Since a few of these have surfaced recently in study hall, let's give ourselves a place to exercise our brains without having to focus on school. I'll start:

If you randomly break a broom stick into 3 pieces, what is the probability that it can be made into a triangle?

2/24/2010 4:51:38 PM

50%

2/24/2010 7:42:24 PM

what do I win?

2/24/2010 8:57:14 PM

^Incorrect, and no credit without work or reasoning.

2/24/2010 11:33:34 PM

you can form a triangle as long as the longest piece is no longer than half the original stick length

[Edited on February 25, 2010 at 12:08 AM. Reason : 25%]


[Edited on February 25, 2010 at 12:17 AM. Reason : numerically it works out to ~19% so i guess it's off to google for the real answer]

2/25/2010 12:04:03 AM

Suppose the stick is of unit length, and call the first break point A, so we have lengths of A and 1-A. Conditional on A < 1/2, we'll find the probability that the next break allows a triangle to be made; then, by symmetry, the unconditional probability will be 2 times the conditional probability (since P(A<1/2) = 1/2).

To form a triangle, the longest piece has to be shorter than half the unit length, like qntmfred says. Conditional on A, that means that break point B (chosen uniformly from 0 to 1) has to lie between 1/2 and (A + 1/2). (Now it should be clear we assumed A < 1/2 to ensure that (A + 1/2) is less than 1; the symmetry is that if A > 1/2, that means B should be between 1/2 and A - 1/2.)

Thus,

P(1/2 < B < A + 1/2 | A < 1/2) = (A + 1/2) - 1/2 = A.

Then the unconditional probability is the integral of the conditional probability over the possible values of A. So integrate A from 0 to 1/2, giving (A^2)/2 = (1/4)/2 = 1/8. Now recall we have to multiply that by 2 to get our final answer, which is 1/4.

2/25/2010 12:42:54 AM

^That is correct and very thorough, I never thought to do this with conditional probabilities and then integrate back out. Nice Job.

+1 for Shadowrunner

2/25/2010 9:14:26 AM

Next one. This another probability related question, I'll post some less math oriented ones when I can look at my books. This is also more well known so if you've seen it, refrain from spoiling it.

How many people must you have in a group for it to be more likely than not that at least 2 people were born on the same day of the year?

2/25/2010 9:40:38 AM

The answer to the original question is 100%. You did not say it had to be a perfectly fitting triangle. Any three sides of any length can be made into a triangle. It's just that some of the legs might keep going, but a triangle is still formed.

Your second question, and the answer to it, is going to assume that everyone's birthdays are evenly spread throughout the year. And that's not true. So, unless you're going to average demographic data on real births throughout the year, and specify a country or area in which the answer will generally hold true, the question is unanswerable.


[Edited on February 25, 2010 at 1:02 PM. Reason : a]

2/25/2010 1:00:00 PM

^First question clarification: the triangle must be physically realized by the broomstick pieces.
Second question clarification: This question assumes a uniform distribution of birthday's throughout the year.

These questions aren't supposed to be semantic brain teasers.

2/25/2010 1:20:04 PM

Well then that's no fun These aren't brain-teasers. They're just math problems.

2/25/2010 1:32:56 PM

I won't comment on the birthday thing since I already know it other than: once you consider non-uniform birthday probabilities it takes even less people so it could be worded as the maximum number of people to have a 50/50 chance, but the wording makes it more confusing.

a = b
a² = a*b
a² - b² = a*b - b²
(a + b)*(a - b) = b(a - b)
a + b = b
b + b = b
2b = b
2 = 1

2/25/2010 1:45:09 PM

Well then that's no fun These aren't brain-teasers. They're just math problems.

2/25/2010 3:00:05 PM

Quote :

"Well then that's no fun These aren't brain-teasers. They're just math problems."

Fair enough. These were the only two floating on the top of my head. I'll post some "real" brain teasers later.

[Edited on February 25, 2010 at 7:19 PM. Reason : .]

2/25/2010 7:17:11 PM

shadowrunner came up with 25% for the first one, as did I, but that is not correct. I simulated it and it came out to 19ish%. there's a solution on dr. Math website that explains why that is

2/25/2010 7:37:39 PM

http://www.exampleproblems.com/Solutions-Casella-Berger.pdf

Page 47, problem 4.12.

Do you have a link for this alternate explanation? I have the original manuscript containing this problem at home and the answer might vary depending on how you define "break a stick randomly into 3 pieces".

2/25/2010 8:01:25 PM

here's one that says 1/4 http://mathforum.org/library/drmath/view/70224.html

here's the one that says .19 http://mathforum.org/library/drmath/view/70602.html

here's a c# program i wrote to simulate it

        static void Main(string[] args)
        {
            var rand = new Random();
            int validTriangles = 0;

            for (int attempts = 0; attempts < 100000; attempts++)
            {
                var a = rand.NextDouble();
                var b = rand.NextDouble() * (1.0 - a);
                var c = 1.0 - a - b;

                if (a < .5 && b < .5 && c < .5) validTriangles++;
                Console.WriteLine(String.Format("{0}/{1} = {2}", validTriangles, attempts, 1.0 * validTriangles / attempts));
            }

            Console.ReadLine();
        }

2/25/2010 8:07:57 PM

The problem is in how you break the stick. If you randomly and uniformly pick two independent points, the probability is 1/4. If however, you break it in two, and then break the larger of the two (as you did in your simulation), then the probability is 1/6.

This is a pretty nerdy day, I found a copy of the article sitting on my AFS space:

http://www4.ncsu.edu/~albeam/files/GARDNER02.pdf

I think that the point of the article is well demonstrated by this thread.

2/25/2010 8:17:51 PM

i was just reading them again, and was about to post that distinction as well. ok, how about another one

2/25/2010 8:19:06 PM

Ok, since you asked. Like I said all I have in my head are statistics/probability ones right now. I'll definitely get around to ones that are not math oriented at some point.

Suppose a swami puts m dollars in one envelope and 2m dollars in another. You and your opponent each get one of the envelopes at random. You open your envelope and find x dollars and the swami asks if you want trade envelopes with your opponent. You reason that you will either get x/2 or 2x dollars if you make the switch. Your expected value of switching is (1/2)(x/2) + (1/2)*(2x) = 5x/4, which is greater than the x dollars you currently have. So you accept the trade.

However, your opponent has made the exact same calculation. How can the trade be advantageous for both of you?

Have fun with this.

2/25/2010 8:28:55 PM

The a=b one is not possible when you add the -b^2. Plug in a number for the two.

a=4
b=4

4=4
4^2=4^2 => 16=16
4^2 - 4^2=4^2-4^2 => 16-16=16-16 => 0=0
Everything below becomes not true.
(4+4)*(4-4)=4(4-4) => (8)*(0)=4(0) ---not true.
4+4=4 ---not true
4+4=4 => 8=4 ---not true
2(4)=4 => 8=4 ---not true
(Assume 2(4)/4=4/4)
2=1 ---not true.

It took a bit to figure out which line messed up the logic.

2/25/2010 8:47:46 PM

^ the standard answer to that one is that the step where you divide by (a-b) is invalid, because a-b=0. No dividing by zero.

2/25/2010 9:00:32 PM

If you open the envelope and find $x, then either $x = m or $x = 2m. Once you open the envelope, the contents of the other envelope are no longer 2x with probability 1/2 and x/2 with probability 1/2; the other contents are known with certainty.

This problem is again pretty dependent on the wording of the question, and the knowledge each player has. The above assumes that the value of m is known to the players. If that's not true, so that each person knows the value of their envelope privately is unable to deduce the other's, the simplest explanation is that the outcomes for each player are not independent, and both players know that it's a zero-sum game.

[Edited on February 25, 2010 at 10:16 PM. Reason : ]

2/25/2010 10:16:05 PM

Quote :

"If you open the envelope and find $x, then either $x = m or $x = 2m. Once you open the envelope, the contents of the other envelope are no longer 2x with probability 1/2 and x/2 with probability 1/2; the other contents are known with certainty."

You don't know the value of m and it is completely arbitrary. So when you observe the amount in your envelope = x, you do not know whether the other envelope contains 2x or x/2. However, the expected value calculation as defined, should still be valid. Thus the paradox.

2/25/2010 11:17:09 PM

OK, it wasn't clear from your statement if you intended that the players knew the value of m. At this point, I should make full disclosure that I've seen this paradox before but felt like I could chip in an answer based on the vagueness of your wording, since that allowed an interpretation I hadn't considered before.

I can math it up if you want, but a proper resolution probably involves invoking Bayes' Theorem and drawing a distinction between prior and posterior probabilities. With your setup, it would be better not to open your envelopes at all and just keep swapping it back and forth forever, creating an expected value that diverges to infinity. The problem is that you're not actually dealing in unconditional probabilities; each swap is conditional on the previous envelope having a finite and fixed, though unknown, value. Evaluating the conditional expected value requires some prior belief about the distribution of values in the envelope, but there's no prior distribution that would always give posterior probabilities of 1/2 and 1/2, no matter the value you find when you open your envelope.

It's probably simpler to consider that clearly, swapping twice has an expected gain of $0, since you end up with the same envelope you started with. This is immediately a contradiction of the idea that when considering each switch in isolation, you have a positive expected gain each time. So then it must be that the expected value of swapping three times is the same as swapping once, since the second and third swaps cancel each other. So all we need to think about is the single swap. Here the zero-sum game comes into play; if I end up winning, then I win $x and the other guy loses $x (keeping in mind that my x, x_me, cannot be the same as the x he's thinking about, x_his; if I "win," that must mean that x_me = x_his/2). So the correct expectation based on the *prior* probabilities is 1/2*x + 1/2*(-x) = 0. It's based on the prior probability of choosing the envelope with the high/low amount, not the posterior probability of the other envelope having 2*x conditional on my envelope having x.

I recommend future problems be less mathy as well, or if you give math problems, they should be more about finding a clever insight that makes a problem easy. This one is fun from a logician's point of view, but the actual resolution is too technical to make a successful thread.

2/26/2010 12:18:04 AM

Quote :

"the standard answer to that one is that the step where you divide by (a-b) is invalid, because a-b=0. No dividing by zero."

http://en.wikipedia.org/wiki/Mathematical_fallacy

2/26/2010 7:48:11 AM

math problems go that way =====================>

We're taking over.


1) A man rode into town on a Friday. He stayed a total of three days and left on a Thursday. How is this logically feasible?


2) You are in an enclosed box. No windows, doors, etc. The walls are impenetrable. You have a ball, a bat, and a bed. How do you get out?

[Edited on February 26, 2010 at 9:52 AM. Reason : googling is for cheaters who wanna look smart on the internets]

2/26/2010 9:52:00 AM

One that was posted here previously:


You have two identical eggs that can withstand a drop from some unknown height without breaking. You want to find out the height at which the egg breaks by dropping it off a balcony of a tall building. You are given a building 100 stories tall and are told the egg will break somewhere within that hundred stories. What is the lowest number of drops you can make with these two eggs and still guarantee you will find the right floor?

Example of a bad solution: Go to floor 50, drop an egg. If it breaks, go back down to floor 1 and work up one floor at a time. If it doesn't break at 50, then go to 75 and try again. The maximum number of drops needed in this solution is 50 -- very inefficient. Find a more efficient solution.

[Edited on February 26, 2010 at 11:33 AM. Reason : ]

2/26/2010 11:29:53 AM

My immediate thought is that for each drop, you pick the middle number of the possible range (rounding obviously necessary). This way, you have the most number of floors that are guaranteed to be eliminated each turn.

I'm counting 7 drops at most for this method. For some reason, I think it's because (1/2)^7 is less than 1/100.

Because of rounding, 6 drops is possible, but 7 is guaranteed.

Example (Answer is 78) - 50, 75, 88, 82, 79, 77, 78.

Example (Answer is 45) - 50, 25, 37, 43, 47, 45.

[Edited on February 26, 2010 at 12:14 PM. Reason : a]

2/26/2010 12:13:25 PM

Example (Answer is 78) - 50, 75, 88 [break], 82 [break], 79 [break], 77, 78.

Example (Answer is 45) - 50 [break], 25, 37, 43, 47 [break], 45.

You're using too many eggs. The oscillating method works well if you have an unlimited number of eggs, but for this particular problem, you only have two.

2/26/2010 12:20:45 PM

lol, the guy who was anal about wording on the first two failed to read the question

2/26/2010 12:30:54 PM

Is the right answer 14?

2/26/2010 12:40:00 PM

indeed it is.

2/26/2010 12:41:26 PM

Word. I had an idea for an algorithm I thought was pretty good, and it gave me 14, but I didn't know if there might be something else very different that I hadn't considered.

2/26/2010 4:14:51 PM

We had this one in my networking class when we learned about flood control. The answer is similar to how TCP does it.

2/26/2010 4:41:44 PM

We know that once an egg breaks, we have to go down to one floor above the highest floor that we know won't break. So our worst case in every scenario is going to be at least the height of the floor that you start on. Then if we can make our starting floor and our largest jump the same distance, we know this is an optimal solution. do do this, we need to decrease our jump size by one each move (since we lose a remaining jump each time) Therefore we need a number X such that the sum from 1 to X is > 100, but the sum from 1 to X-1 is < 100. 14 is that number

Therefore we should start on floor 14 and drop the egg. if it breaks, go to the bottom and work your way up. if it doesn't break, go to floor 14 + 13 = 27 and drop, if it breaks go to 15 and work up, else repeat with decreasing spaces on your way up. you'll hit floors 14, 27, 39,50,60,69,77,84,90,95,99,100.


Nice thread

2/26/2010 5:20:39 PM

OR the egg will break after being dropped from the 1st floor, because it’s an egg.

2/27/2010 8:13:03 PM

DON'T TEASE ME BRO !

2/28/2010 7:22:46 PM

Quote :

"(8)*(0)=4(0) ---not true"

you're saying 0 =/= 0?

Quote :

"1) A man rode into town on a Friday. He stayed a total of three days and left on a Thursday. How is this logically feasible? 2) You are in an enclosed box. No windows, doors, etc. The walls are impenetrable. You have a ball, a bat, and a bed. How do you get out?"

1) his horse's name is Friday.

2) i feel like it has something to do with baseball (batters box?) but the bed is throwing me off

[Edited on March 2, 2010 at 11:24 AM. Reason : -]

3/2/2010 11:24:09 AM

rode into town on a Friday.... wouldn't make sense if his horse's name is Friday


He left wherever he was outside of town on a Thursday and rode into town on Friday

[Edited on March 4, 2010 at 7:38 PM. Reason : ..]

3/4/2010 7:36:54 PM

The horse one is he arrives on a Friday, leaves and comes back several times, is there for a total of three days throughout that, and finally leaves for good on a Thursday

3/7/2010 3:47:24 PM

Gator Riders were caught trespassing by the Swamp Dwellers. The ruthless Chief of the Swamp wanted to toy with his prisoners. He told them at daybreak the next morning, he would give them a test and each man who passed the test would be set free and those who failed would be fed to the alligators. He told them he would line them up in a row, face-to-back. He told them he would place a snake skin band around each of their heads and they would only be able to look straight ahead. Some of the snake skins would be from a rattler (light colored) and some from a cottonmouth (dark colored). He told them they would not be able to communicate in any way – verbally, body motions, touch, or any other way – or they would all be fed immediately to the alligators. Also, they would not be able to see the snake skins on their own heads and they could only look straight ahead. He said he would ask each to guess the color of the snake skin on their own head – light or dark - starting with one in back who could see the snake skins of each of the 19 men in front of him (but not his own). If he guessed correctly, he would be set free but if he guessed wrongly, he would be fed to the gators. Then Chief Swampy would ask the next man who could see the skins of the 18 men in front of him (but not his own) and so forth until all had their guess.



The 20 men had all night to make a plan to save as many as possible. How many could you make a plan to save?

3/9/2010 6:54:24 PM

19. The first guy shouts out in code how many black and how many clear snake skins. Then he dies. Everyone else counts how many of each are in front of them

3/9/2010 9:13:32 PM

^the code can be as simple as saying "light" or "dark" to signify the even number of that particular snake skin. For example:

Guy 20 sees 7 darks and 12 lights so he says "light" (he has a 50/50 shot)
So if guy 19 sees an odd number of light skins, he knows his is light (or if he sees an even number of light skins he knows his is dark)
Guy 18 can count the number of light hats he can see and after hearing guy 19's color can use the same logic. Ex: let's say guy 19 was light. If he sees 10 lights plus 1 light behind him, he knows his is light (to make an even number). If he sees 11 lights plus 1 light behind him he knows his is dark (the extra light would make an odd number).

etc. for 17 through 1.

So it's basically a 50/50 chance that the last one in line will live, but the rest have a 100% chance assuming they can follow this strategy.

[Edited on March 10, 2010 at 10:38 AM. Reason : ]

3/10/2010 10:35:19 AM

I think the first guy lives. The first guy up doesn't want to go down alone. He says one code word per ^, but incorrectly. He gets lucky and it's right, but it starts the vicous cycle of killing everyone else. He rejoices!

In all seriousness though, I think the first guy is toast no matter what, but the others can live.

3/10/2010 12:48:31 PM

How do you solve this one? Sounds like it will be a non-complex one, but at the same time, seems impossible.

Quote :

"A woman is standing in front of the house of a neighbor couple who just moved in. The husband says "We have three daughters. Interestingly, the sum of their ages is the number on our house." To which the wife adds "And the product is 36." The woman replies that she needs one more piece of information in order to determine the ages of the girls. The wife responds "Our oldest daughter has red hair." What are the ages of the three daughters?"

5/7/2010 10:48:07 PM

1, 4, 9

5/7/2010 11:43:11 PM

How so? There are many possibilities.

Somehow that red hair clue is supposed to nail one possibility, but I can't see how.

5/8/2010 4:58:26 AM

I looked up the solution. Want a hint?

5/8/2010 10:24:50 AM

^yes

5/8/2010 3:43:24 PM