Several commenters correctly solved yesterday’s puzzle…..
Archive for the 'Puzzles' Category
(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?
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?
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:
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:
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.
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?
My 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.
Or, if you prefer to skip the video, start here:
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:
The 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:
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.
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!)
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?
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.
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?
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.
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?
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.
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!
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:
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.
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.
Today’s puzzle is specifically for the econo-geeks. Less geeky fare will follow in the near future.
Two of the main lessons that our undergraduates typically take away from their introductory classes are these:
- To minimize distortions, all goods should be taxed at the same rate.
- To minimize distortions, inelastically demanded goods should be taxed most heavily.
What is the correct response to this pair of apparently contradictory lessons?
- Economics is large. It contains multitudes. Get over it.
This has been making the rounds lately; I’m not sure where it first came from.
You’re in a rectangular room. Elsewhere in the room is a man with a gun, who shoots a bullet in a random direction. The bullet careens around the room, bouncing off walls, until it hits either you or one of the various punching bags you’ve placed around the room for purposes of absorbing the bullet. The punching bags must be positioned before you know the random direction of the bullet (though you do know both your own location and the bad guy’s location, neither of which you can change). How many punching bags do you need to guarantee your survival?
This being a math problem, you should treat the room as two dimensional, and yourself, the bullet and the punching bags as points.
Greg Mankiw, with a hat tip to his son Nicholas, asks for a plot of the function x^{x}, where x is a real variable. The answer he points to (provided by Pedagoguery Software) gives this picture/expanation:
Our most recent nightmare scenario triggered some great discussion. (So did the earlier nightmare scenarios, which I’ll review in a day or two.) The conundrum was this:
In front of you are two childless married couples. For some reason, it’s imperative that you kill two of the four people. Your choices are: A. Kill one randomly chosen member from each couple. All four people agree that if they die, they want to be well remembered. Therefore all four ask you, please, to choose A so that anyone who dies will be remembered by a loving spouse. If you care about the four people in front of you, what should you do? |
Commenters, interestingly, split pretty much 50-50 (though it’s hard to get an exact count because several equivocated). Many bought the argument that unanimous preferences should be respected (leading to A). About equally many bought the argument that the preferences of the dead don’t count, and it’s better to leave two happily married survivors than two grieving widows/widowers (leading to B).
A few highlights:
I’m eager to summarize the (largely excellent) discussion of last week’s nightmares and to talk about what it all means. I’ll surely get to that in the next few days. But meanwhile, we have another loose end to tie up.
I recently asked what comes next in the following series:
The answer, of course, is
Please raise your hand if you found this intuitively obvious.
In case your hand didn’t go up, consider the following sequence:
Back in the 1930′s, Kurt Godel proved two amazing facts about arithmetic: First, there are true statements in arithmetic that can’t be proven. Second, the consistency of arithmetic can’t be proven (at least not without recourse to logical methods that are on shakier ground than arithmetic itself).
Yesterday, I showed you Gregory Chaitin’s remarkably simple proof, of Godel’s first theorem. Today, I’ll show you Shira Kritchman and Ron Raz’s remarkably simple (and very recent) proof of Godel’s second theorem. If you work through this argument, you will, I think, have no trouble seeing how it was inspired by the paradox of the surprise examination.
Today, I’m going to give you a short, simple proof of Godel’s First Incompleteness Theorem — the one that says there are true statements in arithmetic that can’t be proven. The proof is due to Gregory Chaitin, and it is far far simpler than Godel’s original proof. A bright high-schooler can grasp it instantly. And it’s wonderfully concrete. At the end, we’ll have an infinite list of statements, all easy to understand, and none of them provable — but almost all of them true (though we won’t know which ones).