Brain teaser thread

[Urban Ranger]

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

[Dauphin]

Nope.

[Urban Ranger]

Pah

[Dauphin]

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

[Capt Dizle]

One

[Ramo]

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]

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]

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]

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]

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]

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

[Dauphin]

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]

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

[Dauphin]

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]

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.