Brain teaser thread - Page 16

November 20, 2002, 02:20

ok no "real" posted for a long time
I like this one:
"In a rectangular array of people, which will be taller, the tallest of the shortest people in each column, or the shortest of the tallest people in each row?"
(with an explanation plz as there are only 2 choices)

Im going to sleep within the hour, so if you feel relatively confident in your answer, or by popular demand, post your own...

One_Brow · November 20, 2002, 13:41

Quote:

Originally posted by LulThyme
ok no "real" posted for a long time
I like this one:
"In a rectangular array of people, which will be taller, the tallest of the shortest people in each column, or the shortest of the tallest people in each row?"
(with an explanation plz as there are only 2 choices)

Im going to sleep within the hour, so if you feel relatively confident in your answer, or by popular demand, post your own...

The shortest person amon the tallest peope from each row will be at least as tall as the tallest person among the shortest person in each column.

Let's say we have m rows and n columns. We will use "i" to denote rows and "j" to denote columns. Each position is identified by (i,j).

Definitions:
t(i,j) is the height of the person in (i,j).
maxr(q) is the largest value of any of t(q,j) for a specific q.
The min-max is smallest value of maxr(q).
minc(q) is the smallest value of any of t(i,q) for a specific q.
The max-min is the largest value of minc(q).
Note the answer above is saying that the max-min <= min-max.

Proof:
We must have some individual (m,n) where the min-max = maxr(m) = t(m,n). This means that for all j, t(m,j) <= t(m,n)

Assume max-min > t(m,n). This means there exists some p such that minc(p) > t(m,n). Thus, there exists some p such that for any i, t(i,p) > t(m,n). In particular, t(m,p) > t(m,n). This is a contradiction.

Thus, the max-min <= min-max.

One_Brow · November 20, 2002, 15:46

I don't reaaly have time to look up one right now, so here's a quickie:

You start of game of Civ3 with 7 opponents, all civs chosen randomly (including yours), no cxultural groupings (so every Civ has an equal chance of appearing independently of the others).

Without regard to which civ is yours, how many different combinations of civs are possible in the original (16 civs) and in PTW (24 civs)?

Winston · November 20, 2002, 15:52

Those last two sound more like high school assignments than brain teasers.

What's the matter, dwarfs with multi-coloured hats not good enough for ya' anymore?

Zero-Tau · November 20, 2002, 16:10

Quote:

Originally posted by One_Brow
You start of game of Civ3 with 7 opponents, all civs chosen randomly (including yours), no cxultural groupings (so every Civ has an equal chance of appearing independently of the others).

Without regard to which civ is yours, how many different combinations of civs are possible in the original (16 civs)?

16*15*14*13*12*11*10*9/(8*7*6*5*4*3*2*1) = 12870

Simple mental arithmetic...

Quote:

and in PTW (24 civs)?

24*23*22*21*20*19*18*17/(8*7*6*5*4*3*2*1) = 735471

Needed a calculator for this one...

November 21, 2002, 00:59

OK the another fun one

family
In Apolytonia, boys and girls are as likely results of a pregnancy. All males are called Ming.
There is also a custom that states, that every family must have exactly one Ming, so each family has kids until they have exactly one boy which they call Ming.
IN the long run, what will be the ratio of Ming to Non-Mings in Apolytonia.

Dauphin · November 21, 2002, 01:06

50:50

I did the maths too, but I'll let someone else show their summation of probability dens... yada yada.

summation [x * 0.5^x+1] for x =0 to infinity

Zero-Tau · November 21, 2002, 11:24

Quote:

Originally posted by Sagacious Dolphin
50:50

I did the maths too, but I'll let someone else show their summation of probability dens... yada yada.

summation [x * 0.5^x+1] for x =0 to infinity

Actually, there's a much simpler argument: Each birth has a 50% chance of being a boy, and a 50% chance of being a girl. It stays this way, regardless of whether some families stop having children. Therefore the answer is 1 to 1.

November 21, 2002, 12:21

Yes to both of you...
Of course Zero-Tau's argumetn is much nicer...
No kids get discarded, all newborns are kept, and there are 50% of each... why would the ratio be any different in the population?
This one was easy but it was fun to check if someone would post the easy way or hard way first...
I guess you want another one?

Dauphin · November 21, 2002, 12:24

It wasn't that hard a way. I did it whilst drunk.

Plus it is very similar to puzzles involving coin tosses for which the answers are pretty much imbedded in my brain.

November 21, 2002, 12:47

I did say it was sorta of easy and I can sum geometrical series in my head too...
But being a math major myself, I like trying to ask puzzles that anyone can do, and sometimes layman even do better on...
Ok then how about this one...

Sagacious Dolphin has just made a daring escape from Mingapulco. If he can just reach the secret "Rebel Against Ming the Tyran" (or wasnt I suppose to mention it?) base at the North Pole, he surely will be safe. Mingapulco is of course on the equator at an undisclosed location. He decides he has better chance of evading would be pursuers by heading straight NorthWest for the duration of the trip.
What is the length of his path?
Assume a spherical Earth and that the distance from the equator to the North Pole is about 20000km.

Dauphin · November 21, 2002, 13:01

[20,000² + 20,000²]^0.5 = ~28,300km

Better start now, may take a while.

Zero-Tau · November 21, 2002, 13:08

Quote:

Originally posted by Sagacious Dolphin
[20,000² + 20,000²]^0.5 = ~28,300km

Better start now, may take a while.

Not that simple, I'm afraid. That formula only works for a plane, not for a sphere. The actual distance would be shorter. Not that I know the correct formula, though...

Oh and LulThyme, what planet are you living on? On mine, the distance from equator to the North Pole is about 10000 km, 1/4 of the circumference of the entire planet, 40000 km...

November 21, 2002, 14:29

Of course the actual number used is not very important...
But you are correct...
Oh maybe Mingapulco is not where we think it is

Now about Dolphin's answer.
At any point on Earth, if your speed NW is Speed, then your Speed North IS Speed / square root of 2
If you think about it, the fact that globally you are travelling on a sphere makes no difference.
So dolphin's answer is correct.
Whats interesting, is that you will circle around the pole an infinite amount of times, but the length of the path and thus the time take is still finite (a sort of round version of Zeno's paradox I would say).

Go ahead Master Escapee, your turn

Dauphin · November 21, 2002, 14:46

Quote:

is that you will circle around the pole an infinite amount of times

My alternative answer to your question would have involved Archimedian spirals, but the maths was less easy.

Let me find one and tailor it.

Dauphin · November 21, 2002, 19:02

Ming and rah both discover a spamfest - the near infinite horde of Finns have re-descended upon Apolyton. Not being able to ban all of Finland, Ming and rah decide to split the Finns into three groups - those who will visit Mingapulco, those who will visit Rahlcatraz and those who get off with just a warning.

To do this, Ming and rah both roll a fair 20-sided dice and each Finn rolls a fair 20-sided die. If the Finn rolls a number on his/her die equal or less than both the numbers Ming and rah rolled, then the Finn goes to Mingapulco. If the Finn rolls a number on his die equal or more than both the numbers Ming and rah rolled, then the Finn goes to rahlcatraz. Otherwise (i.e the Finn rolls a number between the two numbers rolled by rah and Ming), the Finn gets off with a warning.

What is the split of the fates of the Finnish horde?

The Vagabond · November 21, 2002, 22:01

Quote:

Originally posted by Sagacious Dolphin
Ming and rah both discover a spamfest - the near infinite horde of Finns have re-descended upon Apolyton. Not being able to ban all of Finland, Ming and rah decide to split the Finns into three groups - those who will visit Mingapulco, those who will visit Rahlcatraz and those who get off with just a warning.

To do this, Ming and rah both roll a fair 20-sided dice and each Finn rolls a fair 20-sided die. If the Finn rolls a number on his/her die equal or less than both the numbers Ming and rah rolled, then the Finn goes to Mingapulco. If the Finn rolls a number on his die equal or more than both the numbers Ming and rah rolled, then the Finn goes to rahlcatraz. Otherwise (i.e the Finn rolls a number between the two numbers rolled by rah and Ming), the Finn gets off with a warning.

What is the split of the fates of the Finnish horde?

2850/8000 Finns going to Mingapulco.
2850/8000 Finns going to rahlcatraz.
2280/8000 Finns getting off with a warning.
And the last but not least:
20/8000 Finns going to Mingapulco, rahlcatraz, and getting off with a warning SIMULTANEOUSLY.

Disclaimer: If you don't like the latter part of my answer, then SD's problem is not well-posed.

Dauphin · November 21, 2002, 22:14

What's wrong with dismembering Finns?

The Vagabond · November 21, 2002, 22:21

So you did this deliberately, Sagacious? Then this counts as an atrocity, and you should be given up at the mercy of Finnish crowd.

JohnM2433 · November 22, 2002, 01:00

The Vagabond,

The odds of a random Finn getting off with a warning are
the same as the odds of a number randomly picked from 1 to 20 being between two other numbers randomly picked from 1 to 20.

The odds of three numbers randomly picked from 1 to 20 being in forward order (no duplicates) are

(1/20)(19/20)(18/20) + (1/20)(18/20)(17/20) +
(1/20)(17/20)(16/20) + ... + (1/20)(2/20)(1/20) =
(19*18 + 18*17 + 17*16 + ... + 2*1)/20^3 =
2280/8000

if I keyed everything in that last summation into my calculator right. The odds of them being in either forward order or reverse order are twice that, or 4560/8000. That's the odds of the middle number being between the two other numbers. So (since all three numbers are chosen randomly) that should be the odds of a Finn getting off with a warning, I think.

Also, while you are correct that 20/8000 of the Finns will go to both Mingapulco and rahlcatraz, I disagree that they will also get off with a warning, with the problem as stated.

And of course the remaining 3420/8000 will be divided equally between Mingapulco and rahlcatraz, with 1710/8000 going to each. So

57% get off with a warning
21.375% go to to Mingapulco
21.375% go to to rahlcatraz
0.25% go to both

Is that right? The first number seems a tad high, but my math all seems to be correct.

The Vagabond · November 22, 2002, 01:39

Quote:

Originally posted by JohnM2433
The odds of three numbers randomly picked from 1 to 20 being in forward order (no duplicates) are

(1/20)(19/20)(18/20) + (1/20)(18/20)(17/20) +
(1/20)(17/20)(16/20) + ... + (1/20)(2/20)(1/20) =
(19*18 + 18*17 + 17*16 + ... + 2*1)/20^3 =
2280/8000

if I keyed everything in that last summation into my calculator right.

What you actually calculate here seems to be the odds of the first number being smaller than the other two, irrespective of the order of the other two (and no duplicates, of course).

But if you want the other two numbers to also be in the forward order, I believe you should calculate the odds as follows:

(1/20)*((1/20)*(18/20+17/20+...+1/20)+
(1/20)*(17/20+16/20+...+1/20)+ ... +
(1/20)*(2/20+1/20)+(1/20)*(1/20)) = 1140/8000

If you double it as you stated, you get the odds 2280/8000 for getting off with a warning.

Quote:

Also, while you are correct that 20/8000 of the Finns will go to both Mingapulco and rahlcatraz, I disagree that they will also get off with a warning, with the problem as stated.

I believe we are both correct here (or, better to say, both not incorrect), since ill-posed problems often result in such kind of ambiguity.

The Vagabond · November 22, 2002, 01:59

Well, actually what I did is equivalent to dividing your result by two, of course.

Dauphin · November 22, 2002, 14:08

The easiest (to me at any rate) way to solve that puzzle was as follows:

If any two dice have the same number, the Finn is banned. This does not happen 19*18/(20*20) of the time. If the three numbers are different, then the chances of the Finns dice being the middle one is 1/3. Therefore the Finn is free 19*18/(20*20*3). = 57/200.

If the three numbers are the same 1/(20*20), the Finns are dismembered. = 1/400

The remaining probabilities are spliut between rahlcatraz and Mingapulco. = 57/160

The Vagabond · November 22, 2002, 18:58

A very elegant way to solve the problem, Sagacious.

Mine was much more complicated.

Nonetheless, dismembering the Finns like that is an atrocity. Not that I am much against it though.

Just leave Kristalli out of it.

JohnM2433 · November 22, 2002, 22:23

OK, I see what I did wrong. My math assumes that the third number can be any number greater than the first other than the second, but of course it also has to be greater than the second. I think when I was figuring the odds for the third number, I must have somehow assumed that the second number was only 1 greater than the first. Of course, that's only the case a small fraction of the time. Whoops. So it makes sense that the answer I gave was twice the correct one, since the expected number of numbers greater than the second number would be half the maximum, since it varies randomly from the maximum to zero. (I hope that all made sense.)

JohnM2433 · November 22, 2002, 22:30

Quote:

Originally posted by The Vagabond
A car smashes into the tree. The driver gets out, looks around, and says: "How great it's halved! Otherwise I'd be dead now".

Question: What did he mean?

Disclaimer: Don't take it too seriously. It's half joke.

No one answered this one yet. Is it a well-known joke? I don't get it.

November 25, 2002, 23:32

time to resurrect this thread again

We have A:Sagace, B:Vaga, and C:ZTau 3 logicians and friends.

Ming The Moderator takes a set of 8 stamps, 4 red and 4 green, known to the logicians, and loosely affixes two to the forehead of each logician so that each logician can see all the other stamps except those 2 in the moderator's pocket and the two on her own head. He asks them in turn if they know the colors of their own stamps: A: "No" B: "No" C: "No" A: "No B: "Yes"
What are the colors of her stamps, and what is the situation?

November 26, 2002, 02:09

bump for the night crew

The Vagabond · November 26, 2002, 04:13

Quote:

Originally posted by LulThyme
We have A:Sagace, B:Vaga, and C:ZTau 3 logicians and friends.

Ming The Moderator takes a set of 8 stamps, 4 red and 4 green, known to the logicians, and loosely affixes two to the forehead of each logician so that each logician can see all the other stamps except those 2 in the moderator's pocket and the two on her own head. He asks them in turn if they know the colors of their own stamps: A: "No" B: "No" C: "No" A: "No B: "Yes"
What are the colors of her stamps, and what is the situation?

It seems to me the following configuration works out nicely:

A has 2 red stamps; B -- 1 red and 1 green stamps; C -- 2 green stamps. (Or else, A has 2 green stamps while C has 2 red stamps.)

The Vagabond · November 26, 2002, 04:21

Quote:

Originally posted by JohnM2433
No one answered this one yet. Is it a well-known joke? I don't get it.

I don't believe it's a well-known joke. But as no one made a single guess yet, it wouldn't be interesting to give up the answer for the moment.