Brain teaser thread

[Pekka]

I don't have the energy to think but quickly going through options it must be 3 or 4.. I'd go for 4.

[Urban Ranger]

More than that.

[Urban Ranger]

Balls are not replaced: twice

Balls are replaced:

There are 5 configurations (I use letters instead of colours, it's the same) -
ABCD (original, config 0)
AACD (config 1)
AAAC (config 2)
AACC (config 3)
AAAA (config 4, goal)

config 0 -> config 1 : 1 turn
config 1: 1/2 chance remaining config 1, 1/3 chance becoming config 2, 1/6 chance becoming config 3
config 2: 1/2 chance no change, 1/3 chance becoming config 4 (done), 1/6 chance becoming config 3
config 3: 1/3 chance no change, 2/3 chance becoming config 2

Then I forgot how to convert this into expectations

[Dry]

Originally posted by Urban Ranger
[...]config 1: 1/2 chance remaining config 1 [...]

So you may never reach config 4.

[Urban Ranger]

That's true.

[Dauphin]

When you toss a coin there is a possibility that it will never come up heads, but the expected number of throws to get a head is still finite at 2.

[JohnM2433]

Originally posted by Zero-Tau
All rooms with square numbers have their lights flicked on. Their prime factorization would be a1^e1*a2^e2*a3^e3... with all exponents being even. Using the formula for the number of divisors (e1+1)(e2+1)(e3+1)... we see that the square numbers have an odd number of divisors, meaning that they'll end up with the lights on. Conversely, if the number of divisors is odd, then each of e1+1, e2+1, e3+1,... would be odd, meaning that all the exponents would be even, making the number a square.

Woah, I didn't even follow all that! As soon as Sagacious Dolphin asked what property a number needs to have an odd number of factors, I just figured: "For every number, each factor will be matched with exactly one corresponding factor so that the two together multiply to the number. A factor will be matched with itself if and only if it is a root of the number, so square numbers and only square numbers will have an odd number of factors, since the factors will come in pairs, except if the number is square, in which case the root will be by itself."

My way seems a lot simpler to me. (My explanation is a bit more wordy than my actual reasoning is complicated, trust me.)

[JohnM2433]

Originally posted by Urban Ranger
Balls are not replaced: twice

Balls are replaced:

There are 5 configurations (I use letters instead of colours, it's the same) -
ABCD (original, config 0)
AACD (config 1)
AAAC (config 2)
AACC (config 3)
AAAA (config 4, goal)

config 0 -> config 1 : 1 turn
config 1: 1/2 chance remaining config 1, 1/3 chance becoming config 2, 1/6 chance becoming config 3
config 2: 1/2 chance no change, 1/3 chance becoming config 4 (done), 1/6 chance becoming config 3
config 3: 1/3 chance no change, 2/3 chance becoming config 2

Then I forgot how to convert this into expectations

I think config 2 should be: 1/2 chance no change, 1/4 chance becoming config 4 (done), 1/4 chance becoming config 3. The rest appear to be right.

Assuming that's correct...

The expected number of turns to get to config 4 from config 0, 1, 2, or 3 is 1 plus, for each possible next state, its probability of coming next times the expected number of turns to get to config 4 from that new state. Representing the expected number of turns to get to config 4 from config x by cx, we get

c0 = 1 + c1
c1 = 1 + (1/2)c1 + (1/3)c2 + (1/6)c3
c2 = 1 + (1/2)c2 + (1/4)c3 + (1/4)c4
c3 = 1 + (2/3)c2 + (1/3)c3
c4 = 0

After a lot of algebra, we can conclude that c0 = 9, c1 = 8, c2 = 5.5, c3 = 7, and of course c4 = 0.

So the expected number of turns after the first draw is 8.

[JohnM2433]

OK, most of you should find this one fairly easy:

Three missionaries and three canibals are traveling together and come to a river. None of them can swim, but luckily, there is a boat tied to their side. It can only hold two people. For obvious reasons, the missionaries would prefer that no group of canibals ever outnumber a group of missionaries (where a group can be one, but not zero) at any one place. Figure out a way that they can maintain that condition and get everybody across the river.

If that's too easy for you, you might want to try the more general case of x missionaries and x canibals (I don't actually know whether that can be solved or not).

[Jaguar]

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
1 M goes back
1 M and 1 C go over

[Dauphin]

Originally posted by JohnM2433

So the expected number of turns after the first draw is 8.

9 draws in total.

I'm glad you got the same answer as I did, because I had to work it out myself too. It would have been embarrasing if my calculated answer was wrong.

[Jaguar]

What do these words have in common:

Calmness
Canopy
Deft
First
Hijack
Laughing
Stupid

[Jaguar]

rowr, new avatar!

[Dauphin]

They have 3 consecutive letters that are also alphabetically consecutive.

Calmness
Canopy
Deft
First
Hijack
Laughing
Stupid

[Dauphin]

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?