A math riddle for you. Feel free to submit your own.

Related to the riddle in the OP

A man asks his son: Right now I’m three times as old as you. in 10 years I will be only twice as old as you. How long before we’re the same age?

Ah, yes! That’s another variation I’ve heard. Forgot about that one.

There are lies, there are damned lies, and there are statistics.
Nonetheless, choosing between three doors yields a 33% success rate. Choosing between two doors gives a 50% success rate.

Only if the two doors have an equal probability of having the car behind it. Which is not the case in the classic example (The Monty Hall Problem) Mr Shine riffed off of.

This is the infamous “Monty Hall Problem” named after the original Let’s Make A Deal and it’s host. There is an existing thread that went on for a thousand pages.

“The only reason someone would add this option in after the fact is if you chose the car straight away” is not part of the Monty Hall Problem - in it he always offers you a switch after showing one of the doors (and he knows what’s behind the doors and will never show you the good prize).

And it’s actually quite simple: switching will improve your odds to 2/3 - your original pick is wrong 2 times out of three, in those cases switching will make you a winner. In other words, switching is only wrong if you originally had the correct door.

A stupid dad joke my dad used to tell me:

A forest covers 500 acres, how far can a tiger walk into the forest?

half way, after that the tiger is walking OUT of the forest

Right, which is why his answer differs from the original. I think Mr Shine is aware of how it usually goes. :smiley:

There’s at least a little bit of ambiguity in the phrasing of the question in the OP that might lead people to misunderstand.

“how old will the younger brother be if the older brother is 100?”

Could be interpreted as “Ok, now here’s another scenario where an older brother is 100 and a younger brother is half his age”.

I’m not saying that’s the best interpretation, but it’s at least a conceivably rational one. People are used to getting multiple “math questions” where part of the premise of the question stays the same (half his age) and another part varies (the age that is to be halved). It’s not 100% clear that the second question means “this same pair of brothers 96 years later”.

Here is one from Car Talk, abridged:

RAY: Potatoes are 99 percent water. So say you take 100 pounds of potatoes and you set them out on your back porch to dry out. After a while, enough water has evaporated so that they are now 98 percent water. If you were to weigh those potatoes at that moment…

TOM: 99 pounds.

RAY: You are wrong.

RAY: Now, unencumbered by the thought process as usual, my brother guessed 99 pounds.

TOM: Yeah.

RAY: Now, when I guessed, off the top of my head, I guessed about 90 pounds.

TOM: 'Cause it just feels right.

RAY: But if you do the math, 1 percent of 100 --which is what the potato is-- is one pound. As we told you, that’s 1 percent. So 2 percent, when it’s 98 percent water, two percent of the new weight of the mass is still going to be equal to that one pound, and 2 percent of 50 pounds is a pound. So the potato weight is now 50 pounds, not 100.

I agree. It can be read to be saying “One brother is twice as old as the other. If the older brother is four, then the younger brother is two. If the older brother is one hundred, then the younger brother is fifty.”

I think the second part of the question should be “how old will the younger brother be when the older brother is 100?” to eliminate that ambiguity.

A variation of a joke from “Hee Haw”:

At the age of 30, a woman gives birth to a daughter.

Ten years later, the daughter is 10, and the mother is 40, or four times as old.

Five more years later, the daughter is fifteen, and the mother is 45, or only three times as old.

Fifteen more years later, the daughter is thirty, and the mother is 60, or only twice as old.

Question: how long until they’re both the same age?

ISTM that you’d be on solid ground if the sentence read “when the older brother is 100.”

But “**if **the older brother is 100” - OK, we’re tossing out the notion that he’s 4 now, and replacing that by his being 100 now, not 96 years from now, but keeping everything else the same. In that case, we’ve got to choose between little brother being half his age (in the sentence proper) or two years old (in the parenthetical). So either 50 or 2. I’d insist on 50 as well.

Here’s one that I came up with myself (well, took an extremely simple puzzle and complicated it):

My first job was for a bottled tea company. We sold tea in cases of 100, and we sold three kinds of cases: sweet tea, unsweet tea, and assorted (50 each sweet and unsweet). The tea looked the same, only the labels were different.

I was the delivery driver, and my first day, I had a case of each to deliver. Just as I was getting ready to leave, my boss got a call from the bottling factory. “We got a problem,” the bottle-labeler said. “I noticed that the labeling machine was broken, so none of the bottles got labels.”

“Crap,” my boss said. “Okay, we’ll send them back. Put sweet labels on the ones in the sweet case, put unsweet labels on the one in the unsweet case, and I dunno, throw away the ones in the assorted case, no fixing that one.”

“Uh, there’s another problem. I noticed that the case labeler mixed up the labels. There is definitely one case of each, but it put the wrong label on each case.”

“Dammit. Which label went on which case?”

“Uh, I forget?”

My boss hung up. “Okay, first thing we do, we fire that guy, that guy is terrible. But now we’re gonna have to start tasting bottles of tea until we figure out which case has both sweet tea and unsweet tea in it, and then we’ll be able to figure things out, but who knows how many bottles we’ll have to go through and waste? This is a disaster!”

“Actually,” I said, “I can fix it by tasting a single bottle: one sip, and I’ll tell you which label to put on which case. As a bonus, I can tell you how to salvage the case of assorted tea.”

How did I do it?

Here is one that’s a bit harder. You core an apple and the cored apple that remains is 4 inches high. What is its volume (the cored apple). Assume the apple started out as a perfect sphere.

You could do this using calculus (I have checked the answer that way), but if you simply assume that there is an answer independent of the original size of the apple, you conclude that it must be the same if the core is 0 and the apple was originally 4" high. In that case its radius is 2" and its volume is 4pir^3/3 = 32*pi/3

The first part is a classic problem I’ll just leave alone. For the second,

[spoiler]Mix them all together and market the bottles as a special “Sweet tea lite”?

Edit: No, just got it, mix them all together and then add the right amount of sugar to bring it up to sweetened[/spoiler]

Do we read the “wrong label on each case” explicitly. Not, like, the labels were put on in an unknown manner, but for sure, each case is something other than it’s labeled?

If so,

taste a bottle from the case labeled “assorted”. If it’s sweet, that’s the sweet case. If it’s not, that’s the unsweet case. Put the correct label on it, put the assorted label on the other labeled box, put the other label on the box you took a label off to put on the box you tasted from.

One of my favorite math puzzlers is with a standard 5-card hand from a standard 52-card deck.

One player looks at the 5 cards, removes one card face down. Passes a pile of the other 4 cards to the second player. The second player looks at the 4 cards and names the face-down card he hasn’t seen. The two players are allowed to come up with a strategy before playing, but they don’t know which 5 cards are in the hand.

How do you do it?

I came up with a different solution.

Label the assorted case as mystery tea and sell it as is.

“When you hand me the four cards, say the name of the card you kept out loud.”

Decide on an order of the deck (eg. A-K of hearts, then A-K of diamonds, then clubs, then spades). Thus each card has an index number (1 = ace of hearts, 2 = deuce of hearts, etc.) When the first player gets the 5 cards, he removes a card which has index number N, waits N seconds, then pass the four cards to the other player.

Well, if you’re going to go to the trouble of numbering all the cards in the deck, you can Have the first player carefully reorder and position the cards - there are 24 different ways to order four cards with different values, and 16 different ways to position four cards where either can be upright or sideways. That means there’s 384 different ways you can pile up four arbitrary cards, so with some hardcore preplanning and memorization the identity of the removed card can be encoded by the first player and interpreted by the second.