Category: mathematics

A tricky SAT geometry question

June 27, 2019 • 2:30 pm

Just to show you how hard the SATs were, here’s a question (sent by reader Bryan) that I got wrong. I guessed three, assuming that the circumference of the smaller circle (1/3 of the larger), would translate into rotations needed to get around the circumference of the larger circle. How I got a perfect score in 1966 on the math part of the SAT defies me!

At any rate, the correct answer was not one of the SAT’s choices, so nobody got the question right.

Stop the video at 55 seconds in, make your guess, and then watch on.

An explanation of the correct answer for two circles is in the Wikipedia entry for “coin rotation paradox”. You can demonstrate this for yourself by using two coins of equal size.

This trick will make you the hit of any cocktail party—that is, if the cocktail party is full of academics.

ADDENDUM: This gif shows you that a coin rotates twice going around a coin of equal radius. Follow the blue dot from where it starts: you’ll see that it’s gone around once when it’s halfway around the central coin.

Alternative truths: math

July 15, 2018 • 12:31 pm

Here’s a humorous but not completely unimaginable video about what would happen if elementary schools were taken over by the view that there are “multiple truths”. This attitude already infects some of the social sciences and much of the humanities in universities, but math is not completely immune to the “different ways of knowing” infection.

Anyway, it’s just a bit of humor after France’s victory over Croatia in the World Cup. It was a diverse and engrossing game, even featuring an own goal, and the best team won, though Croatia didn’t go down easily. Congrats to France, and we’ll see you again in four years!

h/t: Lesley

A reasoning puzzle– *the* answer

April 17, 2018 • 8:33 am

by Greg Mayer

So, here’s the answer given by Manil Suri to the puzzle he posed in the New York Times on Sunday. First, restating the puzzle:

Four cards are laid in front of you, each of which, it is explained, has a letter on one side and a number on the other. The sides that you see read E, 2, 5 and F. Your task is to turn over only those cards that could decisively prove the truth or falsity of the following rule: “If there is an E on one side, the number on the other side must be a 5.” Which ones do you turn over?

And here is his answer:

Clearly, the E should be turned over, since if the other side is not a 5, the rule is untrue. And the only other card that should be flipped is the 2, since an E on the other side would again disprove the rule. Turning over the 5 or the F doesn’t help, since anything on the other side would be consistent with the rule — but not prove it to be true.

In the article, Suri points out that this is the Wason problem, which many readers recognized it as (I’d never heard of it). He goes on to point out that studying math improves a person’s ability to answer such problems correctly, ans argues that the Wason problem is an especially good way to teach critical thinking, and its use should be encouraged.

He notes that on average 10% of people get it right. I haven’t done a count, but many readers got it correct in the comments– far more than 10%, I would venture. One thing I learned from the responses is that many readers read the problem as referring to just these 4 cards, and that was useful in finding the answer. It never occurred to me– indeed, I think it would never have occurred to me– that the problem referred to just 4 cards. I assumed they were a random sample from a potentially infinite universe of cards. And that, to me, is where the real interest of the problem lies. Different people will read the same puzzle, and think the set up or the question are quite different. This matter of interpretation also occurred to many readers, and it is one to which I will return in a later post.

The Steve Pinker video I posted yesterday is the exact same problem, and Pinker gives Suri’s answer. My apologies to readers who were looking for my reveal yesterday, but Pinker was the reveal. But, as I mentioned above, Suri and Pinker’s answer was not the item of interest to me, and reading through the many comments yesterday made me realize my own interpretation of the problem was one of many, and so I have needed to think through my interpretive analysis further.

A reasoning puzzle

April 15, 2018 • 3:08 pm

by Greg Mayer

In today’s New York Times, there is an opinion piece by Manil Suri, a mathematician at the University of Maryland, Baltimore County, entitled “Does math make you smarter?”

Don’t go and read the piece– that’s why I’ve left out the link! I ask that readers answer the following puzzle he poses in it first.

Four cards are laid in front of you, each of which, it is explained, has a letter on one side and a number on the other. The sides that you see read E, 2, 5 and F. Your task is to turn over only those cards that could decisively prove the truth or falsity of the following rule: “If there is an E on one side, the number on the other side must be a 5.” Which ones do you turn over?

Don’t look at the comments here first, either. Try to answer the question, and then, once you’ve formulated the answer, post it here in the comments. If readers give multiple answers, feel free to debate them, but figure out and post your answer first.

I’ll post tomorrow (Monday) the answer and a discussion, along with the link.

It’s Pi Day!

March 14, 2018 • 10:15 am

Having consumed my share of Costco pies the last few weeks (they’re good, too!), I’m happy to report that it’s Pi Day, celebrated with the following Google Doodle (click on screenshot to go to the Doodle site):

Google itself explains the Doodle here, adding a video and, at the link, a recipe for a scrumptious salted-caramel apple pie. I hope at least one reader makes it:

Happy Pi Day!

Celebrated each year on March 14th (3.14), Pi Day is dedicated to the mathematical constant, Pi. First recognized 30 years ago in 1988 by physicist Larry Shaw, Pi Day observers often celebrate with a slice of their favorite pie in honor of the number’s delicious sounding name.

Notated by the Greek letter “𝛑”, pi represents the ratio between a circle’s circumference (perimeter) to its diameter (distance from side to side passing through the center), and is a fundamental element of many mathematical fields, most significantly Geometry. Though modern mathematicians have calculated more than one TRILLION decimal places beyond the standard “3.14,” pi is an irrational number that continues on to infinity! It’s an important ingredient in the formula for the area of a circle, A=𝛑r².

Today’s delectable Doodle – baked & built by award-winning pastry chef and creator of the Cronut® Dominique Ansel – pays homage to this well-rounded mathematical constant by representing the pi formula (circumference divided by diameter) using — what else — pie!

Go behind-the-scenes of today’s Doodle below!

By the way, has anybody ever proven that pi must be an irrational number? Or did it just work out that way?

Note that Pi Day can only apply in parts of the world where they write March 14 as 3/14/20-; most of the world does it wrong, using 14/3/20. Such parts of the world can have NO Pi day, and therefore they don’t deserve pie.

Note, though, that this Doodle’s reach extends much wider than that of a regular Doodle. Only sub-Saharan Africa, South Asia, and, curiously, Norway and Finland lack the Doodle and and, apparently, pies:

Here are the favorite pies of Professor Ceiling Cat (Emeritus); I am leaving out savory pies:

Malgorzata’s fresh cherry pie made with walnut crust

Pear cream-cheese pie

Peanut-butter/chocolate cream pie

Key lime pie, but only when made with real Key Limes rather than bottled juice. The pie is not nearly as good when made with regular (“Persian”) limes.  There are only a few places in America where you can get the real thing in a restaurant (Manny and Isa’s in Islamorada Florida, on the Keys, used to be one of them, but I don’t know if it’s still there); but you can make it using the tiny Key limes available in many high-class markets.

Blueberry pie, especially when made at Helen’s Restaurant, in Machias, Maine, where they use lowbush blueberries (the small ones) and heap a mixture of cooked and fresh blueberries into an open-top crust, slathering a thick layer of whipped cream all over the top.  Here’s a piece. Hungry?

What’s yours? Anybody eating pie today? I doubt I’ll get the chance.

 

Bertrand’s Box paradox: The answer is 2/3!!!

February 20, 2018 • 10:15 am

There are almost 200 comments now on my post about Bertrand’s Box Paradox yesterday. Let me reprise the problem and then give the solution the way I hit on it:

There are three boxes:

  1. a box containing two gold coins,
  2. a box containing two silver coins,
  3. a box containing one gold coin and a silver coin.

The ‘paradox’ is in this solution to this question. After choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, what is the probability that the next coin drawn from the same box will also be a gold coin?

In the discussion below, I’ll call the boxes #1, #2, and #3 in the order from left to right in this diagram, and use “coins” instead of “balls.”

The answer. People calculated this in various ways, some using Bayesian statistics, some, like me, a simple intuitive multiplication. But the two most common answers were 1/2 (50%) or 2/3 (67%). The latter answer is correct. Let me explain how I thought it through:

One you’ve drawn a gold coin, you know you’ve chosen from either the first or second box above. The third box, containing only silver coins, becomes irrelevant.

You’ve thus chosen a gold coin from the first two boxes. There are three gold coins among them, and thus if you picked a gold coin on your first draw, the probability that you chose from box #1 is 2/3. The probability that you chose it from box #2 is 1/3.

If you chose from box #1, the probability that you will then draw a second gold coin is 1 (100%).

If you chose from box #2, the probability that you will then draw a second gold coin is 0 (0%) for there are no gold coins left.

Thus, the overall probability that if you got a gold coin on the first pick then you will get another one if drawing from the same box is (2/3 X 1) + (1/3 X 0), or 2/3 (probability 67%).  The answer is thus 2/3 (probability 67%). 

If you don’t believe it, first check the Wikipedia explanation. It also explains why people think the probability is 50%, but fail to comprehend that it’s more likely, if you drew a gold coin on the first go, that the box you drew from is #1 than #2.

Then, if you still don’t believe it, try it yourself using either two or three boxes (you don’t really need three). That is, you can do an empirical test, though the explanation above should suffice. You will find that once you’ve drawn a gold coin on your first pick (you can just use two types of coins, with one box having like coins and the other unlike coins), the chance that you will draw another from the same box is 2/3. In other words, you’ll see that outcome 67% of the time. Remember, we are talking outcomes over a number of replicates, not a single try! You’d be safe betting against those who erroneously said 50%.

If you think Wikipedia is wrong and you’re right, good luck with correcting them!