I’m at logic camp.  Here’s a simple, fun puzzle that I learned:

You and your friends Andy and Brandy—two friends who happen to be pretty good at reasoning—are all wearing hats.  A third friend, Candy, has such a deft touch that she can place a feather in one’s hat without them knowing.  The hats are brimmed, so if there is a feather in it, the hat wearer would not be able to see.

Candy has only two feathers, and can do with them what she pleases.

You are standing single filed: You in front, Andy behind you, and Brandy behind her.  In such a configuration, Brandy can see whether or not Andy’s hat is feathered, and both Andy and Brandy could see a feather in your cap, were there to be one.  You can see the wall in front of you, which won’t help you solve the puzzle.

Candy—after maybe placing some feathers into hats, but maybe not—asks Andy and Brandy if they know whether their respective hats have feathers.  They both say “No”.  Candy repeats her question.  This time, Andy says “Yes, I know whether or not there is a feather on my head”.

Is there a feather in your hat?

I made a pretty big discovery at the supermarket today: E.L. Fudge Original (henceforth ELF-1) and E.L. Fudge Double Stuffed (ELF-2) sell for the same price at the supermarket.  My first thought was that the fudge stuffing must be both free to produce and completely weightless.  I quickly discovered that the ELF-1 package actually weighed more than the ELF-2 package, suggesting that the fudge was not weightless but rather some sort of anti-gravity substance that caused the package to weigh less.

It wasn’t until I got home, and opened and carefully examined my newly-purchased ELF-1 and ELF-2 packages, that I understood that there were actually more ELF-1 cookies in their package than ELF-2s in theirs.  Assuming there is indeed twice as much fudge stuffing in an ELF-2* and that the shortbread portions of the cookie weigh the same in ELF-1 and ELF-2, some very basic linear algebra will show that rather than having negative weight, a unit of fudge stuffing weighs 1.42 grams.

As far as the the price of producing the fudge stuffing: given the equality of the prices of the packages, my calculation is that the elves can produce 2 units of fudge per single piece of shortbread (there are two shortbread pieces per cookie).  Of course, that is only if the price is supply-driven.  More likely, it is demand-driven: if the prices ELF-1 and ELF-2 diverged, demand for whichever was cheaper would increase so heavily that it would preclude the feasibility of producing both versions.

The other fascinating bit of data in the above table: the number of calories in an ELF-1 is the same as in an ELF-2.  I don’t know enough about food to make any sense of this, but it might be evidence that the shortbread portions of ELF-1 and ELF-2 are actually different. Otherwise the fudge would have to be calorie-less while still contributing fat.

Previous posts: choosing which side of the street to buy your chicken wings from based on how many you want, E.L. Fudgehenge

*Allison has done some (methodologically questionable) research on the related question of “Is there twice as much stuffing in Double Stuf Oreo than in a regular Oreo?” and concluded there was not.

A thought, continued

Today in class, we were doing some crazy stuff with higher order logic (probably not too crazy as far as HOL goes, but the stuff is just hard to wrap your head around).  We were dealing with three separate levels of objects that were interacting, so obviously I thought about Inception.  I mention this mostly to stress how I definitely was not stoned when I came up with the following revision to yesterday’s Day Break/Groundhog Day idea (I mean, I wouldn’t be stoned anyways, obviously, but that fact just seems worth making explicit):

Alright, let’s say in the universe of Inception, someone spends 24 hours in a dream.  You’ll recall that in the universe of Inception, five minutes in “real” time take an hour in dream time.  That’s a multiple of 20, for 20 days of dream time.  Just assume this happens within the movie, so if you watch Inception, this 20 day dream occurs.

Now, inside that dream, for reasons that probably don’t make a great deal of sense, there is a television that continuously loops the entire season of Day Break.  There are 12 episodes.  Let’s assume there are commercials, so that’s 2 seasons of Day Break per day.  So if you watch Inception, 40 seasons of Day Break are contained within.

Now, yesterday we assumed Taye Diggs lived the same day for ten straight years.  Maybe that was too high.  Let’s call it one year.  So then, each day in the Inception dream, Taye Diggs lives the same day 730 times (or 14,600 days each time you watch Inception).

As per yesterday, I think it’s within reason to assume someone is watching Groundhog Day on the repeated day in Day Break.  And as per yesterday, Bill Murray repeats Groundhog Day for ten years, or 3,650 days.  That is about 1.3 million days per season of Day Break, so 2.6 million per day of Inception dream time, and 53 million Groundhog days per viewing of Inception.

One last step.  On the Groundhog Day that Bill Murray relives 3,650 times, there probably is a kid reading The Lion, The Witch, and The Wardrobe from cover to cover.  That story, of course, takes place in a few days of earth time, but Peter, Lucy, Susan, and Edmond spend something like 20 years in Narnia.

So what I’m trying to tell you is that every time you watch Inception, 1.07 billion years pass in Narnia.

Useless math is my favorite kind of math

Looking at the powers of 26 in my last post, I realized something: The product of two numbers than end in six also ends in six.  It’s because 6*6=36 ends with a 6, so any product will equal 36+X where X is divisible by ten (and hence has a zero in digits column).  Isn’t that neat? (It’s also true for numbers ending in 1 and 0, since 1*1 and 0*0 end in 1 and 0 respectively.  In fact, the 1 and 6 versions actually depend on the 0 version being true).

And because induction proofs are so much fun, let’s show that if X ends in six, then X^i also ends in six for all positive integers i.  Since X=X^1 ends in six, our base step is trivial.  And if we assume that X^i ends in six, then X^(i+1)=X*X^i is the product of two numbers that end in 6, and hence will end in six by the reasoning in the first paragraph (which ought to be formalized).  qed.

“The Unicorn of Probability Theory”

I’m not sure where I came across it first, but I found myself recalling a Bayesian probability quirk (or maybe quirk is the wrong word) recently:

If you are betting on coin flips, and the first flip comes up heads, you want to bet that the second one comes up heads too.

This runs counter to recognizing that the gambler’s fallacy is a fallacy, but there is a simple rationale for betting heads the second time:  the coin might be biased.  If you know a coin has in the past landed on heads, but never on tails, then maybe there’s something up with the coin, and you should follow the data you’ve got.

If you’re like me, before you run out and start betting on coin tosses, you are going to ask: Do biased coins really exist?  I’ll admit to knowing little physics, but for a coin flipping in the air to be biased (say, weighted in such a way that it lands hands 75% of the time), it would need to rotate pretty peculiarly in the air (like constantly accelerating and decelerating with each half turn).

Thankfully, Andrew Gelman (whose blog Statistical Modeling, Causal Inference, and Social Science is just as awesome as that name would imply) and Deborah Nolan flipped an assload of coins and found that—with a high toss and no bouncing—biased coins are a myth.  If you spin them or let them land on the ground and bounce or other things, then you can find a bias, but not through flipping.

This result is important for two reasons:

1. I was right.
2. Someone else bothered to answer the pointless question I had.

Of course, if you fall into that peculiar subset of people who distrusts journal-published science yet still believes in Bayesian inference, then go ahead and pick heads since you’ll be at 50% in the case of “biased coins are unicorns” and above that in the case that they are real.

Gains From Trade

I have recently entered into an arrangement to tutor geometry in exchange for home-cooked meals and wine.  Obviously this is pretty awesome.  Geoemetry—for those that are unaware—is pretty much the best math they bother to teach in high school.*  Far fewer pesky numbers to distract you from all the exciting theorems.  And triangles!

Beyond all the fun I get to have with Geometry (last night we went over rigid motions and the beginnings of what a proof is.  My tutee was disappointingly un-amazed when I explained that a rotation was the composition of two reflections through non-parallel lines, but amazing it is.), this barter system is great for two reasons:

1. There’s some nontrivial probability that I’m a really shitty tutor.  This way, I don’t need to feel guilty of taking the money of some family-friend that I might be hoodwinking.  I think I did okay, but I don’t have a lot of experience with tutoring, so I probably suck.
2. Tutee’s dad is a serious food and wine person.  Veal, mashed potatoes, and green beans for dinner.  And he’s in the wine business, so all high quality stuff on the beverage front.

Seriously, geometry and fancy food every week.  I can’t think of any exchange in my life that has worked so well in my favor.

*The second best math was using trig identities to prove equalities in 11th grade.  Much less geometry, most people hated this.