Archive for the 'Puzzles' Category

Boys, Girls and Hot Hands

This is a post about hot hands in basketball. But first, some relevant history:

The single most controversial topic ever broached here on The Big Questions was not Obamacare, or tax policy, or the advantages of genocide, or the policy treatment of psychic harms. It was this:

The answer, of course, is that you can’t know for sure, because (for example) by some extraordinary coincidence, the last 100,000 families in a row might have gotten boys on the first try. But in expectation, what fraction of the population is female? In other words, if there were many such countries, what fraction would you expect to observe on average?

The “official” answer — the answer, for example, that Google was apparently looking for when they posed this as an interview question — is that no stopping rule can change the fact that each birth has a 50% chance of being either male or female. Therefore the expected fraction of girls in the population is 50%.

That turns out to be wrong. It’s true that no stopping rule can change the fact that each birth has a 50% chance of being either male or female. From this it does follow that the expected number of girls is equal to the expected number of boys. But it does not follow that the expected fraction of girls in the population is 50%. Instead, that expected fraction depends on the country size, but is always less than 50%.

If you don’t see why, I encourage you to browse the archive of relevant blog posts. If you still don’t get it, I encourage you to keep browsing. Whatever your objections might be, you’ll find them addressed somewhere in the archive. I’m not interested in relitigating this. I will, however, happily renew my offer to take $5000 bets on the matter, on the terms described here. Last time around, all takers changed their minds before putting any money on the table.

Now let’s get to the hot hands.

Continue reading ‘Boys, Girls and Hot Hands’


Are You Smarter Than Google? — Part Three

To review the bidding:

Two days ago I posed a puzzle about 10 pirates dividing 100 coins.

Yesterday, I presented what appears to be an airtight argument that the coins must be divided 96-0-1-0-1-0-1-0-1-0.

But yesterday I also told you that the “airtight argument” is in fact not airtight, and that other outcomes are possible. I challenged you to find another possible outcome, and to pinpoint the gap in the “airtight argument”.

Our commenter Xan rose to the occasion. (Incidentally, his website looks pretty interesting.) Here’s his solution:

Continue reading ‘Are You Smarter Than Google? — Part Three’

Are You Smarter Than Google? — Part Two

treasureToday I’ll offer the “official” solution to yesterday’s puzzle — that is, the solution that Google has apparently expected from its job candidates. This is also the solution I gave when I first saw the puzzle, and the solution I usually get from my best students, and the solution given yesterday by some astute commenters.

But this solution has a gaping hole in it. Can you find it?

Continue reading ‘Are You Smarter Than Google? — Part Two’

Are You Smarter Than Google?

pirateI’m not sure where this problem originated. I heard it first from John Conley, and have often assigned it to my classes. Google has used it to weed out job candidates. The answer that Google expects is the same answer I gave John Conley, and the answer I usually get from my best students. That answer is wrong. (Long time readers might feel a sense of deja vu.)

Can you get it right?

Here’s the problem:

Continue reading ‘Are You Smarter Than Google?’

The Big Winners

The winners of our crossword puzzle contest are:

—Todd Trimble (3 mistakes)

—Eric Kehr (4 mistakes, but he corrected them all by email almost immediately)

—Serge Elnitsky (5 mistakes)

—Paul Epps (5 mistakes)

(There were supposed to be three winners, but since there’s a tie for third place, we have four.)

For all those who struggled and want to see the answers, I’m temporarily posting the solution here, but might take it down after a little while in case others want to try the puzzle without being tempted to peek.

Each winner is entitled to a copy of one of my books, with a personal inscription acknowledging your brilliance. If you’re a winner, please send me your mailing address by email and book choice by email or by commenting below.

The choices are:

The Armchair Economist — the principles of economics, applied to everyday life. Available both in the original (1993) edition and in the updated (2012) version. The latter is (I hope) a little better and a lot more up-to-date, but available only in paperback. The Wall Street Journal review is
here. You can read the preface to the 2012 version here.

Fair Play. The argument of this book is that we tend to think most seriously about issues like fairness when we’re explaining them to our children — so we should listen to things we say to children, draw lessons from them, and take those lessons into the marketplace and the voting booth. The Washington Post review is here. You can read a sample chapter here.

More Sex is Safer Sex. A compendium of surprises from economic theory, including how you can do your part to fight STDs by having more sex, and why you should contribute to only one charity. The Financial Times review is here. You can read an excerpt here.

The Big Questions — tackling the problems of philosophy, beginning with “Why is there something rather than nothing?”, using ideas from economics, mathematics and physics. Some reviews are here.

Continue reading ‘The Big Winners’

Puzzle Contest Update

As of now, I’ve received exactly one completely correct answer to this week’s crossword. (The submission actually contained four errors, but it was followed almost immediately by an email from the submitter with the requisite four corrections, so I’m giving full credit.) Congratulations to our frequent commenter EricK.

The contest, however, remains open. I’ll be sending free autographed books (your choice of The Big Questions, The Armchair Economist, Fair Play, or More Sex is Safer Sex) to EricK and the three runners-up, where the runners-up will be determined by some yet-to-be-determined combination of accuracy and timeliness. The contest will close on a date still to be determined, but I plan to keep it open for at least another week. Keep those submissions coming!

Click here to comment or read others’ comments.

Monday Puzzles

Click image to solve puzzle.

So it turns out that if you take a notion to create a crossword puzzle, put it on your blog, and include a “submit” button so that solvers can send you their answers, then — at least if your skill set is similar to mine — writing the code to make that “submit” button work will be about twice as difficult and three times as time-consuming (but perhaps also several times as educational) as actually creating the crossword puzzle. I certainly learned some hard lessons about the difference between POST and GET. But it’s done and (I think) it works.

To do the puzzle online click here. For a printable version, click here. If you do this on line and want to submit your answer, use the spiffy “Submit” button! (And do feel free to compliment the author of that button!). The clues are subject to pretty much the same rules that you’d find in, say, the London Times or the Guardian.

I will gather the submissions and eventually give proper public credit to the most accurate and fastest solvers. Feel free to submit partial solutions; it’s not impossible that nobody will solve the whole thing.

Let’s try to keep spoilers out of the comments, at least for a week or so.

I have one very geeky addendum to all this, leading to a second Monday puzzle — one that might be easy to solve for a reader or two, but most definitely not for me. Unless you’re a very particular brand of geek, you’ll probably want to stop reading right here. But:

Continue reading ‘Monday Puzzles’

Thursday Solution

Wednesday Puzzle

(Source omitted to discourage Googling; acknowledgements will come next week).

You have a sealed lockbox about a cubic yard in volume, containing $100,000 in hundred dollar bills. Your balance scale tells you that the box (with the money inside) weighs 100 pounds. You give the box to your friend Al, who flies it to the moon, while you, along with your balance scale, follow in a separate vehicle. Upon arrival, you retrieve the sealed box, put it on the balance scale and verify that it still weighs 100 pounds. You then give the box to your friend Barb, who loads it into her all-terrain vehicle and drives it to your moonbase, with you following along, again in a separate vehicle. When you get to the moonbase, Barb returns your lockbox. You open it and it’s empty.

Who stole your money, Al or Barb?

Click here to comment or read others’ comments.

Thursday Solution

Tuesday Puzzle

Here I have a well shuffled deck of 52 cards, half of them red and half of them black. I plan to slowly turn the cards face up, one at a time. You can raise your hand at any point — either just before I turn over the first card, or the second, or the third, et cetera. When you raise your hand, you win a prize if the next card I turn over is red.

What’s your strategy?

Click here to comment or read others’ comments.

Friday Followup

Tuesday’s puzzle was hard, though our commenter Bennett Haselton nailed it. In case Bennett has nothing else to work on this weekend, here’s a much harder version.

Once again, Alice, Betty and Carol each has a postive integer stamped on her forehead. They know that two of the numbers add up to the third. This dialogue ensues:

Continue reading ‘Friday Followup’

Tuesday Puzzle

Here’s a puzzle I hadn’t seen before. I’m concealing the source to discourage Googling, but will give credit where it’s due in a couple of days.

Alice, Betty and Carol each has a positive integer stamped on her forehead. They know that one of their numbers is equal to the sum of the other two. They proceed alphabetically around the table, each one either announcing her own number (if she’s managed to figure it out) or announcing that she doesn’t know it.

The game proceeds as follows:

Continue reading ‘Tuesday Puzzle’

A Bayesian Solution

There were many excellent comments on yesterday’s Bayesian Riddle. Here’s what I believe is the simplest and most natural analysis.

First, let’s recall the problem:

A murder has been committed. The suspects are:

  • Bob, a male smoker.
  • Carol, a female smoker.
  • Ted, another male smoker.
  • Alice, a female non-smoker.

You are quite sure that one (and only one) of these suspects is the culprit. Moreover, after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker.

Now your crack investigative team, in which you have total confidence, reports that, on the basis of new evidence, they’ve determined that the culprit is definitely female.

Who’s the most likely culprit, and with what probability?

Notice that if you considered all the suspects equally likely, your estimate would have been three to one for a smoker. Since you estimated only 2-to-1, you must have believed that the individual smokers were less likely than average to be guilty. So when you find out the culprit is female, it’s the female non-smoker — that is, Alice — who is now your prime suspect.

Continue reading ‘A Bayesian Solution’

A Bayesian Riddle

A murder has been committed. The suspects are:

  • Bob, a male smoker.
  • Carol, a female smoker.
  • Ted, another male smoker.
  • Alice, a female non-smoker.

You are quite sure that one (and only one) of these suspects is the culprit. Moreover, after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker.

Now your crack investigative team, in which you have total confidence, reports that, on the basis of new evidence, they’ve determined that the culprit is definitely female.

Who’s the most likely culprit, and with what probability?

Click here to comment or read others’ comments.

Hi, Mom!

MomMy mother, who reads this blog, reports that she’s lost a few nights’ sleep lately, tormented by thoughts of Knights, Knaves and Crazies. Serves her right. Once when she and I were very young, she tormented me with a geometry puzzler that I now know she must have gotten (either directly or indirectly) from Lewis Carroll; you can find it here. If she remembers the solution, she should be able to sleep tonight.

Herewith, a proof that a right angle can equal an obtuse angle. The puzzle, of course, is to figure out where I cheated.

But wait! Let’s do this as a video, since I’m starting to fool around with this technology and could use the practice. Consider this more or less a first effort. If you prefer the old ways, you can skip the video and read the (identical) step-by-step proof below the fold.

Get the Flash Player to see this content.

Or, if you prefer to skip the video, start here:

Continue reading ‘Hi, Mom!’

Hard and Harder

If you failed to solve Wednesday’s problem on Knights, Knaves and Crazies, take comfort from the fact that this has circulated among philosophers under the title “The Hardest Logic Problem Ever”. MIT philosopher George Boolos discussed it in the Harvard Review of Philosophy back in 1996. In that version, Crazies are never silent. But Oxford philosopher Gabriel Uzquiano soon observed that this can’t be the hardest logic problem ever, because it gets harder if the Crazies can be silent. Uzquiano’s new “hardest logic problem ever” was solved by the philosophers Gregory Wheeler and Pedro Barahona — and then solved again, substantially more elegantly, I think, in Wednesday’s comments section right here.

A few more thoughts, on the problem, its solution, and how to make it harder:

Continue reading ‘Hard and Harder’

Knights, Knaves and Crazies

SmullyanThe best dozen or so puzzle books ever written are, without a doubt, the works of Raymond Smullyan. If you’ve never encountered these, stop right now and order yourself a copy of What is the Name of This Book?, which is brilliant on multiple levels. On the surface, it’s a book of particularly amusing little brain teasers. One level down, those brain teasers contain a proof of Godel’s Incompleteness Theorem — solve all the riddles and you’ll have painlessly understood the proof!

Smullyan’s books are heavily populated by Knights who always tell the truth, Knaves who always lie, and bewildered travelers trying to distinguish one from the other via their cryptic utterances. Today’s puzzle is Smullyan-like in its set-up but considerably more difficult than most. It’s been proposed and discussed in philosophy journals, but I’m suppressing the sources (and rewording the problem) to make it a little harder to Google. I’ll of course pay appropriate homage to the authors when I post solutions in the near future. Meanwhile, if you’ve seen this before, or if you’ve found the answer on line, please restrain yourself from posting spoilers. But do post whatever you manage to come up with on your own.

And now to the puzzle:

Continue reading ‘Knights, Knaves and Crazies’

I Too Have Riddled Boxcars Boxcars Boxcars

Okay, Tuesday’s boxcar problem has gotten pretty interesting. I thought I knew the answer, but the comments on Wednesday’s followup post have sowed major seeds of doubt. There are a lot of excellent comments there.

I am thankful that I acknowledged in advance (at the bottom of Wednesday’s post) that I’m less sure of this one than I am of many others. I’d cheerfully bet $1000 (subject to agreement on a suitable referee) that I’m right about this relativity puzzle. (My answer is here.) And as far this old chestnut goes, my answer is here and I hereby cheerfully renew my offer to bet up to $15,000 on the outcome of a computer simulation. (Or any other amount, as long as it’s over $1000, to make this worthwhile.) Email me if you’re interested.

(This is on my mind because I’ve just had a very unpleasant encounter with a troll in another venue, who, like other trolls, is happy to bluster but runs away when you offer to put money on the line.)

For the first time ever, I am turning off comments on this post, because I don’t want to dilute yesterday’s interesting discussion by allowing it to take place half over there and half over here. Go there to participate. Many thanks to the commenters who have forced me to think harder about this, and thanks to anyone else who can help resolve the controversy.

About That Boxcar

Yesterday’s puzzle was this: A boxcar filled with water sits on a frictionless train track. A mouse gnaws a small hole in the bottom of the boxcar, near what we’ll call the right-hand end. What happens to the boxcar?

(Spoiler warning!)

Continue reading ‘About That Boxcar’

Boxcar Willie

I’ve just been pointed to this notice of a conference in honor of the topologist Tom Goodwillie‘s 60th birthday.

This reminded me of several things, not all of them related to the relentless march of time.

For example, once a very long time ago (though it sure doesn’t seem that way) Tom asked me a simple physics question that troubled me far more than I now think it ought to have:

A boxcar full of water sits on a frictionless train track. A mouse gnaws a hole through the bottom of the boxcar, in the location indicated here:

The water, of course, comes gushing out. What happens to the boxcar?

Click here to comment or read others’ comments.

Cats, Dogs and Coin Flips

The solution to yesterday’s rationality test:

This one is much much simpler (and much less infuriating) than some of our earlier rationality puzzles (e.g. here and especially here), but it has a good pedigree, having come to me from my student Tallis Moore, who found it in a paper of Armen Alchian, who attibutes it to the Nobel prizewinner Harry Markowitz.

Several commenters got it exactly right, but whenever possible, I prefer an explanation that invokes cats and dogs. So: Suppose I give you a choice between A) a cat, B) a dog, and C) a coin flip to determine which pet you’ll get:

It’s perfectly rational to prefer the cat to the dog, and perfectly rational to prefer the dog to the cat, but (according to the traditional definition of rationality) quite indefensible to prefer the coin flip to either.

Continue reading ‘Cats, Dogs and Coin Flips’

Another Rationality Test

In front of you lie three urns, labeled A, B and C. Each contains 2000 balls. Urn A has 2 reds and the rest black; Urn B has 20 blue and the rest black; Urn C contains 1 red, 10 blue and the rest black. Like so:

You can reach into the urn of your choice and remove a ball. If you draw red, you get $1000; if you draw blue, you get $100; if you draw black, you get nothing. Which urn do you pick and why?

Continue reading ‘Another Rationality Test’

Mortgage Solution

The solution to yesterday’s mortgage puzzle:

Commenters pointed to several reasons why biweekly payments of (say) $500 will pay your mortgage off so much faster than monthly payments of $1000, but of these by far the most important is that biweekly payments of $500 add up to 500 x 26 = 13,000 dollars, whereas monthly payments of $1000 add up to 1000 x 12 = 12,000 dollars. With the biweekly payments, you make the equivalent of 13 monthly payments every year.

In other words, the key observation is that two weeks is not half a month.

My colleague Michael Wolkoff posed this puzzle to me many years ago, and I’m embarrassed to admit I failed to solve it before Michael gave me the solution. I was reminded of it yesterday when I got a biweekly-plan offer in the mail.

Click here to comment or read others’ comments.

Mortgage Puzzle

I have recently acquired a 30 year mortgage.

Today I’ve received a letter offering to let me make payments on a biweekly basis instead of a monthly basis. If I accept this offer, I will make a biweekly payment equally to exactly half my current monthly payment — and my mortgage will paid off in 23.6 years instead of 30.

Question: How can such a small change in the timing of my payments shave a full 6.4 years off the life of my mortgage?

Click here to comment or read others’ comments.

Wednesday Solution

Monday’s puzzle was open to various interpretations, but under what seems to me to be the most straightforward interpretation, if the number of runners you pass is the same as the number who pass you, you’re the mean runner, not the median.

You can find plenty of correct analysis in Monday’s comment section (see in particular Harold’s perfect comment #39), but here’s a more longwinded explanation:

First, suppose you randomly sample a large number of other runners and discover that half of them are faster than you and half are slower. Then you’re entitled to conclude that you’re the median runner (or, if we’re being careful, you’re entitled to conclude that you’re probably close to the median, since there’s always a chance your sample was unrepresentative).

Now in the problem as given it’s certainly true that half the runners you encounter are faster than you and half are slower. So you might be tempted to use the above reasoning and conclude that you’re the median runner. But that won’t work, because the runners you encounter are not a random sample.

So let’s start over. We might as well assume that you’re the center of the universe, so you’re completely motionless. Everyone who’s faster than you is running forward and everyone who’s slower than you is running backward. People “pass” you when they run past you in the forward direction, and you “pass” them when they run past you in the backward direction.

Continue reading ‘Wednesday Solution’

Monday Puzzle

It’s a lovely morning, and you are jogging along the lakeshore, along with many others (all in the same direction). Albert is the median runner (that is, he runs at the median speed). Betty is the average runner (she runs at the average — i.e. the mean — speed.)

You notice that the number of runners you pass is exactly equal to the number of runners who pass you.

Can you determine whether you’re running faster or slower than Albert? What about Betty?

Edited to add: When I said you were running “along the lakeshore”, I was envisioning the shore of Lake Michigan; i.e. I meant to say that you’re running, effectively, in a straight line, not a circle. Obviously I should have made this clearer. But it’s a good puzzle either way!

Click here to comment or read others’ comments.

Tuesday Puzzle

There’s an old puzzle popular among a certain type of schoolchildren that challenges the solver to write as many positive integers as possible using exactly four 4′s, together with some set of mathematical operations. (As is often the case with school children, the exact rules tend to get negotiated in real time as the puzzle is being solved.) Some examples are:

But when I became a man, I put away childish things. So here’s the grown-up version of the problem, which I got from Mel Hochster over 20 years ago, and still don’t know how to solve:

This being a grown-up problem, the rules are carefully specified:

Continue reading ‘Tuesday Puzzle’

Thursday Solution

Last week, I challenged readers to reconcile two apparently contradictory statements, both of which are frequently made in economics textbooks:

  • To minimize distortions, all goods should be taxed equally.
  • To minimize distortions, inelastically demanded goods should be taxed more heavily. (This is sometimes called the Ramsey rule, after Frank Ramsey, who plays a major role in the final chapter of The Big Questions).

I’ll give you the answer in a minute. The executive summary is that a) “Inelastically demanded goods should be taxed more heavily” is true only in very special circumstances; in general a much more complicated formula is needed, b) When all goods can be taxed, that complicated formula does in fact tell you to tax them all equally, and c) a lot of textbooks give incredibly misleading accounts of all this.

The more detailed answer follows; if you prefer a more mathematical account, click here. To keep things manageable, I’ve assumed all supply curves are perfectly elastic.

Continue reading ‘Thursday Solution’

Tuesday Solution

Remember the bullet problem from two weeks ago? If not, I’ll give you a few moments to refresh your memory.

Okay. Are we ready?

I am pretty well convinced that 16 bags suffice, as argued by Mike H, Jeffrey, Brian, and Jacopo, and generalized by Categories+Sheaves.

The explanation I liked best was Jeffrey’s. Let me try to illustrate it. (Warning: There’s no way, I think, to make this instantly clear. It will take a little work to understand it. Only you can assess the opportunity cost of your own time!)

The gray rectangle is the room you’re in.

The blue dot is you. The red dot is the shooter.

With mirrors on all the walls, you’ll perceive yourself as standing in an infinite grid. The graph shows 16 of the grid rectangles, but the pattern continues forever in every direction.

You’re standing at some point (a,b). The shooter is at some point (p,q). You’ll see copies of that shooter at every point with coordinates (2m ± p, 2n ± q) where m and n range over all integers. Sixteen of those infinitely many points are shown in the picture.

Continue reading ‘Tuesday Solution’