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In one of the stories I'm designing, in Middle Age time, the main character, Mr. P, is a good mathematician serving the King and other lords in different works (army numbers and suppliers counts, ballistic calculations, civil and militar buildings design, etc.).

He is very good making precise calculations.

The problem is that another character, Mr. E, not evil but deceiver after all, tries to get over him.

Beyond his traps, in order to get choosen for the contracts, Mr. E offer always his results earlier than Mr. P.

To get these response times, Mr. E always makes estimations and he adds/substract a safe margin, depending the situation.

Mr. P knows that most of the time Mr. E gets good enough results. For that, he wants to challenge Mr. E in a public meeting. His idea is to link some number related questions, for example:

  1. How many people there is in the meeting/party?
  2. How many houses can be found in all the kingdom?
  3. ... even how many stars can be observed in the sky.

Mr. E will find a good answer for everything. At that point, I need Mr. P to ask Mr. E something (known in this age) that is extremely difficult/not possible to estimate but could be calculated by Mr. P with his calculations.

The problem is that I cannot imagine something calculable and not estimable, that could be demostrated/validated by audience.

If this is not possible, I could settle for a calculation which estimation would be a long way from the real result, and how to demonstrate quickly to any audience.

TL;DR: I need something extremely difficult/not possible to estimate which can be well calculated using equations and able to be checked (in a Middle age scenario, so advanced physics and that stuff is not useful).

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  • $\begingroup$ In the examples you are giving you refer to how much of something exists. He could also ask for how much something will cost and how long something will take, which seem to be more applicable tasks in the job. $\endgroup$ – Backup Plan Sep 26 at 10:24
  • $\begingroup$ Can you give a rough time estimate? Around the 10th century the compass were considered 'Black Magic' and measuring time was basically not possible while moving. Some hundred years later compass where well know and they at least had hourglass and early mechanical clocks. $\endgroup$ – PSquall Sep 26 at 11:23
  • $\begingroup$ A rough time estimate for Mr. E to make their estimations? Let's suppose he elaborates his answer to the contractors in two days, meanwhile Mr. P takes a five days lapse to calculate the perfect solution. $\endgroup$ – Gerifalte Sep 26 at 13:42
  • $\begingroup$ Do you need just a math problem that is very challenging for the time, like Cubic equation, or the problem with a solution that should be well understood be the lay people? $\endgroup$ – Alexander Sep 26 at 17:13
  • $\begingroup$ I see what you are asking, @Alexander , and yes, I should include the need to show/demostrate the real solution in my question. Thank you! $\endgroup$ – Gerifalte Sep 26 at 17:49

18 Answers 18

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Date of the next lunar/solar eclipse. See “Antikythera mechanism”: https://en.m.wikipedia.org/wiki/Antikythera_mechanism

Eclipses follow such a complex pattern, they cannot be estimated from previous events, but as the Greek mechanism shows, they could be calculated.

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    $\begingroup$ This is a fine example of 'calculable, but not estimable." Of course, determining who's most correct is a long, slow process almost all the time... $\endgroup$ – Zeiss Ikon Sep 26 at 12:06
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    $\begingroup$ Mr. P can offer the challenge when he happens to know an eclipse is coming in a few days. $\endgroup$ – SRM - Reinstate Monica Sep 26 at 12:11
  • $\begingroup$ I really like that one. I'm going to leave a little time to people to give another one. The points I like are the easy to fail the estimation and the easy to demostrate. Thank you! $\endgroup$ – Gerifalte Sep 26 at 13:14
  • $\begingroup$ Couldn't Mr. E just say "10000 years from now, give or take 10000 years?" $\endgroup$ – Michael Seifert Sep 26 at 21:20
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    $\begingroup$ @michaelseifert sure, but that’s as pointless as a non answer, and everyone present will know it. $\endgroup$ – SRM - Reinstate Monica Sep 26 at 21:34
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Estimate exponential growth using grains of rice and a chessboard.

The story goes:

The ruler or India was so pleased with one of his palace wise men, who had invented the game of chess, that he offered this wise man a reward of his own choosing and he said to the man: “Name your reward!”

The man responded: “Oh emperor, my wishes are simple. I only wish for this:

-Give me one grain of rice for the first square of the chessboard, two grains for the next square, four for the next, eight for the next and so on for all 64 squares, with each square having double the number of grains as the square before.“

The emperor agreed, amazed that the man had asked for such a small reward – or so he thought. After a week, his treasurer came back and informed him that the reward would add up to an astronomical sum, far greater than all the rice that could conceivably be produced in many many centuries!

The total number of grains would in the Emperor in the story's estimation be quite manageable - but with precise calculation turns out to be: 18,446,744,073,709,551,615 - purported to be sufficient rice to cover the whole of India's landmass a meter deep in rice.

The math is straightforward, it involves the addition of a series of numbers from 1, then 2, then 22, then 23, 24... and so on up to 263. (Ie. the first square = 20 aka 1).

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    $\begingroup$ In actuality, there are 204 squares on a chessboard. There might only be 64 1x1 squares, but there's still the 2x2s, 3x3s, 4x4s, and so on. $\endgroup$ – Draco18s Sep 26 at 22:02
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    $\begingroup$ @Draco18s Well then, you'd end up even more bloated on rice than the wise man or I would. +1 for being pedantic. $\endgroup$ – We are Monica. Sep 26 at 23:09
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    $\begingroup$ Depending on how exact the estimation needs to be to be accepted... this is not a good example. It can be easily seen, that the result must be $2^{64} - 1$. so use $16 * 10^{18}$ and add a safety margin for the 1024 vs 1000 progression to the 6th power. It's not unreasonable to get to a value close to 18e18. $\endgroup$ – MilConDoin Sep 27 at 22:10
  • $\begingroup$ The version I heard of this had the wise man lose his life for unwisely putting his ruler in such debt. $\endgroup$ – Michael Richardson Sep 27 at 22:15
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    $\begingroup$ In my retelling, the wise man was asked to hold the chessboard while they piled on the rice, and the greedy bastard suffocated in rice. $\endgroup$ – SRM - Reinstate Monica Sep 28 at 5:09
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Where a cannon ball will land. A problem of extreme importance in the Middle Ages, why Newton and Galileo were studying gravity, and (because of the scale involved) nearly impossible to estimate with the precision desired by commanders.

Because of the great mass of a cannon ball, the effect of wind and complicating factors of air resistance are significantly reduced.

The equations of motion are a calculation that provides an exact result. Because, as others have pointed out, the calculation is incomplete, there is a built-in error.

Since you've drawn a distinction between 'estimating' and 'calculating', I think estimation would involve spotter balloons and a network of relay flagmen to communicate where a first shot landed, and the spotters giving guidance to 'walk' the fire towards the desired target.

You can maybe see why estimation wasn't beloved by commanders. It required being the first to the field with enough lead time to set up such a network, the chain didn't cover a great deal of field, was easily disrupted by weather and enemy action, and was tremendously difficult to quickly move if the action was happening somewhere else.

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  • $\begingroup$ yep, ballistics was the first thing that came to my mind as not something that can be estimated. Navigation is another, you don't want to estimate that either. $\endgroup$ – Kilisi Sep 26 at 10:56
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    $\begingroup$ Ballistics can be estimated. In fact, they always are, when you're dealing with something like cannonballs. The concept of drag coefficient wasn't even understood in medieval times -- and computed fluid dynamics was beyond anyone's ability. This is the opposite of what's needed. $\endgroup$ – Zeiss Ikon Sep 26 at 12:03
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    $\begingroup$ I find ballistics easy to estimate with experience, what could be acceptable for Mr. E to have, according to his job and the context they live. Thanks anyway! $\endgroup$ – Gerifalte Sep 26 at 13:44
  • $\begingroup$ Until post- World-War One, there weren't any cannon or mortars with sufficient repeatability to get any sort of useful test data. Your Middle-Ages folks would have to develop all the statistical math of mean, Variance, etc. to get anything useful out of such tests. $\endgroup$ – Carl Witthoft Sep 26 at 14:53
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Decrypt an RSA-style Message

Although the specifics of RSA encryption and decryption - or even more generally public key cryptography didn't come around until the 1900s, the ideas of factorization go back to the Greeks.

Furthermore, RSA itself works on rather simple mathematics: multiplication and modulus operations. Wikipedia even has a simple example that you can do by hand yourself. These were definitely available in the middle ages. There's no reason why this wasn't done in the middle ages except that (a) it's tedious when done by hand, and (b) no one came up with it. If Mr. P is thoughtful and ahead of his time, there's no reason why he could not have come up with a scheme like this. As an aside, there were many attempts in the middle ages to hide messages (ways of folding paper, using secret codes, created locks and boxes, etc), so there was definitely a desire for this type of thing.

Anyway, Mr. P could explain clearly how the algorithm works, give the private key and encrypted message, and then ask Mr. E to uncover the original source. If Mr. E is not exactly accurate, then the result will be garbled text / a wrong number. This demands precision.

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    $\begingroup$ I actually asked a question on Crypto.SE about if RSA style asymetric encryption was possible earlier on. It could be relevant. $\endgroup$ – Captain Man Sep 26 at 20:07
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    $\begingroup$ +1 Cryptography has a major goal of making it so that "almost right" is not any better than "completely wrong." You get it completely right, or you are completely off. Of course, this also implies that the person on the other end is also extremely precise. $\endgroup$ – Cort Ammon - Reinstate Monica Sep 27 at 6:32
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I, the King, wish to share the Kingdom's wealth with the People. If the Kingdom's population keeps growing, how long before they collectively are richer than the Royal Family?

An estimate would say 'Probably 100 years'. An exact formula says never.

Stick with me here.

Let's say this is a verrrryy nice king. What goes around comes around- he shares his wealth with his people. For every sum of cash(for ease, we're going to call this amount $2C$) that comes into the kingdom, he takes $\frac{1}{2}C$ for himself. He then decides to group the kingdom by population and closeness to himself: every group has one more person than the one above it. He's on the top by himself, his Queen and heir are one below, his three knights below that, four nobles, etc., all the way down to his thousand peasant farmers. Everyone in each tier gets the cash divided up like this:

enter image description here

So, the fraction of C you get is one over two to the power of the number of people in your tier.

The rich stay rich and the poor stay poor, but everyone's grateful to the king because he gives each of them enough to sustain their own lifestyles, but critically not enough for anyone to move up or down. When you're born, you're shoved into the 'lowest bottom pile'. If someone dies higher up and you have a right to that space, it's yours and you get the cash. For this reason, this system can scale towards infinity.

His Royal Highness is also very intelligent. He knows money talks, and is aware of the meltdown that could occur if he suddenly became very unpopular and the people had more wealth than he did. Or he's an egomaniac. Either way, he wants his family to be #therichest.

With this system, what's the maximum population size you can have before the people have more wealth than the king?

An estimate would say "Well, given the current population growth, I'd guess 20 years your Highness?". However, a Medieval mathematician (Oresme) proved that this sequence converges to 2. Hence why I used $2C$ at the start. Read that proof. It's truly brilliant.

Finally, some simple math:

$Wealth_{King} = \frac{1}{2}$

$Wealth_{Queen} + Wealth_{Heir} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

$Wealth_{Royal\,Family} = 1$

$Wealth_{Infinite\,population} = 2-1 = 1$

$Wealth_{Infinite\,population} = Wealth_{Royal\,Family}$

So, as long as the population is less than infinite, the King can share the Kingdom's wealth with his people and his family will always be richer than everyone else put together. Neat, right?

I'm aware this is such a botched explaination of pretty much everything, so let me know if something's unclear and I'll try to clean it up in an edit.

Hope it helps!

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    $\begingroup$ upvote from me. Very creative! I think this also works because it's counter-intuitive. Until you sit down and do the calculations, there's no way you'd estimate the result. Also, as an aside, I feel like this may be a type of fallacy even nowadays in stock markets, taxation strategies, or other financial institutions across the known world, especially in developing countries $\endgroup$ – cegfault Sep 26 at 15:22
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    $\begingroup$ Counter-intuitive is the name of the game. I'm not sayingggg there's a parallel to be drawn with the real world, buuuuut... $\endgroup$ – mcRobusta Sep 26 at 15:43
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    $\begingroup$ Letting the multiplication factor aside, this reminds me of the fractional reserve banking $\endgroup$ – Adrian Colomitchi Sep 27 at 10:54
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    $\begingroup$ This basically boils down to a feudal pyramid scheme... $\endgroup$ – Darrel Hoffman Sep 27 at 13:35
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The Archimedes eureka problem.

Have a number of strangely shaped objects made out of different materials.

The challenge is to work out which of them is made of the densest material.

You can't estimate that as they are strangely shaped. You can't just weigh them as they are different volumes.

The solution is to weigh them, then sink them in water and see how much the water rises. This lets you work out the volume. You can then divide the weight by the volume for each to get the density.

This was how Archimedes worked out whether a crown was made of gold or not.

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The Sand Reckoner of Archimedes

Archimedes of Syracuse was a Greek mathematician who lived in the 3rd century before the common era. He was probably the greatest mathematician of the antiquity

Among many other things, he was interested in devising a notation for very large numbers. In order to present his suggestion for a system to represent very large numbers, he proposed the following problem:

How many grains of sand would fit inside a sphere as big as the Universe?

Archimedes of Syracuse, Psammites (The Sand Reckoner), 3rd century BCE

For the size of the universe, he used the heliocentric model of Aristarchus of Samos, and estimated (well, took a wild guess) that the sphere of fixed stars has a diameter of (what we would call today) about 2 light years. He then assumed that a sphere with a diameter of one Greek inch (about 19 mm, or 3/4 of an English inch) can fit 640,000,000 grains of sand. (He was interested in a system for representing very large numbers, after all.)

He finally reached the conclusion that the Universe could fit no more that what we would write today as 1063 grains of sand.

The interesting thing is that he actually devised a system for writing such a large number. In the 3rd century before the common era.

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    $\begingroup$ Kinda hard to prove who is right, though, especially if the audience is uninterested in the working. $\endgroup$ – Starfish Prime Sep 26 at 10:36
  • $\begingroup$ @StarfishPrime: The thrust of the research was inventing a system for naming and writing large numbers. It is not that hard to prove that multiplication works. The problem is quite well suited to the Middle Ages... $\endgroup$ – AlexP Sep 26 at 11:42
  • $\begingroup$ I find the proposal quite interesting, but in fact, difficult to make clear to the audience (lords focused in other things). Anyway, although the question was made in the Ancient Greek, the way to check was difficult and mistaken in Archimedes time and also in the Middle age. Thank you for the idea! $\endgroup$ – Gerifalte Sep 26 at 13:54
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    $\begingroup$ It's interesting to note also that ancient India was big on, well, big numbers. According to the Font of All Knowledge, a paro was 10 ^ 400000000000000000 infinities. $\endgroup$ – elemtilas Sep 26 at 18:12
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    $\begingroup$ @elemtilas: Funny thing is, that's neither larger nor smaller than one infinity. $\endgroup$ – AlexP Sep 26 at 18:25
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Length of a Coastline

This is a bit of an underhanded cheat, but it's one way you can guarantee a win for Mr. P. Ask for the length of a coastline around an island or lake, or from one port to another on a seashore, etc. Thing about this is that due to the Coastline Paradox, the length can be almost anything you want it to be depending on how closely you measure it. The more precise your measurement, the longer the coast. So Mr. P can have a number in mind, and due to his meticulous measurements, he would know exactly what granularity of measurements to use in order to reach that exact number. He might even calculate multiple such coast lengths in order to make Mr. E's guess as wrong as possible.

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The Birthday Paradox (simplified).

Have six jars, each with twenty numbered balls (the ten jars can hold differently coloured balls to make sorting simpler), 1 to 20.

Say we extract one ball from each jar, thus obtaining a collection of six balls.

How often will (at least) two of those balls share a number?

Solution:

Calculate the probability of this not happening. The first ball can be any number. The second ball must get one of the remaining 19 numbers, which will happen 19/20ths of the time. The third ball will come up right 18/20th of the time. The fourth will do so 17/20th of the times. The fifth and sixth, 16/20 and 15/20. So all of them will come up right with a chance of (19*18*17*16*15)/(20*20*20*20*20), which is 0.43605, or 43.605% of the time.

Therefore, 56% (more than half) of the time two balls among six will sport the same number.

In the original formulation, "how often in a class of 23 children at least two will share their birthday" (p1=1/365.2544) will also yield around 50%.

To prove who's right, they can design a chance game - extract six balls, bet that there will be two with the same number; if you're right, you get X times the bet. Both the mathematician and his opponent can "offer" a value for X, whoever offers the smallest value of X and manages to be ahead after 20 tries is the winner (the more the tries, the less luck enters into the matter).

For example, the mathematician offers 1.80 dollars to the dollar (which is 1/0.43). 43% of the times he will lose, 56% of the time he will win 180% of the bet; in the end he will have gained S*56%*180% = S*0.56*1.80 = a bit more than S. His opponent might offer double the stake, which will make him lose since 1.80 is less than 2.0; or he might offer one dollar and a half per dollar, which will make him lose on average 56%*150% = 16% of the time. After twenty tries, the chances of him being ahead are negligible (0.84 to the 20th power is about three per cent).

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    $\begingroup$ The Birthday Paradox is always baffling, even though its just probability. You would assume that the chance was always much much lower than it really is. Its also very easy to make a mistake when deriving the solution, which makes it that much better. $\endgroup$ – Shadowzee Sep 27 at 4:19
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Tides

Anything that can be calculated can be estimated. However something with a sinusoidal function like the tides requires significant precision even on an estimate, if you're betting on ebb and you get flow, or you're betting on a high tide and you get low, you're in a reasonable amount of trouble. There's no slack for adding a margin for error, you're right or wrong.

This has an added complication in that time is also a fairly approximate matter in the period. So asking what time high tide is wouldn't work, but asking for the state of the tide at sunrise on the morning of a specific date should be good enough for anyone as it requires both the patterns for the movement of the sunrise as well as the tide.

Anything with which neither party has experience

The key to a good estimate is that you have to know a fair amount about the situation. You might know how much food an army needs for a week or how many cannonballs are required for a campaign. You might know how long it takes to paint a wall of any size and how much paint is required. You might know how many litres of water are in an arbitrary swimming pool or the fuel consumption of an average small car. But if you know nothing about the subject then it's not possible to make an estimate.

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  • $\begingroup$ But you can't calculate tides, even now. The tables are based on decades of observation, which allows us to predict the future based on location of the moon relative to both the local Earth point and the sun. $\endgroup$ – Carl Witthoft Sep 26 at 14:55
  • $\begingroup$ [grin] I think you mean "...then it's not possible to make [a good] estimate." I can estimate everything, I just can't accurately estimate everything. $\endgroup$ – J. Chris Compton Sep 26 at 20:57
  • $\begingroup$ Tides are very estimate-able for any given area. Next high tide is a little more than 12 hours after this high tide, and low tide is a little more than 12 hours after the last low tide. $\endgroup$ – ivanivan Sep 29 at 0:36
  • $\begingroup$ @ivanivan, so by that method, what is the state of the tide at sunrise in 3months? If you have an error around a minute a day you'll be totally wrong. $\endgroup$ – Separatrix Sep 29 at 6:39
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A simple, trivial, one would even say childish. Calculation of how much of a square foots the kingdom is. While knowing it's two dimensions.
Your Mr. P is John Napier. Who wrote a book called

Description of the Wonderful Rule of Logarithms

Using word Wonderful was just to smear in the face of Mr. E (of whom history forgot) that John found a way to calculate faster and with very high precision.

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A lot of good solutions have been given already, I think there are a couple that could be really nice because they can be shown to the public immediately and can be computed very accurately with relative little ease:

  • the period of a pendulum
  • the failure load of a truss structure
  • the biggest fish the fishermen in the next town over have caught based on a small sample (using bell curves, they could have a fisherman present to testify)
  • the volume of an odd shape (perhaps a vase that can be poured out to be measured?)
  • how much weight a small hot air balloon can lift (it's much lower than you would imagine)
  • the 1000th prime number
  • the time it takes for a barrel of water to empty if a hole of a specified size is drilled in the bottom
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Something concerning Benford's law.

Benford's law is an observation about the frequency distribution of leading digits in many real-life sets of numerical data.

In short, smaller digits appear to more frequently lead a number.

For example, if asked for an estimation about how many times 1 would be the leading digit in the whole 90000 tax records of the kingdom, or the population of each village etc, Mr E would probably estimate around 10000, which is wrong by Benford's law.

It is very easy to calculate, very easy to estimate wrong (if not 100% sure) and not very difficult to check (depending on Mr P's authority).

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  • $\begingroup$ It actually doesn't matter what base is used. That's why logarithms work. If you take all those numbers and multiply them all by the same value you still end up with more 1s than any other digit. Sure, all your 1s became 2s, but your 5s, 6s, 7s, 8s, and 9s, all became 1s. $\endgroup$ – Draco18s Sep 26 at 22:07
  • $\begingroup$ The base comment was reffering to the example I gave, which obviously uses the decimal system. Maybe I should delete it, it can be confusing. $\endgroup$ – liakoyras 87 Sep 27 at 6:21
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Anything noteworthy that E discovers and knows P is not working on.

This happened in our history - in 1535, Antonio Maria Fiore challenged Niccolò Tartaglia to a public contest. He gave problems that lead to solving depressed cubic equations, something which was considered difficult if not unsolvable and he (Fiore) found a solution to the problem.

Unfortunately for him, Tartaglia worked on the problem of different cubics, and given the prospect of public humiliation, he managed to find out the general solution before the contest. And won.

(this is somewhat simplified - there were more people and discoveries involved)

source: https://arxiv.org/abs/1308.2181

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  • $\begingroup$ Very good example of a real story. I liked so much the idea about twisting the duel, but I am clearly at Mr. P side. $\endgroup$ – Gerifalte Sep 27 at 11:39
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The Problem of Proving

Any test they would have to do needs a proof. Its nice that MR. P would calculate the exact right value and Mr. E guesses exactly the same value, both might be wrong and nobody would know it. So you need something that is not know, hard to estimate but possible to measure or reassure. Problem is that some calculations might require information that nobody has or can aquire and so the expected result is different. So everything with random elements like natural occurences are out. All we can do are inarguable facts, that can be meassured.

The Size of the Earth

Back in around 200 BC Eratosthenes calculated the circumference of the earth. He knew that a certain time in the year the sun would not cast a shadow in a well at noon in Syene. He knew the distance to Alexandria, that was directly north and asked a friend to measure the angle a shadow would cast from a stick. With that angle and the know distance, he was able to calculate the earths circumference.

Problem is, Mr. P or Mr. E could know that (although unlikely). On top of that, in the middle ages, noone could really measure the distance, so whatever.

But this value could also be used to calculate different values given the angle of shadows at their current position. For example what the angle another sticks shadow would show that is 100 miles further north or south from their current position. Or the distance to a to a random location north or south of their current position given the angle of the shadow there and their current location. If it is a place both dont know, but the distance can be measured, then Mr. P should be able to calculate the distance, but Mr. E should not be able to guess it.

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Digits of pi

There are many estimates for pi, starting from 22/7 and working from there. Your Mr E may even have memorised it to some number of digits.

But Mr P has access to the Newton-Raphson iterative method. (Newton is very much on the border between medieval and Renaissance; indeed he is one of the people who created the Renaissance. So we can justify his methods.) Calculating any precision of pi is therefore simply a matter of time.

And the absolute kicker? However accurately Mr P has learnt it, Mr E can better him by using that number as the starting seed for his own calculation, to immediately get a better value in one simple sum.

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    $\begingroup$ How would clueless public know whether numbers are correct? And why would they care? pi=3.14 is good enough for any practical purpose. $\endgroup$ – Zizy Archer Sep 27 at 11:16
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Anything involving mathematical chaos or general sensitivity to initial conditions will be difficult for the estimator. The trick is to find something which will be reasonably easy for the calculator, which means something where the initial conditions can be reproduced.

My favorite would be showing some rows of Rule 30 and asking for a row a few rows on.

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What if instead of having to calculate something very precisely, you proposed a MCQ (Multiple Choice Questions) instead? A MCQ is a list of questions with (most of the time) 4 proposed answers given for each.

The sky is :

  • A. Blue
  • B. Solid
  • C. Brown
  • D. Above our head

To make it even trickier, sometimes MCQs have malus points if you answer something wrong. That way you can't just randomly pick something when you don't know the correct answer.

You can't estimate the answers of a MCQ. It's either A, B, C or D. If Mr P is really that much greater than Mr E, he'll get only right answers, whereas Mr E will lose points by lacking accuracy. You could even ask each one of them to make a test for the other one; as long as he's better, Mr P will always win :)

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