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June 12

A question of sociopathic probability...

The last week or so, Swedish media has reported on a horrible murder, meticulously planned and executed by two 16-year-old kids against one of their classmates. Much of the reporting has emphasized that while it is exceedingly rare for two kids that young to plan and commit cold-blooded murder, what makes this case absolutely unique in recent Swedish crime-history is that one of the culprits is a girl.

Yesterday, I was having a conversation about this case with a friend of mine, and the conversation made me regret I didn't listen more closely when my high school teacher tried to teach us probability theory. My friend was asking how a person could do something like this, why some basic human instinct didn't stop them during their weeks of planning, to which I responded that both of them are probably psycopaths. My friend then asked "Yeah, but what are the chances that two people in the same classroom are psycopaths?" I thought that was a very good question, and one I'd like you fine people's help with answering (for the record, my answer at the time was "I have no idea, but surely in the hundreds of thousands of Swedish classrooms, it's not so strange that there are two psycopaths in at least a few of them").

Wikipedia states that it is estimated that about 1% of people are psycopaths (it seems to be a good average from the other sources I've looked at). So what is the probability that in a group of 30 people, two of them have that particular affliction? And how do you make that calculation? 195.58.125.52 (talk) 08:27, 12 June 2009 (UTC)[reply]

If one has a group of 30 people sampled from a particular population, and in that population it is estimated that 1% of people have some property X, then the probability that 2 people have property X in the sample of 30 can be computed as follows:

I am assuming that you want to know the probability that exactly 2 people have property X and not at least 2 or more than 2. To compute this probability one uses the binomial distribution. Practically, you want to know the probability of at least 2 people having property X in the sample of 30. In this case, we compute the probability that either 1 person, or 0 people have property X, and subtract the sum off from 1. This probability is:

1 - [(0.99)³⁰ + (0.99)²⁹ x (0.01) x 30]

= 0.03614799831

Thus, 0.03614799831 is the probability that at least 2 people in the given sample of 30 have property X. If you can not follow the above, I am sure that other Wikipedians can explain but the relevant article is binomial distribution. In percentage, there is an approximately 3.6% chance that at least two people in the class are psycopaths. --PS T 09:31, 12 June 2009 (UTC)[reply]

That was both A) very understandable and B) incredibly scary (seriously, 3.6 percent???). Now that you said it, I had a flashback to learning about the binomial distribution using dice and coins. Thanks! 195.58.125.52 (talk) 10:37, 12 June 2009 (UTC)[reply]

This calculation assumes that the probability of a person being afflicted is independent of the class in which he studies. However, it is reasonable to assume that it actually depends on factors like geographic location, socioeconomic status, field of study and so on. So the quoted 1% figure is actually a mixture of classes where the probability is higher and those where it is lower. This can make the probability (of at least 2 psychopaths in a given class) even higher. -- Meni Rosenfeld (talk) 11:00, 12 June 2009 (UTC)[reply]

Just a note that that sort of calculation applies when the property X is exactly specified beforehand. If instead we were to ask "what the chance that at least two people share some property" (where the details of the property are left unspecified), we have to calculate it slightly differently. This is explained more in depth at the Birthday problem article. -- 128.104.112.114 (talk) 17:50, 12 June 2009 (UTC)[reply]

The way of computing the probability given here assumes you're talking about just one specified classroom. Michael Hardy (talk) 01:51, 13 June 2009 (UTC)[reply]

There are also different types of psychopathy. Not all psychopaths have the ability to meticulously plan and execute a murder. (Igny (talk) 17:33, 14 June 2009 (UTC))[reply]

And there is also an unlikely event of two psychopaths recognizing and trusting each other. However if there is only a chance of

5\cdot 10^{-5}

for a particular pupil to be just such a psychopath, the probability of two such psychopaths in at least 1 class out of 1 million classes (of ~30 pupils per class) worldwide will exceed 50%.

crazy hard math question

How many unique ways can you arrange the tiles on a standard Scrabble board?

The board is 15x15 tiles and uses the following numbers of letters: A - 9 B - 2 C - 2 D - 4 E - 12 F - 2 G - 3 H - 2 I - 9 J - 1 K - 1 L - 4 M - 2 N - 6 O - 8 P - 2 Q - 1 R - 6 S - 4 T - 6 U - 4 V - 2 W - 2 X - 1 Y - 2 Z - 1 Blank - 2

The letters do not have to be in an order that makes sense. Not all tiles need to be used at once. E in spot 1,1 with the remaining spaces empty is valid (though a second setup that looks identical but uses a different E tile is not counted a second time). I am not looking for an exact answer, just an order of magnitude. 65.121.141.34 (talk) 14:45, 12 June 2009 (UTC)[reply]

Ok, you've got 100 letter tiles, and 225 places to put them. So, if we ignore the markings on tiles, the number of subsets of the board that can be covered would be

\sum _{n=0}^{100}{225 \choose n}

. We know that

\sum _{n=0}^{225}{225 \choose n}=2^{225}

, and only going as far as 100 would be roughly half of that, by symmetry (a bit less, because 100 < 112, where the actual middle is). Thus, I'd say the number of subsets of the board that can be covered is on the order of

2^{224}

. Next, you have to deal w/ the markings on the tiles... -GTBacchus^(talk) 15:20, 12 June 2009 (UTC)[reply]

If I haven't screwed up the calculation, you have 3 × 10¹⁸¹ possibilities with all tiles used. Might give an impression of a lower bound on the total. —JAO • T • C 15:26, 12 June 2009 (UTC)[reply]

I agree with your calculation (well, up to a factor of two, and it's hardly worth chasing that when the numbers are of this magnitude). Algebraist 15:33, 12 June 2009 (UTC)[reply]

Same here wrt the result, but I think the exact answer is still important, if only to show that it indeed can be found. Providing it is free, so what I got is

3456981978217014632075525114928779982118337773021077837923560875149664042972031452007170111...
...5929753351999852423974761359826877638987511529912396851416083377852241674240000000000000000

This is, again, assuming that all tiles need to be used. If not, I think it's around

10^{190}

but I'll run an exact calculation tomorrow (if nobody beats me to it). -- Meni Rosenfeld (talk) 15:56, 12 June 2009 (UTC)[reply]

Ok, if my calculations are correct (a huge if), the answer to the original question is

7722117868441671924993195080976319850952236039054771466584888315955839702441216546371222929...
...7029197060990587627600742490611073623755475870190820272102817641617988346426872233220965236

Which is roughly

7.72\cdot 10^{181}

. -- Meni Rosenfeld (talk) 19:12, 13 June 2009 (UTC)[reply]

Differentiating the Jacobian determinant, need some help!

Hi everybody, I'm trying to solve this problem and need some hint indeed:

File:Jacobian001.JPG

where the hints are:

File:Jacobian002.JPG

and,

File:Jacobian003.JPG

thanks in advance! --Re444 (talk) 17:19, 12 June 2009 (UTC)[reply]

I do not think that we are permitted to answer questions from you unless you have provided sufficient evidence that you have attempted the question yourself. Could you please cite the book from which this question originates? It is then possible that we may be able to answer the question for you should you provide us the required evidence. --PS T 22:46, 12 June 2009 (UTC)[reply]

It appears to be from "Introduction to Tensor Calculus and Continuum Mechanics" By J. H. Heinbockel [1]. I think it is ok to supply hints in this situation, just not answer the whole question. 207.241.239.70 (talk) 00:55, 13 June 2009 (UTC)[reply]

The referenced book is true : "Introduction to Tensor Calculus and Continuum Mechanics" By J. H. Heinbockel (a free e-book). And I'm a self-learner of that (not a homework then). --Re444 (talk) 19:23, 13 June 2009 (UTC)[reply]

Where is the book online for free? I'd like to try reading it. Thanks. 67.122.209.126 (talk) 03:02, 14 June 2009 (UTC)[reply]

80% of the book is available free, here (at the end of the page) —Preceding unsigned comment added by Re444 (talk • contribs) 08:16, 14 June 2009 (UTC)[reply]