Brain teaser thread

[Dauphin]

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]

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]

Oops, forgot that option

What's wrong with dismembering Finns?

[The Vagabond]

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]

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]

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.

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]

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

[Dauphin]

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]

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]

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]

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.

[Guest]

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?

[Guest]

bump for the night crew

[The Vagabond]

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]

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.