I want to be a bugbear too!
But mr Perkins is no bugbear, he’s a hugbear.
Exactly, and so I am looking forward to sharing this twist with my class, thus advertising calculus.
As an aside, about 100 years before Newton and Leibniz, Galileo figured out that if acceleration is constant, then the distance moved is proportional to the square of the time that has passed. So he essentially integrated a linear function using only the techniques of algebra.
Key to his discovery was that the sum of the first N odd natural numbers is the square of N.
He was also well known as a voracious hugbear.
You, sir, are quite clearly an umberhulk, not a bugbear!
Umberhulk sleepy! You no like umberhulk sleepy. zzzz
MikeJ
1766
If we want them to use calculus, it should be a cannonball problem, where they have to take into account air resistance. Fun times for everyone!
So today at lunch my friend said:
You’re in the jungle and you’re bitten by a poisonous snake! Fortunately, there’s a cure: you have to find this particular frog and suck on it. Unfortunately, you have to suck on a female. Males don’t cure you. Fortunately, males can croak but females can’t.
To your right, you see one such frog sitting on a stump. It isn’t croaking, so it might be a female. To your left, you see two frogs, and one of them croaks. You don’t know which.
You only have time to make your way over to the silent frog or the pair of frogs. If you choose the pair, you have time to suck them both.
Which do you choose? The silent frog that might be a female (50% chance), or the pair of frogs, one of which croaked (and thus is a useless male)?
So he asks me this and I was like lol I know some guys who will want to hear this.
I choose to eat lunch with someone who isn’t going to make me do math.
-Tom
Pair of frogs? On the basis that there is only a 25% chance that both frogs are male?
Probably wrong but what the hell. I blame the poison for my bad choices.
RichVR
1770
I won’t worry about it. Why? A venomous snake is dangerous if it bites me. A poisonous snake is only dangerous if I bite it. :)
For a cartoon presenting the wrong answer, see this TED talk: https://www.youtube.com/watch?v=cpwSGsb-rTs
For a stackexchange post presenting a correct answer, click here: http://math.stackexchange.com/questions/1683658/the-frog-puzzle/1683796#1683796
Yes, this is essentially the same puzzle as the fabled Creepy Window Boy.
Yup! :)
And Rich, I was just quoting my friend. When he said “poisonous”, I heaved a great sigh and made eye contact with various people, looking for someone like you to commiserate with. But I found no one.
I clicked on that link and my head literally exploded. Literally.
-Tom
Hmm. The ‘correct’ answer is based on the fact that one of the two frogs in the pair has not (yet) croaked.
But the frog on your right also has not (yet) croaked. So “The silent frog that might be a female (50% chance)” is wrong. There is exactly the same > 50% probability that the frog on the right is female. How much more than 50% depends on how often the males call - but that calculation is identical to the one based on “one of the pair to my left has not (yet) croaked”.
I think.
Well presumably the males croak to attract females, so the correct answer, after hearing the croak for yourself, is to mimic the call and have all the ladies come to you!
The stackexchange answer leaves the probability that neither are male frogs, where the riddle as posited by Dave says “one of them croaks, you don’t know which”. I think it’s important to say “You hear a frog croak” rather than “one of the frogs definitely croaked”. In fact, right at the beginning of the stackexchange answer, he says
I would argue that the way it was posted here implies that one of the frogs is male.
OK, this doesn’t work. I tried it in the office and the results were not forthcoming. Quite the opposite, in fact, I think some of the ladies moved further away.
That proves only that the ladies in your office are not frogs. Just as well, as licking one of them would have been your last act in your job. .
MikeJ
1780
Well, it enumerates the possibility, but the conditional probability assigned is zero, so that’s fine.
It seems that way to me as well. If I expand out the right-hand frog in the same way, I also get (1/2-p), which makes sense.
