Carnival of Mathematics: Inaugural Edition

I’ve gotten plenty of submissions that span the entire gamut of math-blogging: education, pure math, applied math, debunking bad math – it’s all there. Only the gender distribution could be made slightly more equal (and that’s an understatement). I’m linking to the posters in roughly increasing order of mathematical difficulty, but don’t let my opinions deter you from reading the posts closer to the bottom.

First, Denise of Let’s Play Math has a long collection of quotes exhorting people to study mathematics. George Washington, Martin Gardner, Henri Poincaré, Abraham Lincoln, Galileo Galilei, Bertrand Russell – they’re all there.

Barry Leiba of Staring at Empty Pages uses modular arithmetic to solve a digit puzzle. The puzzle is based on writing any positive integer – say 8293 – and jumbling its digits and subtracting: 8293 – 3982 = 4311. Then delete a nonzero digit. Based on the resulting number, it’s possible to tell which digit you deleted.

P. Sternberg of Discreet Math links to various companies producing non-orientable material, like Möbius shoes and Klein bottle wool hats.

Heath Raftery writes about a probability paradox. Two dice are rolled, of which one is a 4. Given that, what is the probability that the sum of the two dice is a 7? Without further qualification – for example, “At least one die is a 4″ – it’s obviously 1/6. And yet Heath’s professor maintained that the answer was 2/11.

Charles Daney of Science and Reason gives a brief history of algebra from the Middle Ages till the Renaissance. Whereas most accounts begin with a relatively modern mathematician, like Fermat or Euler, Charles begins far earlier, with Al-Khwarizmi, who invented the word “algebra” and after whom the word “algorithm” is named.

In the bad math department, Tyler DiPietro has a long rant fisking the arguments of Sal Cordova, an intelligent design proponent who makes dishonest arguments from information theory and computability theory in order to try refuting evolution.

JD2718 has a puzzle in plane geometry. A parallelogram is defined as a quadrilateral where two opposite pairs of sides are parallel. But there are other equivalent definitions, and other conditions that look like they define parallelograms but in fact don’t.

Science Pundit Javier Pazos tells an anecdote about John von Neumann and the bee problem. The bee problem involves two trains initially spaced 40 km apart and approaching each other at 20 km/h, and a bee that flies at 40 km/h, starting from one train, flying toward the other, turning around when it meets it, flying toward the first train, etc. How far does the bee fly until the two trains meet?

Suresh Venkatasubramanian of GeomBlog has a long primer for the game theory of author ordering. When publishing a scholarly paper, each contributor wants to be as close as possible to first author in order to get more credit; Suresh explains which ordering strategies are stable and which appear the most utilitarian.

He also notes that the theory of algorithms is undergoing fundamental changes as computing power reaches saturation. In the past, it was sufficient for researchers to say that the run time of a process is proportional to some simple function, like x^2; but now, the constant that accompanies that function is becoming increasingly important.

John Kemeny introduces spigot algorithms, which can be used to systematically generate digits of some irrational numbers, including e and pi.

Jeffrey Shalit of Recursivity writes about the prime game. The initial observation is that the decimal representation of every prime p can be reduced to one of 26 primes by scrubbing off digits. That leads to a more general investigation of the concept of subsequences.

Eric Kidd programs infinity into Haskell, in order to solve such expressions as dividing infinity by 2 or adding 1 to infinity. The trick is to teach Haskell about the finite natural numbers first, and then write into it something strong enough to code for infinite numbers.

David Eppstein of 0xDE writes about subgreedy algorithms for Egyptian fractions. An Egyptian fraction is a representation of a fraction p/q as 1/n1 + 1/n2 + … + 1/n(k). A greedy algorithm is one that tries getting the best-looking result in one step without regard for efficiency in later steps; a subgreedy algorithm is one that is almost greedy, but tends to get far better results in the long run.

He also writes about a local version of the central limit theorem, which differs from the regular theorem in that it says repeated distributions look approximately normal in a small neighborhood of 0 rather than globally; and about coloring tilings in such a way that on the one hand the colors display a regular pattern, just like the tiling, but on the other no two adjacent tiles have the same color.

And finally, Mikael Johansson has multiple great posts about algebraic topology and related subjects; choosing three was a fairly hard decision. Of the three that made it into this edition, the easiest is the Borsuk-Ulam theorem, which roughly states that any given time, there’s a pair of antipodal points on Earth with the same temperature.

Also, A for the Layman explains algebraic topology, beginning with simple definitions and ending with an overview of homological algebra. For the braver souls, there’s his post about carry bits and cohomology; group cohomology is an indispensable tool in algebra, and Mikael applies it to addition and multiplication modulo 10, or in other words carry digits.

The next edition will be posted in two weeks, on 2/23, on Good Math, Bad Math. Send submissions to Mark CC, or to me so that I’ll forward them to him.

On a final note, I should remind everyone I’m still looking for someone talented enough to make a decent logo or banner for the carnival.

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98 Responses to Carnival of Mathematics: Inaugural Edition

  1. [...] at Alon’s place, Abstract Nonsense, the first issue of the biweekly Carnival of Mathematics is [...]

  2. Mr. Person says:

    Didn’t see my post. Saving it for another day?

  3. [...] in Mathematics, Carnivals at 9:31 am by Denise The Carnival of Mathematics: Inaugural Edition is up and running at Abstract Nonsense. Alon writes: I’ve gotten plenty of submissions that span [...]

  4. Axel says:

    Hmm, Internal Server Error on “Killing Mind”. I can’t post a comment…

    The 1/6-solution for the probability “paradox” is wrong and 2/11 is the one and only correct answer without any ambiguity. It’s like the Monty Hall “paradox”.

    Heath Raftery is permanently arguing with additional pieces of information like “the first dice”, “the other dice” or “first appearance”. All of his probabilities like the “1 in 6 chance of guessing the value of the other dice” are based on this idea. He states: “The dice are independent objects and their outcomes are determined completely without regard of each other.” That’s true but completely irrelevant for the probabilities. Outcomes are no probabilities! The latter depend on the knowledge you have.

    Just an example to make this clear: You have an urn with two black and two white balls. One ball is randomly chosen. In the first case, you don’t know the color of the ball, in the second case you do. Both experiments are “physically” identical. Now you choose a second ball. In the first case -no knowledge-, the probabilities for a black or white ball are still 1/2, whereas in the second case, they depend on the color of the first ball. If it was a white ball, the probabilities are 1/3 for white and 2/3 for black, if it was a black ball, 1/3 for a black and 2/3 for a white.

    Back to the two thrown dice. Without any piece of information there are 36 possible outcomes with equal probabilities. Therefore sometimes *none of the dice* shows a “4″ – that’s a fundamental difference to his idea that one die shows a “4″ and *now* the second is thrown or you *now* get the number of the second thrown die! This is another experiment. Therefore his solution is wrong.

    Back to the original experiment. You now know that one die shows a “4″, so only 11 of the equal outcomes are possible anymore. 2 of them have the sum 7. The probability is 2/11.

    If you don’t believe it, just run a simulation:

    1. Roll two dice
    2. Check if one die at least shows a “4″
    3. Only in this case, count if the sum is “7″
    4. Calculate the relative frequency of 3.

  5. rod. says:

    Wow! Quite an extensive list of rather interesting math posts! I kinda feel like a kid in a candy store ;-) So much to read, so little time…

  6. Sony says:

    About the “paradox” by Heath Raftery. I think the word simultaneous does a lot to solve the problem correctly.

  7. [...] writing code with the PFP library, I may as well join in on a discussion that got pulled into the Carnival of Mathematics [...]

  8. Now, inspired by Alex, I have done a simulation of the problem as a small coding exercise, and put the results up over at my own blog:
    Response to Heath Raftery

    I end up with the teacher on this one, since viewing my experimental results — 10 runs with each having 50 000 dice rolls, and all completely without 4s discarded — puts 1/6 at almost 5 standard deviations out, and 1/5 also far out: at about 5.5 standard deviations out.

    In contrast to this, 2/11 lies at about 0.13 standard deviations from the experimental mean. I’m convinced. :)

  9. Axel says:

    Sorry, I was in a hurry so my argumentation was rather cryptic (and very misleading). Another try:

    1. “Two dice are rolled simultaneously” That’s the random process with two physically independent dice which generates the outcome. The following results are possible:

    1-1 with p=1/36
    1-2 with p=1/36

    6-6 with p=1/36

    What does this mean? In the long run, we get 1-1 as often as 3-4 or 5-1 or 6-6 regardless what comes later.

    2. “Given that one die shows a ’4′”

    In the long run, we get 1-1 as often as 3-4 or 5-1 or 6-6 but sometimes, one die shows a “4″ and sometimes not. But IF one die shows a “4″ the number of the other is no more independent but limited. Think of a die with a “5″ and the condition “Given that one die shows a ’4′” – the other die MUST have a “4″. Therefore the possible results are:

    1-4 with p=1/11
    2-4 with p=1/11
    3-4 with p=1/11
    4-1 with p=1/11
    4-2 with p=1/11
    4-3 with p=1/11
    4-4 with p=1/11
    4-5 with p=1/11
    4-6 with p=1/11
    5-4 with p=1/11
    6-4 with p=1/11

    These outcomes are equal probable anymore – now 1/11 instead of 1/36 – because the two numbers are already fixed. It’s somehow comparable to the probability of 1/3 of the initially chosen door in the Monty Hall problem which isn’t influenced by the opening of the second door.

    4-3 and 3-4 have the sum 7, so it’s 1/11+1/11=2/11.

    There’s another similar probability paradox:
    A mother has two children. One of them is a daughter. What is the probability that the other child is a girl? (P_boy=p_girl=1/2)

    It’s 1/3…

  10. Axel, Mikael, thanks for taking the time to comment on the dice probability. Sorry about comments being broken – I’ve just fixed that up again (it’s a on-going race to be one step ahead of the spammers, isn’t it?).

    I’d urge you to read the last paragraph again, where I describe the problem that the professor is solving. The only difference is that in each dice roll, the outcomes where no 4′s appear are discarded. In that case you and him are correct in your analysis, no questions. My argument is that there is no suggestion such discarding occurs in the original problem description.

    By the way, the urn analogy is flawed – that’s probabilities without replacement. The colour of the first ball certainly does affect the colours of the remaining balls (because you are removing one). The value of the “first” die does not affect the value of the “second” die.

    Now what I really want you to read is a followup I made about the dice paradox, after spending sometime on the “Woman with two children” problem – a problem that raises very similar issues to the Monty Hall problem! In this followup (in particular in the update at the bottom) I describe just what conditions are necessary for each solution to be true.

    Dice paradox follow-up

  11. Alon Levy says:

    Didn’t see my post. Saving it for another day?

    It was an email address screwup (everyone else sent posts to my Yahoo address, so I worked from my Yahoo). I’ve forwarded it to Mark for the next edition.

  12. [...] first Carnival of Mathematics is up, after over a week of begging, whoring, bidding, and pimping. Most of the posts are purely [...]

  13. Heath: I’m not entirely certain that I agree with your analysis. Your situation seems to me to be the setup where you put one die on the table, with a 4 facing up, and then roll a die, checking whether this actually is a 3. You fix an ordering on the dice, with the die rolling a 4 ordered first, and I don’t buy that you’re entirely allowed to do that in the question as posed.

    Would you please give me pseudocode that captures the situation you describe? If you manage to write it down in a way that a computer can grok, then we can clearer recover just what the situation you describe is; especially since both me and Alex end up interpreting the original question as your “What did the lecturer really answer”-scenario.

    The dice may be independent objects, but the way the question is posed, you have no means of distinguishing them outside “I saw that one bounce first”; or “that one landed further to the right” — and in this case, arguing about what the blue dice does is flawed. There is a difference in information content between “the blue die is a 4″ and “one of the two dice is a 4″. In the one situation, we have the outcomes
    4 – 1
    4 – 2
    4 – 3
    4 – 4
    4 – 5
    4 – 6
    as in your argument, but in the other, we really do have
    1 – 4
    4 – 1
    2 – 4
    4 – 2
    3 – 4
    4 – 3
    4 – 4
    5 – 4
    4 – 5
    6 – 4
    4 – 6
    for the grand total of 11 cases, out of which 2 are interesting. And no, I don’t buy your argument that we should count 4-4 twice due to a difference in the probability that the observer announces the 4.

  14. Scott says:

    It’s great to finally see a mathematical carnival! Thanks for starting this.

    I’ve already made my submissions for the next carnival to MarkCC.

    As far as the logo goes, how about creating one at a site like:

  15. Axel says:


    thanks for your reply but I’m absolutely not convinced.

    You wrote: “In the dice situation, two independent dice are rolled and their outcome, whatever it happens to be, is presented for analysis. ”

    No. “whatever” absolutely doesn’t make sense here for me. The outcome for analysis is clearly restriced by “Given that one die shows a ‘4′”. If we interpret probabilities as long-run relative frequencies, this condition always must be true for the repetitions. Note that in a frequentist framework single-event-probabilities don’t exist.

    And that’s the problem with your solution p=1/6. For me, the question is now what’s the meaning of “Given that one die shows a ‘4′” in a frequentist context (repetitions!) for you? How can we “measure” p=1/6?

    It also seems to me that you’re confusing “knowledge” with “causal influence” and outcomes with probabilities. A causal influence has a temporary order of cause and effect. Probabilities are based on your actual knowledge, that’s why you can “change” the probabilities (not the outcomes!) of actually unknown but already happend events and why your statement “probabilities don’t change over time nor due to events happening nearby” generally is wrong! I can give a simple example even with a purely “objective” frequentist interpretation if you don’t believe it.

  16. Axel, I see what you’re saying. You’re analysis is quite right, given your interpretation of “given that one die shows a 4″. I’m afraid that’s all we disagree on. Let me explain my interpretation:

    “Two dice are rolled simultaneously” to me suggests that no selection of the roll has been made – this is a purely arbitrary roll.
    “Given that one die shows a 4″ says to me that someone has picked one of the dice and told me what it shows. The other dice is still sitting there, unperturbed.

    My algorithm would simply be to roll two dice many times. The value of one of the dice can be reported, and the times that the total is 7 is counted. Obviously this will result in 1/6 – there’s no need to do the simulation.

  17. [...] Carnival of Mathematics: Inaugural Edition « Abstract Nonsense Aargh! Forgot to submit something to this! (tags: mathematics blog+carnival) [...]

  18. Heath: It is a world of a difference between “One die shows a 4″ and “The blue die shows a 4″. In the latter case, it is very obvious that the probability is 1/6 and your argument applies. The former case, though, says that one of the dice shows a 4, but not which, thus opening up the sample space to the 11 cases we’re discussing.

    The model you’re proposing shows very clearly that you want to fix not only -that- a die shows 4, but also -which- die shows 4. And I don’t see how you reconcile that with the original question. I really don’t.

  19. “The model you’re proposing shows very clearly that you want to fix not only -that- a die shows 4, but also -which- die shows 4.”

    That wasn’t my intention at all. I’m going to roll a couple of dice… okay, one of them (my pick) shows a 5. What do you suppose is the chance that the other one is a 3? I think we are in agreement that it is 1/6, right? That’s the scenario I’m imagining. The alternate scenario (that results in 2/11) is that I’ve already decided I’m looking for a 5. I roll until one appears. Then I ask you about the other dice. That’s 2/11. Notice it is the observer that picks the die to report – in particular she picks the one with the 4 showing.

  20. Heath:

    Notice it is the observer that picks the die to report – in particular she picks the one with the 4 showing.

    And precisely this is what makes this problem so closely related to Monty Hall. We have, in both situations, an independent observer with full information that divulges some of that information in a previously agreed manner.

    The roll is done. And since the roll is with fair dice, all outcomes have equal probability.

    Now, the information that one of the two dice shows a 4 is divulged. This restrains the actual sample space to the 11 cases we’ve all been discussing. It does not, however, change the relative probabilities of the outcomes: each ordered pair of outcomes still carries equal probability.

    Now, of these 11 uniformly distributed cases, 2 are the ones we’re looking for. This gives us the 2/11 already talked about.

    If you roll a couple of dice, and then resets one of them to be a five – without me knowing which you reset – and given that I should assume your meddling to be distributed uniformly, I would expect you to pick the blue die with a probability of 1/2, and the red with the same probability. But this model boils down to rolling two dice, tossing a coin, ignoring one of the dice based on the cointoss and analyzing the other one. So the sample space carries 12 items, namely
    1H, 2H, 3H, 4H, 5H, 6H and 1T, 2T, 3T, 4T, 5T, 6T. And yes, it is clear, that in this case, it really is a 1/6 probability.

    But I still find your experiment description completely irreconcilable with the originally posed question:

    Two dice are rolled simultaneously. Given that one die shows a “4″, what is the probability that the total on the uppermost faces of the two dice is “7″?

    It is not the case that we rolled two dice, and one of them, by equal distributed probability was RESET to a 4. It is the case that we happened to chance upon one of those rolls we keep filtering out – we are told that it actually DID happen that one of our rolled dice shows a 4. Thus we know that the only cases we need to consider in our analysis are those athat DO include a 4, and thus it is the “repeted rolling until we find something that actually agrees with the situation” scenario we need to consider, since we already settled that we won’t be in one of the cases that get tossed.

    It is just as with the Monty Hall problem: the extra information from the independent observer does change the range of possibiliities we have to consider, but not the relative probabilities of the occurring cases. We do not IGNORE one of the dice – but we do get extra information about what happened to one of the dice.

  21. jd2718 says:

    Heath, Axel,

    the problem as recorded states:

    Two dice are rolled simultaneously. Given that one die shows a “4″, what is the probability that the total on the uppermost faces of the two dice is “7″?

    It does not say that “the first die” or “the second die” shows 4. It does not say even that “one of the die is chosen at random” and shows 4. The voice is that of an omniscient observer, and the only ambiguity is if the set-up means “exactly one die” or “at least one die.” Other than that, the language is fairly standard. The introduction of order or of an observer alters the problem.

  22. jd2718 says:

    That started fine, but ended too soon. Michael’s analysis is correct. The problem is posed as conditional probability.

  23. Axel says:


    The only ambiguity I’m accepting is the meaning of “one die”. Seems plausible to me that some people will interpret this as “exactly one die”. In this case, the probability that the total of the two dice is “7″ is 1/5.

    Heath, the problem with your interpretation is the criterion for the truth of the statement “One die shows a ’4′”. According to my layman’s correspondence theory of truth this statement is false if none of the two dice *really* shows a “4″, otherwise it’s true.

    But you argue that the statement is true if “one of them (my pick) shows a 5.” So it’s possible that “One die shows a ’4′” is sometimes false for the case of two dice *really* showing 1-4.


  24. Axel says:

    Sorry, must be “one of them (my pick) shows a 4.”

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  29. Killing Mind says:

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