Brain teaser thread - Page 13

Urban Ranger · October 29, 2002, 22:58

1/9+1/8+1/7+1/6+1/5+1/4+1/3+1/2+1

Dauphin · October 29, 2002, 23:09

Nope.

Urban Ranger · October 29, 2002, 23:23

Pah

Dauphin · October 29, 2002, 23:28

As an empirical clue check out what the expectations would be if there were just 1, 2, or 3 jars.

Capt Dizle · October 29, 2002, 23:59

One

Ramo · October 30, 2002, 00:16

Quote:

I've got one, that I'm now kicking myself over. It is deceptively simple.

There are nine jars, each containing a different type of liquid, but the labels have all fallen off. Knowing nothing about the contents, a passerby reapplies the labels at random. What is the expected number of correctly labeled jars?

1*1/9(7/8)⁸ + 2*1/9*1/8(6/7)⁷ + 3*1/9*1/8*1/7(5/6)⁶ +...

I'm not sure about that, though. I hate combinatronics.

Capt Dizle · October 30, 2002, 00:20

The chance of labeling a jar correctly is always going to be one in nine. or for three jars one in three. You can expect to get one label correctly.

Dauphin · October 30, 2002, 00:57

One is correct. As I said, its deceptively simple.

If you consider all the possible permutations, 1/n will have the first label on the correct jar, 1/n will have the second label on the correct jar, etc

Add up the expectations and you get 1/n x n = 1.

Ramo · October 30, 2002, 01:01

I don't know if your reasoning is sound (though I also have a suspicion that the answer is one, given SD's hints)... If you take a summation of all probabilities in any system, you'll always get one.

Capt Dizle · October 30, 2002, 01:03

I can only get the simple ones. I am deceptively simple.

But this one I knew from studying the statistical probablities of six sided dice.

Urban Ranger · October 30, 2002, 02:21

What's that? All the numbers have an equal chance of coming up?

Dauphin · October 30, 2002, 10:43

Quote:

Originally posted by Ramo
I don't know if your reasoning is sound (though I also have a suspicion that the answer is one, given SD's hints)... If you take a summation of all probabilities in any system, you'll always get one.

The expectation = permutations with x correctly placed * x / permutations; summed for all x.

This is the same as saying

The expectation = permutations with nth jar correctly labelled * 1/permutations; summed for all n

All you are doing is summing in the order of jar 1, jar 2, etc rather than in the order of permutations with 0 correctly labelled, permutations with 1 correctly labelled, etc..

Urban Ranger · October 30, 2002, 23:05

Okay, since jimmy ran away without posting a question, would somebody do that?

Dauphin · October 30, 2002, 23:10

A contractor estimated that one of his two bricklayers would take 9 hours to build a certain wall and the other 10 hours. When the two bricklayers worked together, however, 10 fewer bricks got laid per hour. With both men working on the job it took exactly 5 hours to build the wall. How many bricks did it contain?

SnowFire · October 30, 2002, 23:38

Ramo: But that's an expected value, not a sum of probabilities. We expect 1 hat to be correct each time.

As for the new problem: By rates we can quickly see that one rate that makes sense for the inital constraint (and a small wall with lazy workers) is
10 for the fast guy and
9 for the slow guy. 90 bricks total in the wall.
These numbers are nice. Under the assumption that the answer will be a nice integer, just take two multiples randomly and look at how fast we're "catching up" (inelegant elementary school style of doing it, but it works).

30
27
At rate of 47: 235 bricks laid in 5 hours out of 270. 35 behind.

40
36
At rate of 66: 330 bricks laid out of 360. 30 behind.

So each increase of 10 bricks/hour in the fast guy closes the gap by 5 bricks. We're 30 bricks off with 40/36. So let's increase 6 times, for
100
90
At rate of 180: 900 bricks out of 900.

So 900's the answer.

Ramo · October 30, 2002, 23:45

SD and SF, I was responding to jt's post (I crossposted).

JohnM2433 · October 31, 2002, 00:03

Quote:

Originally posted by Jaguar Warrior
1 C and 1 M go over
1 C goes back
2 C go over (1 stays in boat)
1 M and 1 C go back
2 M go over

Danger! Danger, Will Robinson!

There are now 2 cannibals and 1 missionary on the side of the river they started out on, so clearly this answer does not work. (You can see this easily by using pennies and dimes to represent the cannibals and missionaries and moving them back and forth.)

And you can still solve the problem if you consider all the people on one side of the river to be "in the same place" whether they are in the boat or not. Your interpretation makes sense, of course, but it makes the problem too easy to solve.

C'mon, you guys, this one is easy if you think about it properly...

Oh, and regarding the labels and the jars, it seems to me again that you guys are engaging in a lot of unnecessary complex math. Each label has a 1/9 probability of being on the right jar, and there are 9 labels, so clearly the expected number of correct labels is 1, the chance of a label being correct times the number of labels. Or equivalently, each jar has a 1/9 probability of being correctly labeled, and there are 9 jars, so the expected number of correctly labeled jars is 1, the chance of a jar being correctly labeled times the number of jars.

JohnM2433 · October 31, 2002, 00:16

I got the number of bricks using some simple algebra, but I see that SnowFire posted the answer first. Since we got the same answer using different methods, I think we can say that it's the correct one even before Sagacious Dolphin's response.

My missionaries/cannibals problem is still waiting for a correct answer. Try doing it the hard way, where everyone on the same side of the river is always considered to be "in the same place". It's really not that difficult...

Dauphin · October 31, 2002, 00:24

Quote:

Originally posted by JohnM2433
And you can still solve the problem if you consider all the people on one side of the river to be "in the same place" whether they are in the boat or not.

2C ->
<- 1C
2C ->
<- 1C
2M ->
< - MC
2M ->
<- C
2C ->
<- M
MC ->

JohnM2433 · October 31, 2002, 00:45

Yep! If you realize that no outnumbering can occur in the boat, you can abstract the whole thing to one or two people crossing from each side every other turn. Then it's really fairly simple, since at most points there's only one possibility, and you can backtrack from what few dead ends there are. One just need keep in mind that if you have to make a move that undoes the previous move, you should backtrack, since the result is the same as doing nothing at all, and that can't help to lead to a solution. Soon, the remaining moves are obvious.

Personally, I would have sent a cannibal back for the remaining cannibal at the end, since the missionaries would presumably be even more pleased when they outnumber the cannibals. I don't know why I bother to mention that.

You've posted quite a few problems, so maybe you should give SnowFire a chance to post the next one.

SnowFire · October 31, 2002, 00:54

Yup, can easly be done by algebra, but 'twas lazy. Being that I got one, here's another puzzle... as I'm sure you all know, there's an obvious answer from some calculation, a reverse psychology "but!" possibility if it seems too straightforward, and the reverse-reverse psychology strategy of "wait! the obvious thing was right after all!" So answers without reasons are pointless.

A man is bragging to his friends about his incredible speed as a runner. "No matter how fast I'm running, I can always double my speed if I want to!", he says. Then he adds modestly: "Well, actually, I cannot go faster than the speed of light, of course". Naturally, his friends don't believe him, and decide to put him to the test. They take him to a 60 km long, straight section of road at an abandoned place in the countryside, and tell him to start running, and that they will be right behind him in their car. They have a gun, and tell him that every time he hears the gun firing, he should double his speed, but not look back. Behind his back, his friends have agreed to treat him to a nice dinner if he runs the 60 kilometres in 17 seconds or less. The man starts running at a speed of 2 m/s, and his friends in the car behind him fire the gun once every second. Assuming the man really is capable of doubling his speed with only the speed of light as the limit, and that the car can somehow keep up with him, does he get the dinner?

Urban Ranger · October 31, 2002, 01:11

60km/17s (3529.41m/s) is a lot faster than the speed of sound in air (330m/s). So at some point the man will no longer be able to hear the gun shots.

JohnM2433 · October 31, 2002, 01:47

Ooh! Good catch! That means that the man will never get up to 60km/17s, so of course his average speed will always be below that. Thus he will run the 60 km in more than 17 s, so he won't get the dinner.

I was about to post that his speed would be over 60 km/s during the 17th second, so he would have to have covered the entire distance after that second. But obviously that's not the case, since he stops hearing the gun.

Urban Ranger · October 31, 2002, 02:08

Thanks

I'm digging up a question. If any of you has one handy, feel free.

CapitanGarlic · October 31, 2002, 02:20

A man buys coconuts for 15$ (US, but that's just for the sake of argument) per dozen and sells them at $6 per dozen. (also $US) Doing this, he becomes a millionaire. How?

Urban Ranger · October 31, 2002, 02:26

That's just a cover for his arms smuggling empire

CyberGnu · October 31, 2002, 02:39

He started out a billionare?

SnowFire · October 31, 2002, 03:01

How do you make a small fortune in general aviation? Start out with a large one.

I expect the answer's similar. He started out a billionaire or bigger.

Edit: This is what happens when you leave your reply-to window open for 20 minutes without refreshment before actually posting. A bit late on this one.

CapitanGarlic · October 31, 2002, 03:04

yup.

CyberGnu · October 31, 2002, 03:21

So, in the spirit of things:

A man walks into a hotel on Monday and stays two weeks, but he leaves on Friday. How is this possible?