Bernard Leong » Research & Pedagogy

A Mathematical Solution to Eddington’s Controversy Problem

Bernard Leong — Sat, 23 May 2009 08:22:28 +0000

Sometime back, while reading Fazlollah M. Reza’s “An Introduction to Information Theory“, I revisited a solution that I have sketched out years back on the Eddington’s controvery problem. This problem is interesting to me because it exemplifies the type of confusion that existed in probability prior to the introduction of set theory considerations. Eddington is the same astrophysicist who did the solar eclipse experiment to demonstrate the prediction of light bending using Einstein’s general relativity. I thought I should share the solution which I co-solved with Yen Lee, an old friend from Cambridge who’s now a postdoc in Purdue University. To quote the problem, “If A, B, C, D each speaks the truth 1 in 3 times (independently), and A affirms that B denies that C delcares that D is a liar, what’s the probability that D was speaking the truth?” Historically, this problem was examined by M. Gardner in an article entitled “Brain Teasers that involve Formal Logic” and to everyone’s surprise, some theoretical physicists and mathematicians are embroiled in getting the correct the number of the solution. So, I will discuss the problem in detail, giving my solution to the problem and explain why Eddington’s answer of 25/71 was greeted with so much protests from the thinkers of the time. (Warning: If you are not a theoretical physicist or mathematician, you will be inundated by a plethora of mathematical symbols, hence you are warned before proceeding to the interesting parts).

To simplify the problem without bringing up the controversy, we make the following assumptions (see the last section of the blog article):

All four men made statements.
A, B, and C each made a statement that either aﬃrmed or denied the statement that follows.
A lying affirmation is taken to be a denial and a lying denial is taken to be an affirmation.

Let a, b, c and d be the event when A, B, C and D tells the truth respectively, and let

p be the event that D lied,
q be the event that C declared that D lied,
r be the event that B denied that C declared that D lied,
s be the event that A affirmed that B denied that C declared that D lied.

We are told that A, B, C, D each tel the truth (denote truth as T and false as F) with a probability of 1/3. Thus, we get the following:

where == indicates that two logical statements are equivalent.We set up the following logical operations for the problem. The event p is the event that D lied, i.e that d is false, so

and we use the logical operators && means AND, || means OR and ^ means XOR.Let’s work out q, which is the event that C declared that D lies, i.e. q is the event that C declared that “p is true”. This happens if either “p is true and C tells the truth” or “p is false and C lies”, hence, we get the following:

Then we continue with r which is the event that B denied statement q, i.e. r is the event that B said that “q is false”. This happens if either “q is false and B tells the truth” or “q is true and B lies”, hence

Finally, we evaluate s which is the event that A affirmed the statement r, we run thru the same calculation and find:

where we use the identity (x == y) == !(x^y). Since the XOR operation is commutative and associative, we can just write:

You can refer to this truth table which I have painstakingly constructed:

With the logical relation s, we proceed to ﬁnd the probability that D told the truth (event d) given what A said (event s):

and we calculate the denominator by setting up the truth table: Therefore we calculate the answer:

and yes, my solution is slightly different from Eddington’s solution of 25/71, which I will explain how he come to that solution.

How Eddington got his solution

In this section, we describe Eddington’s original solution to the problem [Eddington (1935)]. Eddington interpreted the question as the following: D has made a statement X, and A stated that B denied that C contradicted X, hence what is the probability that the statement X is true?

There are only two premises made in Eddington’s view: (1) D’s statement, X and (2) A’s statement that “B denied that C contradicted X”. He assumed that we do not know whether B and C made any relevant statements. An example will be if B truthfully denied that C contradicted X, there is no reason that C aﬃrmed X. By this reasoning, the only cases which are inconsistent with what’s known are: (1) A truths, B lies, C truths, D truths and (2) A truths, lies, C lies, D lies.

If A is lying, we don’t know what B said and if A and B both spoke the truth, we do not know what C said. Since case 1 and 2 occur twice and eight times out of 81 occasions respectively, therefore D’s 27 truths and 54 lies are reduced to 25 truths and 46 lies. Therefore the probability becomes 25/(25 + 46) = 25/71
≈ 0.35. Hence that’s Eddington’s answer to the problem.

A Historical Depiction of the Problem from Gardner’s paper “Brain Teasers that involve Formal Logic”

I thought that it may be good to quote the comments on the Eddington problem here from Gardner’s paper (note that while trying to write out the solution, I have read all the papers cited below by spending a few days digging in old stacks library of journals in Cambridge and, yes, they have Nature papers dated back 1900s):

“Eddington’s answer of 25/71 was greeted by howls of protest from his readers touching off an amusing controversy that was never decisively resolved. The English astronomer Herbert Dingle, reviewing Eddington’s book in Nature (March 23, 1935) dismissed the problem as meaningless and symptomatic of Eddington’s confused thinking about probability. Theodore Sterne, an American physicist, replied (Nature, June 29, 1935) that the problem was not meaningless but lacked sufficient data for a solution. Dingle responded (nature, Sept 14, 1935) by contending that if one granted Sterne’s approach, there were enough data to reach a solution of exactly 1/3. Eddington then re-entered the fray with a paper entitled “The Problem of A, B, C and D” (Math. Gaz., October 1935) in which he explained in detail how he calculated the answer.

The difficulty lies chiefly in deciding exactly how to interpret Eddington’s statement of the problem. If B is truthful in making his denial, are we justified in assuming that C said that D spoke the truth? Eddington thought not. Similarly, if A is lying, can we then be sure that B and C said anything at all? Fortunately, we can side-step all these verbal difficulties by making (as Eddington did not) the following assumptions:

All four men made statements.
A, B, and C each made a state that either aﬃrmed or denied the statement that follows.
A lying aﬃrmation is taken to be a denial and a lying denial is taken to be an aﬃrmation. “

Related Links:
[1] David Stirzake, Elementary Probability (presented a solution using Set theory).

Buffon’s Needle and Noodle Problem

Bernard Leong — Fri, 06 Feb 2009 07:00:09 +0000

Recently, while flipping through a research paper made me think about the estimation of π (which we know is 3.141… ). Here is an interesting problem which is associated with a method to estimate π. A needle of length a is thrown on to the plane covered with equally spaced parallel lines with seperation b. What is the probability that the needle will cross a line? How can this be extended to a random curve of length A? I thought I might just share a simple mathematical solution for both the Buffon’s Needle and Buffon’s Noodle problem that I have worked out years back on this problem during my PhD years.

The only way to do this is to illlustrate with a figure to show how a needle might land so that it just touches one of the parallels. Making use of the symmetry of the system, we only need to look at one half sgement in the diagram.

The vertical position of the needle does not matter because shifting it up or down leaves the state of crossing a vertical line unchanged. The only factors which will matter will be: (a) The needle’s angle with the horizontal and (b) the distance of the centre of the needle to the nearest parallel.The centre P is equally likely to fall anywhere between the parallels, which is essentially, the assumption of uniform distribution and for a fixed angle θ, the chance that the line crosses a parallel will be

because the line crosses a parallel if the centre falls within x-units of either parallel. (See the dashed lines in the figure).

The angle θ is thought to be uniforrmly distributed from 0 to (π/2) because of twirling. The crossings that happen for θ also work for π – θ.Now we proceed to compute the probability that θ is between 0 and π/2. We need the mean value of (2x/b) for x = a/2 cos θ and it works out to be (a/b cos θ). Then integrating to find the average

and it gives the following relation

The above relation suggests a geometrical relation between a pure chance experiment and the value of the transcendental number, π, i.e. we can estimate π by tossing a long needle on a grid of squares.

We can extend the Buffon’s needle problem to Buffon’s noodle (a random curve of length A). Suppose we cut up the noodle into several needles, i.e.

Hence, tossing the noodle is like dropping a couple of needles at one go. The equation earlier for Buffon’s needle can be extended to

Therefore, on average, a random curve of length A is expected to intersect the lines 2A/(π b) times.

Related Links:

Firing of BPL Football Managers and Maximal Efficiency

Bernard Leong — Mon, 28 Jan 2008 18:01:48 +0000

Till today, since the beginning of the Barclays Premiere League, eight out of twenty football managers are already replaced. Sometime back, together with two economists (Toke Aidt and Daniel Sgroi) and another fellow physicist in Cambridge (Bill Saslaw), we co-wrote a research paper entitled “Power Laws, Maximal Efficiency and the Turnover of Sports Managers” (which was published in Physica A and got some press attention in the UK after the World Cup 2006). Yes, I have done the “freaknomics” type research by looking at the phenomena of sports managers being fired for all sorts of reasons. It was the research that I have done on the football clubs that eventually led me to watching the Premiere League football till today. It was one of the most enjoyable and multi-disciplinary research collaborations for me. In this short blog entry, I will explain the research work with a “freaknomics” tradition and explain why the current firings will continue to happen because of market efficiency in the Premiere League.

Let me start off with a simple situation that you face. If you believe that life is random, any six sided dice you roll will give you a random probability of 1/6, provided that it is not a loaded dice. Consider the following scenario, if you are into sports betting, is the probability of betting in every football match just 1/3 (since there are only three outcomes – win, lose and draw)? An avid football fan will tell me that there are many factors affecting a football match, for example, the manager’s ability, the injuries sustained by the football team, the media reports which goes both ways like speculating this manager will be sacked if he wins or loses a game, or even two big clubs playing is different from a big vs a mediocre one.

It turns out that a combination of ideas from physics and economics provide an interesting explanation to why the football managers are getting easier to fire in the recent days. The idea stems from the science of complex systems in physics. In this framework, you look at the macroscopic properties (for e.g. the distribution of the tenure of football managers) that arises from relatively simple microscopic interactions (for e.g. two football clubs playing against each other with a 1/3 win, lose or draw and taking the scores of 3 for win, 1 for draw and 0 for loss). Unlike neo-classical economic theory that is built on deductive analytical methods, for e.g., game theory, the theory of complex systems might be use to ask the question of maximal efficiency in any economic system.

So why did we pick the tenures of football managers, if you may ask? There is a background story behind this. Originally, we intend to use the tenures of CEO in Fortune 500 as a dataset to demonstrate our proof of concept. The idea is to demonstrate that the firing of CEOs result in the economic system trying to move towards maximal economic efficiency. However, the problem is that for each data point of a CEO from a certain industry, say IT, I have to pay 5 Singapore dollars for each datapoint. Since we academics are poor people, we have to rely on smarter ways to find data.

The inspiration came to me during a discussion with Toke, Daniel and Bill over a coffee break. Since we are thinking of mathematically modelling CEOs and the way that they got fired, the scenarios can be extremely complex, given that CEOs can lose jobs due to mergers, or by many other factors. There must be a dataset out there which is free and easily obtainable using Google search. For some strange reason, I was thinking of football managers and it was likely because I recalled some friends of mine were watching the champions league game that night. Hence it may be simpler to pick something simpler where the job is determined by win, lose or draw like any simple constructed game. The truth of the matter is that we picked football managers because we thought that it would be easy to gather the data. So, being a young and brash scientist, I told the rest that I will spend the night using Google to collect the tenures of the managers for the top 10 Premiere league clubs and do a plot. If the logarithmic plot is a power law, then we can account using the concept of self organized criticality in the physics of complex systems. So, thanks to those football fanatics who have collected the tenure years of the managers who managed their favourite football clubs, I am able to collect about 200 datapoints and did a first approximate plot and not surprisingly, I got a power law (see the graph below, which I used a simple model vs the actual results from the league, and there is a matching of theory and reality).

Of course, 200 datapoints are too small for the type of idea which I am pursuing. So, over the summer, we got a student intern who helped us to collect all the manager tenure dates and football statistics of all the leagues in Europe and the total dataset is about 7183 managers (datapoints). Instead of telling you the complete story, I will focus on the part which will tell you why we are seeing so much firings recently in the Premiere league.

The first thing is that if we splice, cut or mix the datasets of football managers’ tenure in the different leagues, the power laws are consistent throughout and we can measure how volatile a system in firing managers using the gradient of the power law. For example, I can split the datapoints in the seperate leagues – BPL, Serie A, Prima Liga and Bundesliga, the power law will still turn up. We did some statistical tests to examine whether we are getting a real straight line or a curve. It turns out that the French league did not satisfy a power law unless you start from counting the tenure from 2 years onwards. Well, it did correspond to reality, because the French has a labor law that states that you cannot fire someone unless they have worked for you two years. Hence most French managers don’t get fired so quickly after two years. We also tried on other types of sports, for e.g. American football and baseball. That helps us to feel that our idea is pretty robust. So, to cut the long story short, the Premiere League has an exponent of -2.23 (before 2002) which means that they don’t fire managers that often, while the Spanish La Liga and the Italian Serie A leagues, the exponents were far higher like -2.8. So, we started examining why that is the case.

It turns out that European managers are more mobile as compared to the English counterparts. Most European managers spent two to three years before they got fired for poor performances, and they are flexible in moving to another league easily without much problem. A lot of money has been poured into these leagues and the fans demand performance, and hence if a manager starts losing 5 to 10 games in a row, they got booted out by their presidents with the advice of the director of football, a system which is not common in the Premiere League until recently.

That comes to my favourite part, with more money into the Spanish and Italian leagues, you find that the better clubs will be better in getting better players. The reason why we see a power law is because in such zero sum games, there is a differentiation of good and bad, i.e. the rich gets richer and the poor gets poorer. A lot of small clubs will have more firings than the big clubs, because their performance are consistently bad.

What are all these results telling us? Here is a way to think about the problem. The interactions of the managers within the leagues and the external forces have built up till the point when it hits the critical mass. That pushed the system into a state which the physicists call it self-organized criticality. Avalanches and earthquakes are examples of systems which exhibit this kind of behaviour once they break a certain threshold. So, what does this mean to economics? The answer is simple. Let me use a result we observed from the data. Suppose you take the data for the English managers and cut it into different eras (1890-1920, 1920-1940, 1940-1960, …) , the power law only starts to manifest when it reaches 1960s. So, we went back to the archives and found two interesting things which happened that caused the distribution to move to a power law. The first is that football matches have moved from the stadium to the TV screen, allowing more people to watch the game and indirectly contribute to the money of the game. The second event which is more significant, is that the wages of football players have gone uncapped. Before 1960s, the maximum wage of a English football player is 20 pounds, but now, it went uncapped. Hence, the competition for talent has led to the system moving towards a maximal efficient state. The best managers and football clubs will start to pay higher for talent. With better talent, they win better games with the correct management. Hence the power law starts to manifest, because the football league has reached a “critical state”.

In fact, we predicted a few years back, that if the money into the Premiere League clubs increases, the power law will get steeper. Recently, I added the datapoints in the past few years back to my plot and found that the exponent has increased. Interestingly, the BPL is starting to share similar exponents close to her European counterparts. So, what are we seeing in the Premiere league now is that with more investments, there is a demand for better results, and as a result, we start to see the distinction between good and bad managers and hence, in turn, the power law is a demonstration of how efficient the system is evolved.

One interesting feature is that these results cannot be used to point the accusing finger at any manager’s current performance given that the results are purely statistical. Hence, we can’t apply them to one individual. However, we could use the results to calculate the probability of a manager’s survival in the future.

References:

Toke Aidt, Bernard Leong, William Saslaw and Daniel Sgroi, “A power law distribution for the tenures of Sports Managers”, Physica A, 370(2):697-703.

The Birthday Problem

Bernard Leong — Mon, 15 Oct 2007 07:58:28 +0000

Three days ago, I discovered that one of my colleagues from Singapore Angle shares the same birthday with me. It reminded me of this interesting first year Cambridge undergraduate mathematical problem: What is the least number of persons required if the probability exceeds 0.5 that two or more persons have the same birthday (excluding the year)? So, I will offer the solution to the birthday problem (on my birthday, of course) and examine some interesting implications about the solution to this problem.

Let me start with the assumptions to solve this problem:

We exclude the year of birth from the date of birth in the problem. For example, both John and Harry share the same birthday on the date 28 March.
We omit Februrary 29 as a possible birthday. This is easily generalized if you want to include it. Please also omit the year from the birthday as it will complicate the problem.
We assume that there are 365 days for an Earth year. Let N be the number of days where you can get a birthday, and in the case that we are living on Earth, N = 365 days, and r be the number of individuals, say r = 23, means that there are 23 individuals.

We want to compute the probability of nobody sharing similar birthdays, P (A), hence it is obvious that the probability of getting at least one pair of birthdays, P (B), is just by taking the complement P (B) = [1 − P (A)]. Hence, for the first person, there are N days to choose from the year, and for the second person, there are (N − 1) days to pick from the year so that he or she does not share the same birthday as the first person. By induction, we conclude that for the r-th individual, he or she only has (N − r + 1) days to pick from.

Therefor the number of ways for no matching birthdays for r individuals are: N (N − 1) . . . (N − r + 1)

To work out the probability, we need the total sample space, we need the number of ways r people can have their birthdates without restriction. There are N ways for each person. Using the multiplication principle, the total number of ways that the birthdays can be assigned to r individuals will be: N to the power of r (N^r).

We proceed to compute P (A), the probability of r individuals not sharing the same birthdays:

Therefore the probability of finding at least one pair of birthdays, P (B) is

So, we solve the problem. Knowing the solution is not enough, but understanding the implication of the solution is what matters. So, let’s try to put some numbers into the equations. Let’s take N = 365 days and work out a table for different values of r (refer to table 1) .

If you realize by now, the more people present in the bar, the more likely that you can find someone with the same birthday with you (excluding the year of course). It sounds counter-intuitive, but the solution is true. I thought that since there are about 50 over social-political bloggers in Singapore, the chance of finding someone with the same birthday is definitely greater than 0.5. Guess what, I found one same to mine just right next to my doorstep. Lastly, I used to tell this to my female students in Cambridge. If there is a guy coming to them making such a bet (usually a pub has about 100 people), don’t take the bet.