Tag Archives: Cheryl’s birthday

At least there’s no giant boulder chasing you…

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[Is this the way out or a costly dead end?]

We’ve tackled all sorts of brain teasers in the past. From the Brooklyn Nine-Nine desert island seesaw to several hat puzzles, from Cheryl’s birthday logic puzzle to a diabolical light switch puzzle, we have conquered all challengers thus far!

But never before have we confronted a puzzle with as much backstory as today’s contender. Ladies and gentlemen and PuzzleNationers of all sorts, today we battle the Temple Tunnel puzzle.

Imagine that you’re a professor leading a group of eight grad students on an expedition into a booby-trap-filled temple.

[No, not THAT professor.]

After two of the students bump into an altar, they activate a trap, sending everyone scrambling for the exits before the temple collapses all around you.

The group finds itself in a room with five tunnels and an hourglass detailing how much time you have to escape. One of them leads back to the altar and the other four are possible routes of escape. Unfortunately, you can’t remember which one it is!

All you remember is that it took approximately twenty minutes to get here from the exit. How do you determine which tunnel is the correct one, and get everyone to safety?

Oh wait, there’s one more little complication. That altar the students bumped into? It released the vengeful spirits of the temple’s king and queen, which have possessed two of your students. So you can’t trust what they say.

So how do you figure out which tunnel is the right one without being deceived by your two compromised students?

[Image courtesy of XKCD.com.]

*deep breath* Wow, that’s quite a setup! So let’s summarize:

  • You have an hour to escape, and four corridors to explore.
  • Each corridor will require 40 minutes to explore: 20 minutes to determine if it’s the exit, and 20 minutes back to report your findings.
  • Whatever groupings you break the team up into, you have two possible liars among them, and no way to determine which ones are the liars before sending them down a tunnel.

For a wonderful animated version of this puzzle, as well as its solution, check out the YouTube video below from TED-Ed:

Now, while the solution itself is quite clever, I can’t help but ask certain questions:

It says that the possessed students can’t harm the others, but can they mislead them with actions as well as words?

I’ve seen several proposed solutions that included not only sending groups down the tunnels, but instructing one or more of them to leave the temple immediately if they find the exit (meaning that not seeing them return would confirm they’d found the exit). But if the liars can simply stay at the dead end, that would be a false confirmation of finding the exit.

The video is ambiguous about this, because it says the spirits will lead them to their doom, but then it also says that the curse only affects their communication.

How does the group know you’re not one of the liars?

The solution is entirely dependent upon you being able to explore a tunnel alone, because that determines the groupings for the other three tunnels. If you have to take someone with you (either an honest student or a liar), that affects your ability to draw proper conclusions from the other groupings. And even if you find the exit, the student with you could lie about it, and there’s no way to prove the truth to the group definitively.

Why not just ask each student individually a question the ancient king or queen wouldn’t know the answer to?

Presumably the spirits of ancient royalty wouldn’t know about the latest episode of NCIS or which version of Windows we’re up to.

In any case, this was a delightful mind-bender, one that has stumped many an intrepid solver. How did you do? Tell us in the comments below!

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It’s Follow-Up Friday: Birthday Puzzle edition!

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Welcome to Follow-Up Friday!

By this time, you know the drill. Follow-Up Friday is a chance for us to revisit the subjects of previous posts and bring the PuzzleNation audience up to speed on all things puzzly.

And today, I’d like to return to the subject of birthday brain teasers!

Working on the Cheryl’s Birthday brain teaser a few days ago reminded me of another birthday-fueled puzzle that’s been around forever.

How many people do you need to enter a room before the probability of any 2 or more people sharing a birthday (day and month only, not year) is greater than 50%?

Assume for the sake of the puzzle that birthdays in the population at large are equally spread over a 365 day year.

Now, given that there are 365 days in the year, you’d assume the number of people necessary to get that probability of a shared birthday above 50% would be more than half of 365, or 183 people.

But it turns out that, statistically speaking, you don’t need anywhere near that many people.

Let’s break it down. Person A has a birthday. Person B has a birthday. There’s only one possible pairing, A-B. Person C has a birthday, but creates three possible birthday pairings: A-B, A-C, and B-C.

Person D could have a different birthday, but the introduction of Person D begins escalating the number of POSSIBLE shared birthdays. With these four people, we have SIX possible pairings: A-B, A-C, A-D, B-C, B-D, and C-D.

Our fifth person, Person E, allows for TEN possible pairings: A-B, A-C, A-D, A-E, B-C, B-D, B-E, C-D, C-E, and D-E. The probability of a shared birthday is increasing much faster with each new person.

As it turns out, it only takes 23 people to give us a 51% probability of a shared birthday.

And that would certainly save on catering.

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Are you smarter than a Singaporean student?

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We love brain teasers here at PuzzleNation Blog. Whether we’re dealing with curious parking spaces, men in hats, the crew of the Enterprise playing games, or the seesaw-based conundrum that so baffled Captain Holt on Brooklyn Nine-Nine, we thoroughly enjoy tackling these often diabolical and curious logic problems.

And one has been making the rounds on Facebook, Twitter, and other social media platforms recently. This one comes from a Singaporean classroom, and has made headlines all over the Internet.

Hmmm, doesn’t seem like a lot of information, does it?

So, we have ten possible dates.

Let’s put them in a chart to organize them as best as we can.

Now, let’s analyze each statement in order, since the progression is the key to solving this brain teaser.

Albert says: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.

Since Albert is told the month, and there are multiple options for each month, there is no way he could know. At first. But he does know Bernard doesn’t know either. How?

Deduction. If Cheryl told Bernard 18 or 19 (the only days that appear once), Bernard WOULD know Cheryl’s birthday. So Albert can eliminate those two options.

And for Albert to KNOW that, Cheryl cannot have told him May or June, since those were the only months with days that appear once.

A lot of information in a single sentence. Let’s move on to the next sentence.

Bernard says: At first I don’t know when Cheryl’s birthday is, but I know now.

Bernard is on the same track as Albert. He’s eliminated May and June. And he says he knows Cheryl’s birthday. If you look at our chart now, there are three singlet dates (15 for August 15, 16 for July 16, and 17 for August 17). If he was told 14, he wouldn’t know if it were July or August, so we can eliminate those.

From ten possible days, we’re down to three. And Albert’s final sentence finishes the job.

Albert says: Then I also know when Cheryl’s birthday is.

Since Albert is only told the month, it has to be July, because there are two possible dates left in August.

Therefore, in impressively brisk fashion, both Albert and Bernard have deduced that Cheryl’s birthday is July 16. And so have we!

We’ve also deduced that Cheryl is sort of a pain in the ass. But I suspect that wasn’t much of a brain teaser.

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