Friday Puzzler -- Licking Toads

Re: Friday Puzzler -- Licking Toads

Henry Higginbotham

This sounds a little like Marilyn vos Savant's golden pancake puzzle. I never fully wrapped my head around the logic on that one, but here's an attempt at applying it to this one.

The toad you're looking at has a 1 in 2 chance of being a male.

Of the two behind, one of which croaked and announced himself as a male, there are these possible arrangements:

M=the male that croaked

m=a second, bonus male

Toad 1 Toad 2

M F

M m

m M

F M

So picking either of the two would have a 3 in 4 chance of success.

Close!

Alex Y

But not quite right.

Not familiar with the pancake version; will have to look that one up.

Liking your Answer, Henry

Alex Y

It's not the one given where I saw this posted, but I can't reconcile the difference. Another way of saying your answer is that there is a 50% chance I will pick the croaker and be cured, and a 50% chance I will pick the toad that did not croak, and a 50% chance that one will be male and I am cured, so .5 + .5*.5 = 75%. I can't find fault with that.

The other analysis says there are two toads, one on the left (capitalized) and one on the right (lower case). The equally probable arrangements of those toads are:

Mm

Mf

Fm

Ff

But the fact that one of them croaked tells us that Ff is not the case. Then we are left with three cases, and whether you pick the one on the left or the one on right, there is a 2/3 chance it is male.

There is a very subtle flaw in one of these reasonings, and I'll be darned if I can find it.

Re: I tend to agree with you, but

Henry Higginbotham

If I'm remembering correctly, Marilyn made a distinction as to which side you were looking at of a pancake that was, I think, brown on both sides. As I said, I never fully bought in to her reasoning, but seeing as her IQ is 500 or something, I let it go.

On the other hand, to this day I disagree with her on a puzzle involving when and how much pay raise you'd elect to take.

But nobody's ever offered to let me write a column in the Sunday newspaper magazine, either.

Re: Also, in my answer

Henry Higginbotham

I used "M" to denote the toad that croaked, without regard to left or right position, and "m" to denote a possible second male. If there were only one male, he'd be "M" no matter his position. I'm not sure if that's different from how you summarized it.

Back to liking the 2/3 answer

Alex Y

Henry, I think I have finally figured this out. Let's use your notation. Before any frog croaked, the possibilities for the frogs sitting on the stump are:

Frog1 Frog2

m m

m f

f m

f f

And these are all equal possibilities. If we observed 80 stumps, we would find 20 with two males, 40 with one male and one female, and 20 with two females.

Now if we wait long enough before each stump to hear one croak. we will find that the 20 with two males will have 10 croaks from Frog 1 and 10 from frog 2. So now our distribution of stumps where we heard a croak will be

Frog1 Frog2 Count

M m 10

m M 10

M f 20

f M 20

f f 0

There is a 50% (30/60) chance we will pick the frog that croaked (5 from Mm, 5 from mM, 10 from Mf, and 10 from fM), and it will be male.

There is a 50% chance (30/60) we will pick a non-croaker (5 from Mm, 5 from mM, 10 from Mf, and 10 from fM), but of those 30 picks, only the 5 from Mm and 5 from mM will be males

So, in total, we will have picked a male in 40/60 cases, 67%.

Re: Back to liking the 2/3 answer

David Weaver

in front, 50% chance (but probably less than 50 in a simulation as the toad in front didn't croak male (Since it hasn't croaked yet).

behind assuming equal choices of your case (2/3)*0.5 + 1/3*(1) = 2/3

I wonder if there is some chance that because the back group of toads croaked first (in a simulation) that there's an increased probability that it's the 2 male toad group vs. one of the other two. That's maybe going a little too deep.

I'd drag my feet a little on licking the left toad of the double group just in case the other one croaks, though.