In looking for a job in finance, I have found that many companies like to use brain teasers as interview questions. Some of them are quite tricky, but they always have cute solutions. So, I'd like to open a thread dedicated to the subject. If it proves popular I'll sticky it. If it dies, it dies. Here are the rules: Post a brain teaser, or invent one. You can give hints, or not. You can post the solution, or not.
I'll go first. This one kept my friend and I busy for a while. Don't try to do it the hard way Please Register or Log in to view the hidden image! Suppose you have a bowl full of k noodles. You reach in with both hands, and draw two ends, and tie them together. You repeat this process k times, until no free ends remain. What is the expected number of noodle loops when you're finished? You don't have to evaluate the expression (at least, I couldn't).
It's k if k is 0 or 1. But if k is 2 then some of the time you expect to wind up with only 1 loop, so strictly less than k when k > 1.
See rpenner's response. It's a series of constants: N(k) = 1 + ... + f(k) It's your job to come up with the (...) and the f(k) Please Register or Log in to view the hidden image!
FTW! I think rpenner may have beaten you, even if he didn't post a solution Please Register or Log in to view the hidden image!
Next one: Suppose you and I play a game. We each have an infinite bag of coins and a round table. The game starts with the first player placing a coin on the table. The second player places a coin on the table and so on, until the table is covered. The winner is the last person to place a coin on the table. What is the strategy for winning, and would you like to go first?
ben.. i will confess a lack of knowledge when it comes to math..i didnt do too well in math till i got to algebra, i did have applied math with electronics..got an associate degree in electronics..got a job as a carpenter (paid better)(it has math..just not alot of it)..havent used it much since then..but have always thought i would have done good in it had i stuck with it.. (btw..didnt have to delete my previous post, thought it was funny..) i think i am trying to ask you to explain some references when it come to the higher math..like that e symbol..(epsilon, isnt it? but what does it mean..)
e is the ``natural number''. You can define it in a number of ways. For example, \(e = 1 + \frac{1}{2} + \frac{1}{3\cdot 2} + \frac{1}{4\cdot 3 \cdot 2} + \cdots\) Another way to define e is \(\int_1^e \frac{1}{x} dx = 1\). Either way, I changed the question Please Register or Log in to view the hidden image! I realized that that was more of a math question, and less of a brain teaser, in the spirit of the thread. Note: The question I erased was: Prove \(e^{\pi} > \pi^e\).
Yup, wouldn't surprise me. I was debating about posting the solution, but I figured the answer alone shouldn't give too much away for anyone trying to solve it on their own. <mouse over>You start with k open noodles in the bowl. You grab the end of a noodle. There's a one in 2k - 1 chance you grab the other end of the same noodle and create a loop. Otherwise you knot two noodles together into one (longer) noodle. Either way, there are now k-1 open noodles left in the bowl (and maybe a loop). The second time you knot two ends together you have a one in 2(k-1) - 1 = 2k-3 chance of creating a new loop and either way you're left with k-2 loops. And so on. Since the probability of closing a loop each time you tie a not is independent of which noodles you knotted on the other occasions, the expected number of loops is just the sum of the probabilities 1/(2k-1) + 1/(2k-3) + ... + 1/5 + 1/3 + 1.</mouse over>
Yes you can. If you do that with every noodle, you'll have k loops. Every time you tie two different noodles together, you'll end up with one less loop. And that's how you can find the answer (warning - poorly explained process follows): Tying two different noodles together reduces the number of ends by two. Tying both ends of a single noodle or chain of noodles together also reduces the number of ends by two, and it makes a loop. So, the chance of making a loop at any step is 1/(ends-1) At the first step, that chance is 1/(2k-1) At the second step, it's the same as if you started with (k-1) noodles. So clearly(?), the total expected loops is the sum given by przyk. Somehow, I doubt that makes any sense at all. It was easier to figure out than to explain. Haven't we done that here before? Calculator: \(e^\pi = 23.1...\) \(\pi^e = 22.4...\) 23.1 > 22.4 q.e.d. Go first in the centre, then symmetry is your friend! Anything you can do, I can do oppositely, Anything you can do, I can do too! My turn: A column of soldiers 1km long is marching in a straight line at constant speed. A jeep drives at constant speed from the rear of the column to the front of the column, then back to the rear (assume zero turnaround time). By the time the jeep returns to the rear, the column has advanced 4 km. How far did the jeep travel?
Ehrmmm... not sure there is one. Please Register or Log in to view the hidden image! It seemed to have been easier when I did it 20 years ago, but I'm damned if I remember how! Although I seem to recall that graphs and triangle were involved?
I'm not sure what you're thinking... Say you have five noodles. You tie two ends together, joining two noodles into one. You now have (effectively) four noodles, and no loops. So now the most loops you can make is four, right? Please Register or Log in to view the hidden image! Now that's just crazy talk! Why make it harder than it has to be?
