I haven’t yet looked at the comments about the puzzle I reported this morning that had been proposed by Russian Prime Minister Mikhail Mishustin when visiting a science-oriented “sixth form” class.

Here’s the puzzle again:

Construct a perpendicular from the (red) point on the circle to the diameter, without using any measuring devices.

In other words, given a circle with a diameter marked on it, and a point on the circle, can you find a way to draw a line from the point that hits the diameter at a right angle. (As marked in green above.)

The beauty of this question is the seemingly outrageous restriction not to allow measuring devices, which means that you cannot use a compass or a marked ruler. All you are allowed is an unmarked ruler to draw straight lines.

The Guardian has now published the four-step solution (there may be others); go to the preceding link to see it. Below you can see the PM drawing the solution (he must know his math).

In lieu of Readers’ Wildlife today, we have a puzzle, one posted (as the Guardian reports) by Russian Prime Minister, Mikhail Mishustin when he was visiting a science-oriented “sixth form” (what age of kids are these?) school. Matthew sent me a link.

Here’s the problem, and there’s a clue in both the photo below and in the Guardian article:

Construct a perpendicular from the (red) point on the circle to the diameter, without using any measuring devices.

In other words, given a circle with a diameter marked on it, and a point on the circle, can you find a way to draw a line from the point that hits the diameter at a right angle. (As marked in green above.)

The beauty of this question is the seemingly outrageous restriction not to allow measuring devices, which means that you cannot use a compass or a marked ruler. All you are allowed is an unmarked ruler to draw straight lines.

Matthew couldn’t solve it and, as I haven’t had my coffee, I’m not even going to try.

Here’s, a picture of Mishustin posing the problem (and giving a bit of a solution):

Yesterday I posted this math teaser:

Mate this is hilarious

GO! pic.twitter.com/fK3av04Zj4

— Dr. Jessica Taylor (@DrJessTaylor) March 28, 2020

128 people came up with answers. I said there were two, depending on where one puts the parentheses in the last equation, but the mathies say that there is a convention: one does the multiplication first, and then the addition. The only trick in the piece was the last line: the kid is wearing two sneakers and holding two cones of whatever that stuff is. (What is it?)

Here is my answer, which I think is correct if you use the “multiply first convention”

6 sneakers = 30, ergo 1 sneaker = 5
Two boys + two sneakers = 20.  Two boys + 10 = 20, ergo one boy = 5
4 cones plus one boy = 13. 4 cones + 5 = 13; ergo 4 cones = 8, so that one cone = 2.

In the last picture, we have one sneaker plus (one boy with two cones and two sneakers) times one cone.
Ergo 5 + (5 + 4 + 10) X 2 is the solution. That is 5 + 19 X 2
Using the multiplication rule first, that works out to 5 + 38 = 43.

If you put the parentheses in the last equation around (sneaker plus boy with cones and sneakers) X cone, you’d get 24 X 2 or 48. But the mathies say that this is wrong under the convention.

So the correct answer is 43. (I hope I didn’t screw up!)

Thyroid Planet was the first to post the correct answer(s) 29 minutes after the contest started, saying “48 or 43”.

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.

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.

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!