Who doesn’t love a good mathematical riddle? Challenge yourself to one of these brain teasers, and whip them out at your next social gathering.
The Potato Paradox
Q: Say you have 100 pounds of potatoes, which are 99% water by weight. You let them dehydrate until they’re 98% water. How much do they weigh now? The surprising answer is 50 pounds!
A: For 100 pounds of potatoes with 99% water weight, you have 99 pounds of water and 1 pound of solids. This is a 1:99 ratio. If the water decreases to 98%, then you have 2% of solids. This is a 2:98 ratio, which reduces to 1:49. The weight of the solids never changed, so you have 1 pound of solids and 49 pounds of water, so the new total weight is 50 pounds.
The Birthday Problem
Q: Imagine you have n randomly-chosen people in a room. What is the probability that some pair of them share a birthday?
A: This problem has an interesting and unexpected solution. It follows from the solution that there’s a 50.7% chance that in a room with n=20 people, a pair will share a birthday. And there’s a 99.9999% chance that if you have n=200 people in a room, there will be a pair that shares a birthday. So, if you’re ever in a room with 19 other strangers, there’s a greater chance that you share a birthday with one of them than if you were to flip a coin and get tails.
The Monty-Hall Problem
Q: Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
A: Yes, you should switch! Counterintuitively, you would have a ⅔ chance of winning the car if you switch your choice, and a ⅓ chance if you stick with your choice.
Hilbert’s Paradox of the Grand Hotel
Q: Consider a hypothetical hotel with a countably infinite number of rooms, all of which are occupied. Can the hotel accept any more guests?
A: Yes, in fact, it can accept infinitely many more new guests. You can think about it this way: in order to make room for the new guests, move the guest occupying room 1 to room 2, the guest occupying room 2 to room 4, and in general, the guest occupying room n to room 2n. This would make all the odd-numbered rooms free.
The Barber Paradox
Q: The barber is the “one who shaves all those, and only those, who do not shave himself.” Does the barber shave himself?
A: This is an application of Russell’s Paradox, which deals with sets that contain themselves. The answer to this barber question is contradictory. The barber can’t shave himself as he only shaves those who do not shave themselves. So if he shaves himself, then he is no longer a barber.