Out of a regular grade school classroom, two students are chosen at random. Both happen to have blue eyes. If the odds are exactly 50-50 that two randomly chosen students in the class will have blue eyes: How many students are in the class?I will show you how I arrive at FOUR as my answer:

Our class has Archie, Betty, Charlie and Davy in it. Archie has BROWN eyes, the rest have BLUE eyes. Any pair with Archie in it goes in one group, all Archie-less pairs in the other group. As long as the two groups are precisely the same size, we're good to go.

A-B, A-C, A-D vs. B-C, B-D, C-D

Does it work for other numbers? I have no idea. Here's where someone else can send me the proof and I'll insert it for everyone's edification.

Here are our classmates (plus two more photos, just because they came up when I searched for "brown eye boy"):

Time for

Here are this week's picks:

Fewer than 50 51 - 100 -- Paul 101 - 150 -- Word Woman 151 - 200 -- Joe Kupe 201 - 250 -- Ross 251 - 300 -- Magdalen 301 - 350 -- B.Haven 351 - 400 -- Curtis 401 - 450 -- Phil 451 - 500 501 - 550 -- legolambda 551 - 600 601 - 650 651 - 700 701 - 750 -- zeke creek 751 - 800 801 - 850 851 - 900 901 - 950 951 - 1,000 | 1,001 - 1,050 -- David 1,051 - 1,100 -- Henry B.W. 1,101 - 1,150 1,151 - 1,200 1,201 - 1,250 1,251 - 1,300 1,301 - 1,350 -- Mendo Jim 1,351 - 1,400 1,401 - 1,450 1,451 - 1,500 1,501 - 1,550 1,551 - 1,600 1,601 - 1,650 1,651 - 1,700 1,701 - 1,750 1,751 - 1,800 1,801 - 1,850 1,851 - 1,900 1,901 - 1,950 1,951 - 2,000 | 2,001 - 2,050 2,051 - 2,100 2,101 - 2,150 2,151 - 2,200 2,201 - 2,250 2,251 - 2,300 2,301 - 2,350 2,351 - 2,400 2,401 - 2,450 2,451 - 2,500 2,501 - 2,750 2,751 - 3,000 3,001 - 3,250 3,251 - 3,500 3,501 - 4,000 4,001 - 4,500 4,501 - 5,000 > 5,000 > 5,000 + new record |

*In the event that a single round number is announced with a qualifier such as "about" or "around" (e.g., "We received around 1,200 entries."), AND two separate people picked the ranges of numbers just before and*

*just after*

*that round number, the prize will be awarded to whichever entrant had not already won a prize, or in the event that both entrants had won a prize already or neither had, then to the earlier of the two entries on the famous judicial principle of "First Come First Serve," (or in technical legal jargon, "You Snooze, You Lose")*.

*As of July 2012, this rule is officially no longer obsolete (and also I still just like having fine print).*

## 8 comments:

While 4 is perfectly correct, I don't know that it's the intended answer. The bad grammar in the problem doesn't help. And I can make an argument that 4 is best.

But you'll find that there are a host of numbers (such as 15 blue-eyed students out of 21 students). In fact, all the ridiculously large answers posted in the comments before today work.

This problem is a quadratic Diophantine equation, a case in which an equation with two variable is interesting primarily for the integer solutions. (Look up "Diophantine equation" on Wikipedia for the general points.)

The number of solutions to this one is indeed infinite--except for that "regular" classroom. And here's where the bungled grammar comes in. Notice that the phrase "regular grade school classroom" seems to indicate a grade-school classroom that is regular (as opposed to contrived).

I don't think that WS meant that, though. He said it, but my guess is that he meant a

classthat was of a typical size, thus making 21 the "best" answer. However, I have no idea why you couldn't have four children in a regular classroom. That happens every day.Oh well, I've gone on far too long.

Phil

Phil here again (with apologies). I see, Magdalen, that you asked for proof. So here it is (for the 15 blue-eyed students out of 21 students overall).

The probability that the first student has blue eyes is 15/21. The probability that the second student also has blue eyes is 14/20. Multiply these two numbers, and you get 1/2.

The same approach works for the too-large answers provided in the comments before today.

Phil

"Brown eye boy"?

Why not "Brown Eyed Girl"?

Just for that, I'm requesting "Moondance" as a photo-search word (or have you done that one already?)

It's "brown eyed boy" because I named him "Archie" and with the notable exception of actress Archie Panjabi, that's usually a boy's name.

If the only acceptable answer is 21 students, then we all guessed too high. If you don't know that's the answer, I doubt anyone is going to use the "counting on one's fingers" approach I did to work it out. So the question becomes, how many mathematicians will enter the puzzle.

If Will Shortz had meant "a class size between 15 and 30 students" he should have said so. I was thinking of Little House on the Prairie when have the kids are home with flu.

Moondance it is!

Just a bit of reasonable pique from Magdalen.

She summarizes a feeling I have had too often: "If Will Shortz had meant 'xxxx' he should have said so."

I came up with the nicely satisfactory "4" fairly quickly and, not knowing there were an infinite number of alternatives, stopped there.

I didn't think of the lower answer, but didn't guess many people would send in a correct 15 blue-eyed in a class of 21 students.

I share a birthday with Martin Gardner.

Our answer of four was the class in northern remote Michigan!

Our answer of 21 derives from the following formula:

X=blue eyed

Y=total class size

(X(X-1))/(Y(Y-1))=1/2

Enjoyed this one!

I'm with Will on this one. I got an answer of 4 quickly, but because of the "regular grade school classroom" decided there must be more answers. The next answers I got were 21 and 120, which convinced me that the only answer that Will would accept is 21.

An interesting side note: my "not a robot" number is 120.

Post a Comment