far above 5, the trial does not count toward the tally in part (d). Ten is also exactly what to expect, because $25 \times 0.40 = 10$.

(d) The estimate is the proportion of trials that matched or beat the real result: $\frac{6}{200} = 0.03$.

(d) Read it in context. If the lottery treated renters like a random draw at 40 percent, about 3 percent of 25-winner lotteries would produce 5 or fewer renters. Getting 5 is not impossible under fairness, only uncommon.

(e) What the group can say: $0.03$ is below the 5 percent mark usually treated as surprising, so the outcome is unusual under the fairness assumption, and that is evidence worth pursuing. What they cannot say: the simulation does not show the lottery was rigged, since about 3 in 100 fair lotteries look like this and the simulation identifies no mechanism. A simulation can only tell you how often chance alone produces a result this extreme.

(e) Name the approximation. The 25 winners are drawn without replacement from 500 applicants, so the draws are not truly independent, yet the digit scheme treats each of the 25 as an independent draw at $0.40$. That is defensible here because 25 is 5 percent of 500, inside the 10% condition that the sample be no more than a tenth of the population, since $25 \le \frac{500}{10} = 50$.