# On number sense — Would one day of NYC coffee fill the Statue of Liberty? And other fun questions…

Today I gave a lecture on what I call number sense, using a process of estimation and approximation in order to calculate various unknown quantities, including a few fantastical ones. How much coffee is made per day in New York City? Would it fill up the Statue of Liberty? Approximately how many babies are born in New York City each day? If you made a stack of quarters to reach the distance to the moon, what would the dollar-value be? If you piled those quarters in a heap, would it fit in Central Park? How much does the Empire State Building weigh?

These kinds of back-of-the-envelope calculations, in my view, have at their heart the idea that one can solve a difficult or seemingly impossible problem by breaking it into more manageable pieces. We don’t just pull a final answer out of the air, but rather make simplifying assumptions and informed estimates for related quantities that we are more familiar with or have knowledge about, and then use that information to derive a better estimate for the main quantity. For most of these questions, at the outset we may have little idea what would be a reasonable answer, but by the end, we gain some insight and find ourselves a little closer to the truth.

These kind of calculations are also known as Fermi estimations, in light of Fermi’s remarkable ability to make surprisingly accurate estimations on the basis of little or no hard data. The wikipedia page (thanks to Artie Prendergrast-Smith for mentioning this link in the comments) emphasizes that even in a case where the estimate is significantly off the true value, nevertheless we may still find value in the Fermi calculation, because it focusses our attention to the reasons for the divergence. In discovering which of the assumptions underlying our calculations was wrong, we come to a deeper understanding of the true situation.

In the lecture, I began with some very easy cases. For example, how many seats were in the auditorium? The students estimated that there were approximately 12 seats per row and about 10 rows, so 120 seats in all. How old was one of the students, in seconds? Well, he was 18 years old, and so we could simply multiply each year by 365 days, times 24 hours per day, times 60 minutes per hour, times 60 seconds per minute, to get
$$18\times 365\times 24\times 60\times 60\approx 600,000,000 \text{ seconds}.$$
One student objected about leap days, since 365 should be 365.25 or so. But I pointed out that this difference was not as important as it might seem, since already we had made far larger rounding assumptions. For example, the student was not exactly 18 years old, but 18 years old and some several months; by using 18 years only, we made a bigger difference in the answer than caused by the leap-day issue, which would be a difference of only five days or so over 18 years. For the same reason, we should feel free to round the numbers to make the calculation easier. We are aiming at a ballpark estimate rather than an exact answer.

Let’s now do some more interesting cases.

Coffee in New York. How much coffee is made each day in New York? Would it fill the Statue of Liberty? First, let me say that I really don’t have any definite information about how much coffee is made each day in New York, and I fear that my own coffee-obsessed perspective will lead me to over-estimate the amount, but let’s give it a try. New York City has a population of approximately 10 million people. Some of those people, like myself, drink a large amount of coffee each day, but many of the others do not drink coffee at all. I would think that a sizable percentage of the NYC population does drink coffee, perhaps as much as a third or even half consumes coffee daily. Many of those coffee-drinkers have more than one cup per day, and also surely more coffee is made than consumed. So it seems reasonable to me to estimate that we have approximately one medium cup of coffee per person on average per day in New York. Basically, we’re saying that the heavy coffee drinkers and the made-but-not-sold coffee approximately makes up for those who abstain, making the average about one cup per person. So we are talking about 10 million cups of coffee per day. A medium cup of brewed coffee at Starbucks is I think about 12-16 ounces, a little less than a pint, and so let’s say about 3 cups per liter. This amounts to roughly 3 million liters of coffee.

Would it fill the Statue of Liberty? The statue itself is, I estimate, about twenty stories tall, counting the base, and if each story is 15 ft, or 5 meters, that would mean 100 meters tall, counting the base. But I think that the base is about half the height, so let’s say 50 meters for the actual statue itself. I’ve never been inside the statue, but my students say that it is about 10 meters across inside, a little more at the bottom than near the top. If we approximated it as a rectangular solid, that would give a volume of $10\times 10\times 50$ cubic meters, or 5000 cubic meters. But since the statue tapers as you go up, particularly in the arm holding the torch, it really is more like a cone than a rectangular solid, and so we should divide by three. But let’s divide just by two, because she isn’t quite as tapered as a cone. So the Statue of Liberty has a volume of approximately 2500 cubic meters. One cubic meter can be thought of as a 10 by 10 by 10 array of little 10cm cubes, and each of those is exactly one liter. So a cubic meter is 1000 liters, and therefore the Statue of Liberty has a volume of $2500\times 1000=2.5$ million liters. But since we had 3 million liters of coffee, the answer our calculation arrives at is: Yes, one day’s worth of New York coffee would fill up the Statue of Liberty!

Well, we do not have perfect confidence in our estimates and assumptions — for example, perhaps there are many fewer coffee drinkers in New York than we estimated or perhaps we underestimated the volume of the Statue of Liberty. Since the estimated volumes were of basically similar magnitudes, we aren’t really entitled to say that definitely the coffee would fill up the Statue of Liberty. Rather, what we have come to know is that those two volumes are comparably similar in size; they are in the same ballpark.

Elevator trips. While riding downtown last weekend with my son on the subway, a crowded 4 train, we overheard the group standing next to us talking about elevators. One lady said, “My elevator company serves as many elevator trips in New York City in five days as the population of the entire world,” and the rest of her group, impressed, nodded affirmatively in reply. But my thoughts, upon hearing that, were to make a quick calculation. Suppose all 10 million NYC residents rode an elevator 10 times every day, which is way too high (probably one trip per person per day is more reasonable, since many people live and work in buildings without elevators). Even in this extreme case of ten trips per person per day, it would mean only 100 million trips total per day, or 500 million trips over 5 days. This is much less than the world population, and so no way is that person’s claim true, especially since there are also many elevator companies. I thought of mentioning my calculation to those people on the subway, but decided against it. Walking out of the subway in the East Village, however, I asked my son (14 years old) whether he heard those people talking about elevators, and he replied, “Oh, yes, and when they said that, I calculated it in my head: no way is that true.” He then proceeded to explain his calculation, similar to mine. Yay, Horatio!

The Chicago marathon. In the run-up to the Chicago marathon this year, on a route that would wind through the windy city streets, Newsweek magazine reported, “Chicago Marathon organizers expect 1.7 million fans to line the route this year.” (Thanks to the critical math commentary of Mark Iris for bringing this example to my attention.) The organizers had emphasized the economic impact of these spectators, many of whom would presumably be eating in Chicago restaurants and staying at Chicago hotels. But is this a reasonable number?

Let’s calculate. A marathon is approximately 26 miles, and the route has two sides for spectators, so let’s round it to 50 miles of spectator viewing spots. Each mile is about 1800 yards, so we have $50\times 1800=90000$ yards of viewing spots. Each spectator, standing shoulder-to-shoulder, with all their stuff, takes up about a yard of space. So if the marathon was lined on both sides for the entire route with spectators standing shoulder-to-shoulder, this would amount to about 90,000 spectators. In order to have 1.7 million spectators, therefore, they would have to be lined up behind each other. But even if the spectators were 10 people deep on each side for the entire route, which is a vast crowd, it would still amount only to 900,000 people. We would have basically to double this to get to 1.7 million. So, if the live event really had 1.7 million spectators lining the route, then it would mean that the race was lined 20 people deep on each side for the entire route. No way is this number correct! I have never had the chance to attend the Chicago marathon, but at the New York marathon, which I assume is comparable, I know that while there are thick crowds at the finish line in Central Park and at some of the other prominent or especially interesting race locations, most of the rest of the route is much thinner, and at the typical nothing-special location, one could expect easily to have a front-row spot.

How many bricks on the college campus? Our campus, covered with some lovely woods and green meadows, has a number of brick Georgian buildings. Most of these are the same standard size as the mathematics department, but there are also some larger buildings such as the library, the performing arts center, the campus center and the gymnasium. At my lecture, the students and I agreed that altogether it amounted to about 30 buildings of the size of the math building. This building is approximately 30 meters by 70 meters, which would make a perimeter of 200 meters, but because the building has wings and is not purely rectangular, let’s add another 100 meters for the folds in the walls, so about 300 meters around the base of the building. The building is two stories tall, about 10 meters tall, making the area of the walls about 3000 square meters. Of course, there are windows and doors that cut out of these walls, but let’s say that they are approximately accounted for by the extra bricks used in the trimming flourishes at the corners and top of the building. So we have 3000 square meters of brick. Each brick is approximately 10 cm by 30 cm, and so one square meter of brick would have about ten rows of a little more than three bricks across, or about 20 bricks. So each building has about $3000\times 20=60,000$ bricks. And with 30 buildings, this amounts to $30\times 60,000=180,000$ bricks in the buildings on campus. There are also some bricks in the central fountain, a few small brick walls and some bricks lining some of the walkways. So let’s add another ten percent for those extra bricks, arriving at about 200,000 bricks on campus in all.

Positive test result for a rare disease. Suppose that as part of your check-up, your doctor randomly administers a clinical test for a certain rare medical condition. The test is 99 percent accurate, in terms of false positives and false negatives, in the sense that 99 percent of people taking the test have an accurate result with the test. Suppose also that the disease itself is rare, held by only one in a million people. If your test comes back positive, what is the likelihood that you actually have the disease?

This is a subtle question. It might seem to make sense initially that there is a 99 percent chance that you have the disease, since that is the accuracy of the test. But this isn’t actually right, because it doesn’t account for the fact that the disease itself is extremely rare, and so the total number of false positives will actually far outweigh the true positive results. For example, suppose that one million people take the test. About one of these people actually has the disease, and that person is 99 percent likely to test positive. So we expect about one true positive result. And for the others, who don’t have the disease, 99 percent of them will test negative. In other words, about 990,000 people will test negative. The remaining 10,000 people, however, one percent of the total, will have false positive results! So you are in this group of 10,000 people who tested positive, with only one of them actually having the disease. So the odds that you actually have the disease are only about one in ten thousand.

Quarters to the moon. How many quarters would you have to stack to reach the moon? How much would it be worth? how much would it weigh? More than the earth? More than the Empire state building? If you put all those quarters into a pile, would it fit in Central Park?

Well, OK, the question is a little absurd, and there are all kinds of problematic issues: we couldn’t ever actually make such a tower of quarters, as it would topple over; it doesn’t make sense to build a tower to the moon, since the moon is in orbit around the earth and it is moving; the quarters would begin to orbit the earth themselves, or fall back down to the earth or to the moon. But let’s just try to ignore all those problematic issues, and just try to answer how many quarters it would take to make a pile that high.


Update: What a riot it was! I really had a lot of fun.

# The pirate treasure division problem

In my logic course this semester, as a part of the section on the logic of games, we considered the pirate treasure division problem.

Imagine a pirate ship with a crew of fearsome, perfectly logical pirates and a treasure of 100 gold coins to be divided amongst them. How shall they do it? They have long agreed upon the pirate treasure division procedure: The pirates are linearly ordered by rank, with the Captain, the first Lieutenant, the second Lieutenant and so on down the line; but let us simply refer to them as Pirate 1, Pirate 2, Pirate 3 and so on. Pirate 9 is swabbing the decks in preparation. For the division procedure, all the pirates assemble on deck, and the lowest-ranking pirate mounts the plank. Facing the other pirates, she proposes a particular division of the gold — so-and-so many gold pieces to the captain, so-and-so many pieces to Pirate 2 and so on.  The pirates then vote on the plan, including the pirate on the plank, and if a strict majority of the pirates approve of the plan, then it is adopted and that is how the gold is divided. But if the pirate’s plan is not approved by a pirate majority, then regretfully she must walk the plank into the sea (and her death) and the procedure continues with the next-lowest ranking pirate, who of course is now the lowest-ranking pirate.

Suppose that you are pirate 10: what plan do you propose?  Would you think it is a good idea to propose that you get to keep 94 gold pieces for yourself, with the six remaining given to a few of the other pirates? In fact, you can propose just such a thing, and if you do it correctly, your plan will pass!

Before explaining why, let me tell you a little more about the pirates. I mentioned that the pirates are perfectly logical, and not only that, they have the common knowledge that they are all perfectly logical. In particular, in their reasoning they can rely on the fact that the other pirates are logical, and that the other pirates know that they are all logical and that they know that, and so on.

Furthermore, it is common knowledge amongst the pirates that they all share the same pirate value system, with the following strictly ordered list of priorities:

Pirate value system:

1. Stay alive.
2. Get gold.
3. Cause the death of other pirates.
4. Arrange that other’s gold goes to the most senior pirates.

That is, at all costs, each pirate would prefer to avoid death, and if alive, to get as much gold as possible, but having achieved that, would prefer that as many other pirates die as possible (but not so much as to give up even one gold coin for additional deaths), and if all other things are equal, would prefer that whatever gold was not gotten for herself, that it goes as much as possible to the most senior pirates, for the pirates are, in their hearts, conservative people.

So, what plan should you propose as Pirate 10? Well, naturally, the pirates will consider Pirate 10’s plan in light of the alternative, which will be the plan proposed by Pirate 9, which will be compared with the plan of Pirate 8 and so on. Thus, it seems we should propagate our analysis from the bottom, working backwards from what happens with a very small number of pirates.

One pirate. If there is only one pirate, the captain, then she mounts the plank, and clearly she should propose “Pirate 1 gets all the gold”, and she should vote in favor of this plan, and so Pirate 1 gets all the gold, as anyone would have expected.

Two pirates. If there are exactly two pirates, then Pirate 2 will mount the plank, and what will she propose? She needs a majority of the two pirates, which means she must get the captain to vote for her plan. But no matter what plan she proposes, even if it is that all the gold should go to the captain, the captain will vote against the plan, since if Pirate 2 is killed, then the captain will get all the gold anyway, and because of pirate value 3, she would prefer that Pirate 2 is killed off.  So Pirate 2’s plan will not be approved by the captain, and so unfortunately, Pirate 2 will walk the plank.

Three pirates. If there are three pirates, then what will Pirate 3 propose? Well, she needs only two votes, and one of them will be her own. So she must convince either Pirate 1 or Pirate 2 to vote for her plan. But actually, Pirate 2 will have a strong incentive to vote for the plan regardless, since otherwise Pirate 2 will be in the situation of the two-pirate case, which ended with Pirate 2’s death. So Pirate 3 can count on Pirate 2’s vote regardless, and so Pirate 3 will propose:  Pirate 3 gets all the gold! This will be approved by both Pirate 2 and Pirate 3, a majority, and so with three pirates, Pirate 3 gets all the gold.

Four pirates. Pirate 4 needs to have three votes, so she needs to get two of the others to vote for her plan. She notices that if she is to die, then Pirates 1 and 2 will get no gold, and so she realizes that if she offers them each one gold coin, they will prefer that, because of the pirate value system. So Pirate 4 will propose to give one gold coin each to Pirates 1 and 2, and 98 gold coins to herself. This plan will pass with the votes of 1, 2 and 4.

Five pirates. Pirate 5 needs three votes, including her own. She can effectively buy the vote of Pirate 3 with one gold coin, since Pirate 3 will otherwise get nothing in the case of four pirates. And she needs one additional vote, that of Pirate 1 or 2, which she can get by offering two gold coins. Because of pirate value 4, she would prefer that the coins go to the highest ranking pirate, so she offers the plan:  two coins to Pirate 1, nothing to pirate 2, one coin to pirate 3, nothing to Pirate 4 and 97 coins to herself.  This plan will pass with the votes of 1, 3 and 5.

Six pirates. Pirate 6 needs four votes, and she can buy the votes of Pirates 2 and 4 with one gold coin each, and then two gold coins to Pirate 3, which is cheaper than the alternatives. So she proposes:  one coin each to 2 and 4, two coins to 3 and 96 coins for herself, and this passes with the votes of 2, 3, 4 and 6.

Seven pirates. Pirate 7 needs four votes, and she can buy the votes of Pirates 1 and 5 with only one coin each, since they get nothing in the six-pirate case. By offering two coins to Pirate 2, she can also get another vote (and she prefers to give the extra gold to Pirate 2 than to other pirates in light of the pirate values).

Eight pirates. Pirate 8 needs five votes, and she can buy the votes of Pirates 3, 4 and 6 with one coin each, and ensure another vote by giving two coins to Pirate 1, keeping the other 95 coins for herself. With her own vote, this plan will pass.

Nine pirates. Pirate 9 needs five votes, and she can buy the votes of Pirates 2, 5 and 7 with one coin each, with two coins to Pirate 3 and her own vote, the plan will pass.

Ten pirates. In light of the division offered by Pirate 9, we can now see that Pirate 10 can ensure six votes by proposing to give one coin each to Pirates 1, 4, 6 and 8, two coins to Pirate 2, and the remaining 94 coins for herself. This plan will pass with those pirates voting in favor (and herself), because they each get more gold this way than they would under the plan of Pirate 9.

We can summarize the various proposals in a table, where the $n^{\rm th}$ row corresponds to the proposal of Pirate $n$.

1 2 3 4 5 6 7 8 9 10
One pirate 100
Two pirates * X
Three pirates 0 0 100
Four pirates 1 1 0 98
Five pirates 2 0 1 0 97
Six pirates 0 1 2 1 0 96
Seven pirates 1 2 0 0 1 0 96
Eight pirates 2 0 1 1 0 1 0 95
Nine pirates 0 1 2 0 1 0 1 0 95
Ten pirates 1 2 0 1 0 1 0 1 0 94

There are a few things to notice, which we can use to deduce how the pattern will continue. Notice that in each row beyond the third row, the number of pirates that get no coins is almost half (the largest integer strictly less than half), exactly one pirate gets two coins, and the remainder get one coin, except for the proposer herself, who gets all the rest. This pattern is sustainable for as long as there is enough gold to implement it, because each pirate can effectively buy the votes of the pirates getting $0$ under the alternative plan with one fewer pirate, and this will be at most one less than half of the previous number; then, she can buy one more vote by giving two coins to one of the pirates who got only one coin in the alternative plan; and with her own vote this will be half plus one, which is a majority. We can furthermore observe that by the pirate value system, the two coins will always go to either Pirate 1, 2 or 3, since one of these will always be the top-ranked pirate having one coin on the previous round. They each cycle with the pattern of 0 coins, one coin, two coins in the various proposals. At least until the gold becomes limited, all the other pirates from Pirate 4 onwards will alternate between zero coins and one coin with each subsequent proposal, and Pirate $n-1$ will always get zero from Pirate $n$.

For this reason, we can see that the pattern continues upward until at least Pirate 199, whose proposal will follow the pattern:

199 Pirates: 1 2 0 0 1 0 1 0 1 0 1 0 1 $\dots$ 1 0 1 0 0

It is with Pirate 199, specifically, that for the first time it takes all one hundred coins to buy the other votes, since she must give ninety-eight pirates one coin each, and two coins to Pirate 2 in order to have one hundred votes altogether, including her own, leaving no coins left over for herself.

For this reason, Pirate 200 will have a successful proposal, since she no longer needs to spend two coins for one vote, as the proposal of Pirate 199 has one hundred pirates getting zero. So Pirate 200 can get 100 votes by proposing one coin to everyone who would get zero from 199, plus her own vote, for a majority of 101 votes.

200 pirates: 0 0 1 1 0 1 0 1 0 1 0 $\dots$ 0 1 0 1 1 0

The reader is encouraged to investigate further to see how the pattern continues. It is a fun problem to work out! What emerges is the phenomenon by which longer and longer sequences of pirates in a row find themselves unable to make a winning proposal, and then suddenly a pirate is able to survive by counting on their votes.

It is very interesting also to work out what happens when there is a very small number of coins. For example, if there is only one gold coin, then already Pirate 4 is unable to make a passing proposal, since she can buy only one other vote, and with her own this will make only two votes, falling short of a majority. With only one coin, Pirate 5 will survive by buying a vote from Pirate 1 and counting on the vote of Pirate 4 and her own vote, for a majority.

Even the case of zero coins is interesting to think about! In this case, there is no gold to distribute, and so the voting is really just about whether the pirate should walk the plank or not. If only one pirate, she will live. Pirate 2 will die, since Pirate 1 will vote against. But for that reason, Pirate 2 will vote in favor of Pirate 3, who will live. The pattern that emerges is:

lives, dies, lives, dies, dies, dies, lives, dies, dies, dies, dies, dies, dies, dies, lives, ….

After each successful proposal, where the pirates lives, for subsequently larger numbers of pirates, there must be many deaths in a row in order for the proposal to count on enough votes. So after each “lives” in the pattern, you have to double the length with many “dies” in a row, before there will be enough votes to support the next pirate who lives.

See also the Pirate Game entry on Wikipedia, which is a slightly different formulation of the puzzle, since tie-votes are effectively counted as success in that variation. For this reason, the outcomes are different in that variation. I prefer the strict-majority variation, since I find it interesting that one must sometimes use two gold coins to gain the majority, and also because the death of Pirate 2 arrives right away in an interesting way, rather than having to wait for 200 or more pirates as with the plurality version.

Another (inessential) difference in presentation is that in the other version of the puzzle, they have the captain on the plank first, and then always the highest-ranking pirate making the proposal, rather than the lowest-ranking pirate. This corresponds simply to inverting the ranking, and so it doesn’t change the results.

The puzzle appears to have been around for some time, but I am unsure of the exact provenance. Ian Stewart wrote a popular 1998 article for Scientific American analyzing the patterns that arise when the number of pirates is large in comparison with the number of gold pieces.