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An unusual lottery result made the news recently: on October 1, 2022, the PCSO Grand Lotto in the Philippines, which draws six numbers from ${1}$ to ${55}$ at random, managed to draw the numbers ${9, 18, 27, 36, 45, 54}$ (though the balls were actually drawn in the order ${9, 45,36, 27, 18, 54}$). In other words, they drew exactly six multiples of nine from ${1}$ to ${55}$. In addition, a total of ${433}$ tickets were bought with this winning combination, whose owners then had to split the ${236}$ million peso jackpot (about ${4}$ million USD) among themselves. This raised enough suspicion that there were calls for an inquiry into the Philippine lottery system, including from the minority leader of the Senate.

Whenever an event like this happens, journalists often contact mathematicians to ask the question: “What are the odds of this happening?”, and in fact I myself received one such inquiry this time around. This is a number that is not too difficult to compute – in this case, the probability of the lottery producing the six numbers ${9, 18, 27, 35, 45, 54}$ in some order turn out to be ${1}$ in ${\binom{55}{6} = 28,989,675}$ – and such a number is often dutifully provided to such journalists, who in turn report it as some sort of quantitative demonstration of how remarkable the event was.

But on the previous draw of the same lottery, on September 28, 2022, the unremarkable sequence of numbers ${11, 26, 33, 45, 51, 55}$ were drawn (again in a different order), and no tickets ended up claiming the jackpot. The probability of the lottery producing the six numbers ${11, 26, 33, 45, 51, 55}$ is also ${1}$ in ${\binom{55}{6} = 28,989,675}$ – just as likely or as unlikely as the October 1 numbers ${9, 18, 27, 36, 45, 54}$. Indeed, the whole point of drawing the numbers randomly is to make each of the ${28,989,675}$ possible outcomes (whether they be “unusual” or “unremarkable”) equally likely. So why is it that the October 1 lottery attracted so much attention, but the September 28 lottery did not?

Part of the explanation surely lies in the unusually large number (${433}$) of lottery winners on October 1, but I will set that aspect of the story aside until the end of this post. The more general points that I want to make with these sorts of situations are:

1. The question “what are the odds of happening” is often easy to answer mathematically, but it is not the correct question to ask.
2. The question “what is the probability that an alternative hypothesis is the truth” is (one of) the correct questions to ask, but is very difficult to answer (it involves both mathematical and non-mathematical considerations).
3. The answer to the first question is one of the quantities needed to calculate the answer to the second, but it is far from the only such quantity. Most of the other quantities involved cannot be calculated exactly.
4. However, by making some educated guesses, one can still sometimes get a very rough gauge of which events are “more surprising” than others, in that they would lead to relatively higher answers to the second question.

To explain these points it is convenient to adopt the framework of Bayesian probability. In this framework, one imagines that there are competing hypotheses to explain the world, and that one assigns a probability to each such hypothesis representing one’s belief in the truth of that hypothesis. For simplicity, let us assume that there are just two competing hypotheses to be entertained: the null hypothesis ${H_0}$, and an alternative hypothesis ${H_1}$. For instance, in our lottery example, the two hypotheses might be:

• Null hypothesis ${H_0}$: The lottery is run in a completely fair and random fashion.
• Alternative hypothesis ${H_1}$: The lottery is rigged by some corrupt officials for their personal gain.

At any given point in time, a person would have a probability ${{\bf P}(H_0)}$ assigned to the null hypothesis, and a probability ${{\bf P}(H_1)}$ assigned to the alternative hypothesis; in this simplified model where there are only two hypotheses under consideration, these probabilities must add to one, but of course if there were additional hypotheses beyond these two then this would no longer be the case.

Bayesian probability does not provide a rule for calculating the initial (or prior) probabilities ${{\bf P}(H_0)}$, ${{\bf P}(H_1)}$ that one starts with; these may depend on the subjective experiences and biases of the person considering the hypothesis. For instance, one person might have quite a bit of prior faith in the lottery system, and assign the probabilities ${{\bf P}(H_0) = 0.99}$ and ${{\bf P}(H_1) = 0.01}$. Another person might have quite a bit of prior cynicism, and perhaps assign ${{\bf P}(H_0)=0.5}$ and ${{\bf P}(H_1)=0.5}$. One cannot use purely mathematical arguments to determine which of these two people is “correct” (or whether they are both “wrong”); it depends on subjective factors.

What Bayesian probability does do, however, is provide a rule to update these probabilities ${{\bf P}(H_0)}$, ${{\bf P}(H_1)}$ in view of new information ${E}$ to provide posterior probabilities ${{\bf P}(H_0|E)}$, ${{\bf P}(H_1|E)}$. In our example, the new information ${E}$ would be the fact that the October 1 lottery numbers were ${9, 18, 27, 36, 45, 54}$ (in some order). The update is given by the famous Bayes theorem

$\displaystyle {\bf P}(H_0|E) = \frac{{\bf P}(E|H_0) {\bf P}(H_0)}{{\bf P}(E)}; \quad {\bf P}(H_1|E) = \frac{{\bf P}(E|H_1) {\bf P}(H_1)}{{\bf P}(E)},$

where ${{\bf P}(E|H_0)}$ is the probability that the event ${E}$ would have occurred under the null hypothesis ${H_0}$, and ${{\bf P}(E|H_1)}$ is the probability that the event ${E}$ would have occurred under the alternative hypothesis ${H_1}$. Let us divide the second equation by the first to cancel the ${{\bf P}(E)}$ denominator, and obtain

$\displaystyle \frac{ {\bf P}(H_1|E) }{ {\bf P}(H_0|E) } = \frac{ {\bf P}(H_1) }{ {\bf P}(H_0) } \times \frac{ {\bf P}(E | H_1)}{{\bf P}(E | H_0)}. \ \ \ \ \ (1)$

One can interpret ${\frac{ {\bf P}(H_1) }{ {\bf P}(H_0) }}$ as the prior odds of the alternative hypothesis, and ${\frac{ {\bf P}(H_1|E) }{ {\bf P}(H_0|E) } }$ as the posterior odds of the alternative hypothesis. The identity (1) then says that in order to compute the posterior odds ${\frac{ {\bf P}(H_1|E) }{ {\bf P}(H_0|E) }}$ of the alternative hypothesis in light of the new information ${E}$, one needs to know three things:
1. The prior odds ${\frac{ {\bf P}(H_1) }{ {\bf P}(H_0) }}$ of the alternative hypothesis;
2. The probability ${\mathop{\bf P}(E|H_0)}$ that the event ${E}$ occurs under the null hypothesis ${H_0}$; and
3. The probability ${\mathop{\bf P}(E|H_1)}$ that the event ${E}$ occurs under the alternative hypothesis ${H_1}$.

As previously discussed, the prior odds ${\frac{ {\bf P}(H_1) }{ {\bf P}(H_0) }}$ of the alternative hypothesis are subjective and vary from person to person; in the example earlier, the person with substantial faith in the lottery may only give prior odds of ${\frac{0.01}{0.99} \approx 0.01}$ (99 to 1 against) of the alternative hypothesis, whereas the cynic might give odds of ${\frac{0.5}{0.5}=1}$ (even odds). The probability ${{\bf P}(E|H_0)}$ is the quantity that can often be calculated by straightforward mathematics; as discussed before, in this specific example we have

$\displaystyle \mathop{\bf P}(E|H_0) = \frac{1}{\binom{55}{6}} = \frac{1}{28,989,675}.$

But this still leaves one crucial quantity that is unknown: the probability ${{\bf P}(E|H_1)}$. This is incredibly difficult to compute, because it requires a precise theory for how events would play out under the alternative hypothesis ${H_1}$, and in particular is very sensitive as to what the alternative hypothesis ${H_1}$ actually is.

For instance, suppose we replace the alternative hypothesis ${H_1}$ by the following very specific (and somewhat bizarre) hypothesis:

• Alternative hypothesis ${H'_1}$: The lottery is rigged by a cult that worships the multiples of ${9}$, and views October 1 as their holiest day. On this day, they will manipulate the lottery to only select those balls that are multiples of ${9}$.

Under this alternative hypothesis ${H'_1}$, we have ${{\bf P}(E|H'_1)=1}$. So, when ${E}$ happens, the odds of this alternative hypothesis ${H'_1}$ will increase by the dramatic factor of ${\frac{{\bf P}(E|H'_1)}{{\bf P}(E|H_0)} = 28,989,675}$. So, for instance, someone who already was entertaining odds of ${\frac{0.01}{0.99}}$ of this hypothesis ${H'_1}$ would now have these odds multiply dramatically to ${\frac{0.01}{0.99} \times 28,989,675 \approx 290,000}$, so that the probability of ${H'_1}$ would have jumped from a mere ${1\%}$ to a staggering ${99.9997\%}$. This is about as strong a shift in belief as one could imagine. However, this hypothesis ${H'_1}$ is so specific and bizarre that one’s prior odds of this hypothesis would be nowhere near as large as ${\frac{0.01}{0.99}}$ (unless substantial prior evidence of this cult and its hold on the lottery system existed, of course). A more realistic prior odds for ${H'_1}$ would be something like ${\frac{10^{-10^{10}}}{1-10^{-10^{10}}}}$ – which is so miniscule that even multiplying it by a factor such as ${28,989,675}$ barely moves the needle.

Remark 1 The contrast between alternative hypothesis ${H_1}$ and alternative hypothesis ${H'_1}$ illustrates a common demagogical rhetorical technique when an advocate is trying to convince an audience of an alternative hypothesis, namely to use suggestive language (“`I’m just asking questions here”) rather than precise statements in order to leave the alternative hypothesis deliberately vague. In particular, the advocate may take advantage of the freedom to use a broad formulation of the hypothesis (such as ${H_1}$) in order to maximize the audience’s prior odds of the hypothesis, simultaneously with a very specific formulation of the hypothesis (such as ${H'_1}$) in order to maximize the probability of the actual event ${E}$ occuring under this hypothesis. (A related technique is to be deliberately vague about the hypothesized competency of some suspicious actor, so that this actor could be portrayed as being extraordinarily competent when convenient to do so, while simultaneously being portrayed as extraordinarily incompetent when that instead is the more useful hypothesis.) This can lead to wildly inaccurate Bayesian updates of this vague alternative hypothesis, and so precise formulation of such hypothesis is important if one is to approach a topic from anything remotely resembling a scientific approach. [EDIT: as pointed out to me by a reader, this technique is a Bayesian analogue of the motte and bailey fallacy.]

At the opposite extreme, consider instead the following hypothesis:

• Alternative hypothesis ${H''_1}$: The lottery is rigged by some corrupt officials, who on October 1 decide to randomly determine the winning numbers in advance, share these numbers with their collaborators, and then manipulate the lottery to choose those numbers that they selected.

If these corrupt officials are indeed choosing their predetermined winning numbers randomly, then the probability ${{\bf P}(E|H''_1)}$ would in fact be just the same probability ${\frac{1}{\binom{55}{6}} = \frac{1}{28,989,675}}$ as ${{\bf P}(E|H_0)}$, and in this case the seemingly unusual event ${E}$ would in fact have no effect on the odds of the alternative hypothesis, because it was just as unlikely for the alternative hypothesis to generate this multiples-of-nine pattern as for the null hypothesis to. In fact, one would imagine that these corrupt officials would avoid “suspicious” numbers, such as the multiples of ${9}$, and only choose numbers that look random, in which case ${{\bf P}(E|H''_1)}$ would in fact be less than ${{\bf P}(E|H_0)}$ and so the event ${E}$ would actually lower the odds of the alternative hypothesis in this case. (In fact, one can sometimes use this tendency of fraudsters to not generate truly random data as a statistical tool to detect such fraud; violations of Benford’s law for instance can be used in this fashion, though only in situations where the null hypothesis is expected to obey Benford’s law, as discussed in this previous blog post.)

Now let us consider a third alternative hypothesis:

• Alternative hypothesis ${H'''_1}$: On October 1, the lottery machine developed a fault and now only selects numbers that exhibit unusual patterns.

Setting aside the question of precisely what faulty mechanism could induce this sort of effect, it is not clear at all how to compute ${{\bf P}(E|H'''_1)}$ in this case. Using the principle of indifference as a crude rule of thumb, one might expect

$\displaystyle {\bf P}(E|H'''_1) \approx \frac{1}{\# \{ \hbox{unusual patterns}\}}$

where the denominator is the number of patterns among the possible ${\binom{55}{6}}$ lottery outcomes that are “unusual”. Among such patterns would presumably be the multiples-of-9 pattern ${9,18,27,36,45,54}$, but one could easily come up with other patterns that are equally “unusual”, such as consecutive strings such as ${11, 12, 13, 14, 15, 16}$, or the first few primes ${2, 3, 5, 7, 11, 13}$, or the first few squares ${1, 4, 9, 16, 25, 36}$, and so forth. How many such unusual patterns are there? This is too vague a question to answer with any degree of precision, but as one illustrative statistic, the Online Encyclopedia of Integer Sequences (OEIS) currently hosts about ${350,000}$ sequences. Not all of these would begin with six distinct numbers from ${1}$ to ${55}$, and several of these sequences might generate the same set of six numbers, but this does suggests that patterns that one would deem to be “unusual” could number in the thousands, tens of thousands, or more. Using this guess, we would then expect the event ${E}$ to boost the odds of this hypothesis ${H'''_1}$ by perhaps a thousandfold or so, which is moderately impressive. But subsequent information can counteract this effect. For instance, on October 3, the same lottery produced the numbers ${8, 10, 12, 14, 26, 51}$, which exhibit no unusual properties (no search results in the OEIS, for instance); if we denote this event by ${E'}$, then we have ${{\bf P}(E'|H'''_1) \approx 0}$ and so this new information ${E'}$ should drive the odds for this alternative hypothesis ${H'''_1}$ way down again.

Remark 2 This example demonstrates another demagogical rhetorical technique that one sometimes sees (particularly in political or other emotionally charged contexts), which is to cherry-pick the information presented to their audience by informing them of events ${E}$ which have a relatively high probability of occurring under their alternative hypothesis, but withholding information about other relevant events ${E'}$ that have a relatively low probability of occurring under their alternative hypothesis. When confronted with such new information ${E'}$, a common defense of a demogogue is to modify the alternative hypothesis ${H_1}$ to a more specific hypothesis ${H'_1}$ that can “explain” this information ${E'}$ (“Oh, clearly we heard about ${E'}$ because the conspiracy in fact extends to the additional organizations ${X, Y, Z}$ that reported ${E'}$“), taking advantage of the vagueness discussed in Remark 1.

Let us consider a superficially similar hypothesis:

• Alternative hypothesis ${H''''_1}$: On October 1, a divine being decided to send a sign to humanity by placing an unusual pattern in a lottery.

Here we (literally) stay agnostic on the prior odds of this hypothesis, and do not address the theological question of why a divine being should choose to use the medium of a lottery to send their signs. At first glance, the probability ${{\bf P}(E|H''''_1)}$ here should be similar to the probability ${{\bf P}(E|H'''_1)}$, and so perhaps one could use this event ${E}$ to improve the odds of the existence of a divine being by a factor of a thousand or so. But note carefully that the hypothesis ${H''''_1}$ did not specify which lottery the divine being chose to use. The PSCO Grand Lotto is just one of a dozen lotteries run by the Philippine Charity Sweepstakes Office (PCSO), and of course there are over a hundred other countries and thousands of states within these countries, each of which often run their own lotteries. Taking into account these thousands or tens of thousands of additional lotteries to choose from, the probability ${{\bf P}(E|H''''_1)}$ now drops by several orders of magnitude, and is now basically comparable to the probability ${{\bf P}(E|H_0)}$ coming from the null hypothesis. As such one does not expect the event ${E}$ to have a significant impact on the odds of the hypothesis ${H''''_1}$, despite the small-looking nature ${\frac{1}{28,989,675}}$ of the probability ${{\bf P}(E|H_0)}$.

In summary, we have failed to locate any alternative hypothesis ${H_1}$ which

1. Has some non-negligible prior odds of being true (and in particular is not excessively specific, as with hypothesis ${H'_1}$);
2. Has a significantly higher probability of producing the specific event ${E}$ than the null hypothesis; AND
3. Does not struggle to also produce other events ${E'}$ that have since been observed.
One needs all three of these factors to be present in order to significantly weaken the plausibility of the null hypothesis ${H_0}$; in the absence of these three factors, a moderately small numerical value of ${{\bf P}(E|H_0)}$, such as ${\frac{1}{28,989,675}}$ does not actually do much to affect this plausibility. In this case one needs to lay out a reasonably precise alternative hypothesis ${H_1}$ and make some actual educated guesses towards the competing probability ${{\bf P}(E|H_1)}$ before one can lead to further conclusions. However, if ${{\bf P}(E|H_0)}$ is insanely small, e.g., less than ${10^{-1000}}$, then the possibility of a previously overlooked alternative hypothesis ${H_1}$ becomes far more plausible; as per the famous quote of Arthur Conan Doyle’s Sherlock Holmes, “When you have eliminated all which is impossible, then whatever remains, however improbable, must be the truth.”

We now return to the fact that for this specific October 1 lottery, there were ${433}$ tickets that managed to select the winning numbers. Let us call this event ${F}$. In view of this additional information, we should now consider the ratio of the probabilities ${{\bf P}(E \& F|H_1)}$ and ${{\bf P}(E \& F|H_0)}$, rather than the ratio of the probabilities ${{\bf P}(E|H_1)}$ and ${{\bf P}(E|H_0)}$. If we augment the null hypothesis to

• Null hypothesis ${H'_0}$: The lottery is run in a completely fair and random fashion, and the purchasers of lottery tickets also select their numbers in a completely random fashion.

Then ${{\bf P}(E \& F|H'_0)}$ is indeed of the “insanely improbable” category mentioned previously. I was not able to get official numbers on how many tickets are purchased per lottery, but let us say for sake of argument that it is 1 million (the conclusion will not be extremely sensitive to this choice). Then the expected number of tickets that would have the winning numbers would be

$\displaystyle \frac{1 \hbox{ million}}{28,989,675} \approx 0.03$

(which is broadly consistent, by the way, with the jackpot being reached every ${30}$ draws or so), and standard probability theory suggests that the number of winners should now follow a Poisson distribution with this mean ${\lambda = 0.03}$. The probability of obtaining ${433}$ winners would now be

$\displaystyle {\bf P}(F|H'_0) = \frac{\lambda^{433} e^{-\lambda}}{433!} \approx 10^{-1600}$

and of course ${{\bf P}(E \& F|H'_0)}$ would be even smaller than this. So this clearly demands some sort of explanation. But in actuality, many purchasers of lottery tickets do not select their numbers completely randomly; they often have some “lucky” numbers (e.g., based on birthdays or other personally significant dates) that they prefer to use, or choose numbers according to a simple pattern rather than go to the trouble of trying to make them truly random. So if we modify the null hypothesis to

• Null hypothesis ${H''_0}$: The lottery is run in a completely fair and random fashion, but a significant fraction of the purchasers of lottery tickets only select “unusual” numbers.

then it can now become quite plausible that a highly unusual set of numbers such as ${9,18,27,36,45,54}$ could be selected by as many as ${433}$ purchasers of tickets; for instance, if ${10\%}$ of the 1 million ticket holders chose to select their numbers according to some sort of pattern, then only ${0.4\%}$ of those holders would have to pick ${9,18,27,36,45,54}$ in order for the event ${F}$ to hold (given ${E}$), and this is not extremely implausible. Given that this reasonable version of the null hypothesis already gives a plausible explanation for ${F}$, there does not seem to be a pressing need to locate an alternate hypothesis ${H_1}$ that gives some other explanation (cf. Occam’s razor). [UPDATE: Indeed, given the actual layout of the tickets of ths lottery, the numbers ${9,18,27,35,45,54}$ form a diagonal, and so all that is needed in order for the modified null hypothesis ${H''_0}$ to explain the event ${F}$ is to postulate that a significant fraction of ticket purchasers decided to lay out their numbers in a simple geometric pattern, such as a row or diagonal.]

Remark 3 In view of the above discussion, one can propose a systematic way to evaluate (in as objective a fashion as possible) rhetorical claims in which an advocate is presenting evidence to support some alternative hypothesis:
1. State the null hypothesis ${H_0}$ and the alternative hypothesis ${H_1}$ as precisely as possible. In particular, avoid conflating an extremely broad hypothesis (such as the hypothesis ${H_1}$ in our running example) with an extremely specific one (such as ${H'_1}$ in our example).
2. With the hypotheses precisely stated, give an honest estimate to the prior odds of this formulation of the alternative hypothesis.
3. Consider if all the relevant information ${E}$ (or at least a representative sample thereof) has been presented to you before proceeding further. If not, consider gathering more information ${E'}$ from further sources.
4. Estimate how likely the information ${E}$ was to have occurred under the null hypothesis.
5. Estimate how likely the information ${E}$ was to have occurred under the alternative hypothesis (using exactly the same wording of this hypothesis as you did in previous steps).
6. If the second estimate is significantly larger than the first, then you have cause to update your prior odds of this hypothesis (though if those prior odds were already vanishingly unlikely, this may not move the needle significantly). If not, the argument is unconvincing and no significant adjustment to the odds (except perhaps in a downwards direction) needs to be made.

As I have mentioned in some recent posts, I am interested in exploring unconventional modalities for presenting mathematics, for instance using media with high production value. One such recent example of this I saw was a presentation of the fundamental zero product property (or domain property) of the real numbers – namely, that $ab=0$ implies $a=0$ or $b=0$ for real numbers $a,b$ – expressed through the medium of German-language rap:

EDIT: and here is a lesson on fractions, expressed through the medium of a burger chain advertisement:

I’d be interested to know what further examples of this type are out there.

SECOND EDIT: The following two examples from Wired magazine are slightly more conventional in nature, but still worth mentioning, I think. Firstly, my colleague at UCLA, Amit Sahai, presents the concept of zero knowledge proofs at various levels of technicality:

As math educators, we often wish out loud that our students were more excited about mathematics. I finally came across a video that indicates what such a world might be like:

A friend of mine recently asked me for some suggestions for games or other activities for children that would help promote quantitative reasoning or mathematical skills, while remaining fun to play (i.e. more than just homework-type questions poorly disguised in game form).    The initial question was focused on computer games (and specifically, on iPhone apps), but I think the broader question would also be of interest.

I myself have not seriously played these sorts of games for years, so I could only come up with a few examples immediately: the game “Planarity“, and the game “Factory Balls” (and two sequels).   (Edit: Rubik’s cube and its countless cousins presumably qualify also, due to their implicit use of group theory.)  I am hopeful though that readers may be able to come up with more suggestions.

There is of course no shortage of “educational” games, computer-based or otherwise, available, but I think what I (and my friend) would be looking for here are games with production values comparable to other, less educational games, and for which the need for mathematical thinking arises naturally in the gameplay rather than being artificially inserted by fiat (e.g. “solve this equation to proceed”).  (Here I interpret “mathematical thinking” loosely, to include not just numerical or algebraic thinking, but also geometric, abstract, logical, probabilistic, etc.)

[Question for MathOverflow experts: would this type of question be suitable for crossposting there?   The requirement that such questions be “research-level” seems to suggest not.]

This will be a more frivolous post than usual, in part due to the holiday season.

I recently happened across the following video, which exploits a simple rhetorical trick that I had not seen before:

If nothing else, it’s a convincing (albeit unsubtle) demonstration that the English language is non-commutative (or perhaps non-associative); a linguistic analogue of the swindle, if you will.

Of course, the trick relies heavily on sentence fragments that negate or compare; I wonder if it is possible to achieve a comparable effect without using such fragments.

A related trick which I have seen (though I cannot recall any explicit examples right now; perhaps some readers know of some?) is to set up the verses of a song so that the last verse is identical to the first, but now has a completely distinct meaning (e.g. an ironic interpretation rather than a literal one) due to the context of the preceding verses.  The ultimate challenge would be to set up a Möbius song, in which each iteration of the song completely reverses the meaning of the next iterate (cf. this xkcd strip), but this may be beyond the capability of the English language.

On a related note: when I was a graduate student in Princeton, I recall John Conway (and another author whose name I forget) producing another light-hearted demonstration that the English language was highly non-commutative, by showing that if one takes the free group with 26 generators $a,b,\ldots,z$ and quotients out by all relations given by anagrams (e.g. $cat=act$) then the resulting group was commutative.    Unfortunately I was not able to locate this recreational mathematics paper of Conway (which also treated the French language, if I recall correctly); perhaps one of the readers knows of it?

I was recently at an international airport, trying to get from one end of a very long terminal to another.  It inspired in me the following simple maths puzzle, which I thought I would share here:

Suppose you are trying to get from one end A of a terminal to the other end B.  (For simplicity, assume the terminal is a one-dimensional line segment.)  Some portions of the terminal have moving walkways (in both directions); other portions do not.  Your walking speed is a constant $v$, but while on a walkway, it is boosted by the speed $u$ of the walkway for a net speed of $v+u$.  (Obviously, given a choice, one would only take those walkways that are going in the direction one wishes to travel in.)  Your objective is to get from A to B in the shortest time possible.

1. Suppose you need to pause for some period of time, say to tie your shoe.  Is it more efficient to do so while on a walkway, or off the walkway?  Assume the period of time required is the same in both cases.
2. Suppose you have a limited amount of energy available to run and increase your speed to a higher quantity $v'$ (or $v'+u$, if you are on a walkway).  Is it more efficient to run while on a walkway, or off the walkway?  Assume that the energy expenditure is the same in both cases.
3. Do the answers to the above questions change if one takes into account the various effects of special relativity?  (This is of course an academic question rather than a practical one.  But presumably it should be the time in the airport frame that one wants to minimise, not time in one’s personal frame.)

It is not too difficult to answer these questions on both a rigorous mathematical level and a physically intuitive level, but ideally one should be able to come up with a satisfying mathematical explanation that also corresponds well with one’s intuition.

[Update, Dec 11: Hints deleted, as they were based on an incorrect calculation of mine.]

Given that there has recently been a lot of discussion on this blog about this logic puzzle, I thought I would make a dedicated post for it (and move all the previous comments to this post). The text here is adapted from an earlier web page of mine from a few years back.

The puzzle has a number of formulations, but I will use this one:

There is an island upon which a tribe resides. The tribe consists of 1000 people, with various eye colours. Yet, their religion forbids them to know their own eye color, or even to discuss the topic; thus, each resident can (and does) see the eye colors of all other residents, but has no way of discovering his or her own (there are no reflective surfaces). If a tribesperson does discover his or her own eye color, then their religion compels them to commit ritual suicide at noon the following day in the village square for all to witness. All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).

[Added, Feb 15: for the purposes of this logic puzzle, “highly logical” means that any conclusion that can logically deduced from the information and observations available to an islander, will automatically be known to that islander.]

Of the 1000 islanders, it turns out that 100 of them have blue eyes and 900 of them have brown eyes, although the islanders are not initially aware of these statistics (each of them can of course only see 999 of the 1000 tribespeople).

One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.

One evening, he addresses the entire tribe to thank them for their hospitality.

However, not knowing the customs, the foreigner makes the mistake of mentioning eye color in his address, remarking “how unusual it is to see another blue-eyed person like myself in this region of the world”.

What effect, if anything, does this faux pas have on the tribe?

The interesting thing about this puzzle is that there are two quite plausible arguments here, which give opposing conclusions:

[Note: if you have not seen the puzzle before, I recommend thinking about it first before clicking ahead.]