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June 18

Constant rule of differentiation

Constant rule states that derivative of any constant is zero:

let y=c.

${\frac {dy}{dx}}=\lim _{\Delta \ x\to 0}{\frac {\Delta y}{\Delta x}}=\lim _{\Delta \ x\to 0}{\frac {c-c}{\Delta x}}=\lim _{\Delta \ x\to 0}{\frac {0}{\Delta x}}={\frac {0}{0}}=0$

Why should 0/0 be 0 rather than undefined?--218.102.124.108 (talk) 05:24, 18 June 2008 (UTC)[reply]

Because limit of the ratio of two functions is not necessarily equal the ratio of limits of those functions. Anyway check the limit's definition, and see that values of 0/Δx are arbitrarily close to zero when Δx approaches zero (in fact they are all equal zero), so they satisfy conditions in the limit definition. --CiaPan (talk) 05:47, 18 June 2008 (UTC)[reply]

Your mistake is in the step

\lim _{\Delta \ x\to 0}{\frac {0}{\Delta x}}={\frac {0}{0}}

. Taking a limit is not the same as plugging in a value for the function; that is,

\lim _{x\to a}f(x)

is not necessarily

f(a)\;\!

(if it was, we wouldn't invent limits in the first place). The correct steps are

\lim _{\Delta \ x\to 0}{\frac {0}{\Delta x}}=\lim _{\Delta \ x\to 0}0=0

. If you make this mistake it will happen when you calculate any derivative, not just for a constant. Do read up on the definition of limits. -- Meni Rosenfeld (talk) 08:28, 18 June 2008 (UTC)[reply]

More specifically, you can 'plug in' limits like that, provided the denominator does not equal zero, as it does in your case.--Fangz (talk) 11:51, 18 June 2008 (UTC)[reply]

You can plug in values if the function is continuous at the point. It is true that for rational functions, this happens exactly when the denominator is not zero at the point. Not all functions are rational, though. -- Meni Rosenfeld (talk) 12:54, 18 June 2008 (UTC)[reply]

Thanks everyone. I think I've known what's wrong with me. But what I intended to solve is 0/0:

{\frac {0}{0}}={\frac {0^{1}}{0^{1}}}=0^{1-1}=0^{0}=1?

So what is the result of 0/0?--218.102.124.108 (talk) 15:34, 18 June 2008 (UTC)[reply]

0/0 is undefined in all common structures. See division by zero. -- Meni Rosenfeld (talk) 15:38, 18 June 2008 (UTC)[reply]

{\frac {0}{0}}=x\iff 0=x\cdot 0

, and that's true for any number x (it can also be true for infinity in some circumstances), so basically 0/0 can equal anything. If you end up with a 0/0 somewhere, you have to go back a step and work it out a different way, because it could come out to be absolutely anything. For example, if f(x)=0/x, then, if you want f to be continuous, you have to set f(0)=0, but if f(x)=x/x then f(0)=1, or if f(x)=2x/x, then f(0)=2. (Those are pretty trivial examples, but they should give you an idea of what can happen.) --Tango (talk) 15:52, 18 June 2008 (UTC)[reply]

The string of characters "0/0" can mean two different things. It can mean an attempt to divide 0 by 0, in which case it can't equal anything as it is undefined. It can also mean a certain Indeterminate form, in which case it could indeed evaluate to anything, but is only a mnemonic, not an actual mathematical entity. -- Meni Rosenfeld (talk) 16:05, 18 June 2008 (UTC)[reply]

Yes, I should have been clear I was speaking very informally, just to give an idea of why it is undefined. --Tango (talk) 16:18, 18 June 2008 (UTC)[reply]

cubic metres, cubic kilometers

A recent article about the Niger Delta talked about "10 trillion cubic meters of natural gas reserves" in one region there, and "28,800 cubic km" (kilometres, we assume) of natural gas reserves is Russia. Too many zeroes...will 1000 cubic meters be 1 cubic kilometre, and therefore 28,800 cubic km, will be 28,800,000 cubic meters...which is only 28.8 million cubic meters...which must be way less than 10 trillion (10,000,000,000,000 (is a trillion a thousand billion?)...So the Niger Delta reserves in this one region are millions of times more than the reserves in Russia? Surely we are confused over the math, but maybe the press was wrong. Can anyone confirm or correct our understanding? Thanks if you can help. —Preceding unsigned comment added by 213.84.41.211 (talk) 14:20, 18 June 2008 (UTC)[reply]

There are 1000 meters in a kilometer, but 1000^3 = 1,000,000,000 cubic meters in a cubic kilometer. Wikiant (talk) 14:27, 18 June 2008 (UTC)[reply]

So there are in fact 29 trillion cubic meters of gas in Russia. I think mass is a much more useful measure for the quantity of a gas then volume, though. -- Meni Rosenfeld (talk) 15:36, 18 June 2008 (UTC)[reply]

Agreed. I don't think Nigeria is the same temperature as Russia. Algebraist 15:42, 18 June 2008 (UTC)[reply]

It might be underground, though. At least, there is probably less of a difference. --Tango (talk) 15:48, 18 June 2008 (UTC)[reply]

I think such measures are based on what the real volume would be under certain standardized conditions. See Natural gas#Energy content, statistics and pricing. PrimeHunter (talk) 00:04, 19 June 2008 (UTC)[reply]

I figured as much, but I still find it unnecessarily confusing. -- Meni Rosenfeld (talk) 11:49, 19 June 2008 (UTC)[reply]

In answer to the side question, most people writing about numbers that big in English use a billion to mean 1,000,000,000 (10⁹) and a trillion to mean 1,000,000,000,000 (10¹²). But until recent decades, it was usual in English outside of North America for a billion to mean 1,000,000,000,000 (10¹²) and a trillion 1,000,000,000,000,000,000 (10¹⁸). The cognate words in most other languages have the larger meanings, and some people still prefer them for use in English. See long and short scales (and please do not debate them here). --Anonymous, 02:27 UTC, June 19, 2008.

A billion being a thousand million is pretty much universal these days. There's no need to worry about possible alternatives. --Tango (talk) 11:58, 19 June 2008 (UTC)[reply]

See billion (word). Bo Jacoby (talk) 22:35, 20 June 2008 (UTC).[reply]

Random numbers

Random numbers may be generated on computers, graphing calculators, and even Excel spreadsheets. (Although these are technically referred to as "pseudo" random numbers, that distinction is irrelevant for purposes of this question.) The following scenario came to mind, from which this question arises. Let's say that a teacher asks his students, as a homework practice exercise, to generate 5 random numbers, with each random number ranging from 0 to 9 inclusively. So, for example, one student might generate 3, 7, 2, 1, 9. Another student might generate 5, 6, 3, 5, 2. And so forth. Now, say there is a lazy student who did not do the homework and, when asked, lies to the teacher and fabricates that he generated 7, 7, 7, 7, 7. (Also, indicating a none-too-bright student, as he could just as easily have lied with a less "suspicious" result of another fictitious string of numbers such as, say, 4, 1, 8, 3, 0 or so.) So, the teacher gets suspicious at the result of 7, 7, 7, 7, 7, and suspects that the student cheated and did not, in fact, actually do the homework assignment. So, here is my question. Should the teacher, in fact, get suspicious at such a result? Is a result such as 7, 7, 7, 7, 7 equally likely as a series of 5 different digits (say, 3, 8, 0, 2, 6) … or is such a result not as equally likely in the generating of random numbers? My understanding is that any (and all) of the 5 separate digits is randomly generated and, thus, a result such as 7, 7, 7, 7, 7 should not be a surprising result. But, my gut instinct also tells me that it is highly unlikely to generate a result such as 7, 7, 7, 7, 7 and that a mixture of different digits is much more likely / probable. What is the truth of the matter? If all 5 digits are randomly generated, why does a result like 7, 7, 7, 7, 7 seem so improbable? Any insights? Thanks. (Joseph A. Spadaro (talk) 23:23, 18 June 2008 (UTC))[reply]

There are more than one way to answer this; I'll pick the one that seems most in line with your motivation. It all comes down to Bayes' law. Let A be the event "student cheated", B be "Student chose 38026", C be "student chose 77777". Assume P(A)=10% and that a cheating student can only come up with a constant sequence, thus P(C|A)=10%. Then

P(A|C)={\frac {P(A)P(C|A)}{P(C)}}={\frac {0.01}{0.010009}}\approx 1

, while

P(A|B)=0

. Of course you can replace the assumptions with whatever you find reasonable to get more realist estimates. -- Meni Rosenfeld (talk) 23:36, 18 June 2008 (UTC)[reply]

Theoretically a computer can generate a random number, not just a pseudorandom number. Although in my opinion and ignorance of physics, I am under the belief that of what’s on the market, only the quantum random number generators can be truly random, and those will cost as much as the computer to run them.

As for the actual question… I would try to find some way of measuring how random a number “looks”. Very subjective, but 2 4 9 5 2, 9 5 2 3 1, and 7 7 7 7 7 are all equally likely, but like most normal people my first reaction wouldn’t be to question the first two, but would be the third: because there are a lot of numbers that “look like” the first two, but only a few if any numbers that “look like” the third. GromXXVII (talk) 23:57, 18 June 2008 (UTC)[reply]

(e/c)Random is as random does - but human brains are very good at spotting patterns, even when they're spurious. Write down what you think would be 50 perfectly random coin tosses, and you're not likely to write down any sequence with more than 3 or 4 heads in a row. And yet, you're actually quite likely to get a run of 5, which you will find if you actually do the coin tosses a few times. The run of 5 heads is a pattern, and since we don't expect to see patterns in "random" data, it throws us.

That said, if you see something "odd", then you can perform a hypothesis test to see if it's statistically unlikely. First, you get an idea of what the results would look like if they were selected randomly - say, with each digit uniformly distributed between 0 and 9. Then you quantify just what is "odd" about Johnny's 7, 7, 7, 7, 7 (i.e. "it's the same digit 5 times). Finally, you calculate the probability that you would find a result equally as odd as or more odd than Johnny's, and from that probability determine whether it still seems unlikely (5% is a common cutoff, 1% if you're being more lenient). The probability of getting 5 digits the same is 1/1000, or 0.1%, so Johnny might be looking a little nervous right now.

And for the silly answer, both students fail. The perfectly random selection would be 4, 4, 4, 4, 4[1]. Confusing Manifestation(Say hi!) 00:06, 19 June 2008 (UTC)[reply]

[ec] Some more thoughts:

77777 has a lower Kolmogorov complexity than 38026. This is why a cheater with no access to an adequate RNG is more likely to generate 77777 than 38026.
While 77777 is as likely as 38026, there are simple associated variables that are not equilikely. For example, you can define a variable "how many different digits are in the sequence". This will be 5 with probability ≈30%, and 1 with probability 0.01%.
Occam's razor says the simplest explanation to an observation tends to be the correct one. The sequence 38026 is not likely to occur, but if it does occur, the only explanation is that it happened randomly (again assuming that without a RNG it's difficult to come up with it). If, however, we are presented with 77777, the simplest explanation is that the student cheated.

-- Meni Rosenfeld (talk) 00:14, 19 June 2008 (UTC)[reply]

Here are some approximate odds (I didn't want to use a calculator to find the exact odds):

Chances of a given student picking 7 7 7 7 7 exactly: 1 in 100,000.

Chances of a given student picking any number 5 times in a row: 1 in 10,000.

Chances of any student in a class of 25 picking any number 5 times in a row: 25 in 10,000 or 1 in 400.

Chances of any student in a 8 classes of 25 each picking any number 5 times in a row: 200 in 10,000 or 1 in 50.

Chances of any student in a 8 classes of 25 each picking any number 5 times in a row or any sequence of 5 numbers in a row (0-4, 1-5, 2-6, 3-7, 4-8, 5-9, 6-0, 7-1, 8-2, 9-3): 200 in 5,000 or 1 in 25.

Add in reverse sequences like 9 8 7 6 5 and even sequences like 0 2 4 6 8 and odd sequences like 1 3 5 7 9 and now reverse those, too, and there's a good chance that one student would hit one of those. StuRat (talk) 00:17, 19 June 2008 (UTC)[reply]

I don't think that's a good argument. There's a lot of reason to think that a student without a RNG can just make up a number like 38026, and it's arguable that a number like 77777 is unlikely to arise out of a student told to work out a random number generation and thus *more* likely to be random. (In fact, IIRC, there's good evidence that in human generated sequences, usually runs of numbers are *less* common than in actual random numbers, especially in the binary case.) Trying to find ways that 77777 is special is also very dangerous, since the arguments previously put forward are equally applicable to any finite sequence. Even something like Kolmogorov complexity isn't helpful, because you can easily define a post-facto language where 38026 is easy to generate while 77777 is not. (E.g. 38026 can be someone's birthday, while 77777 cannot)

I think the probably best way to deal with the problem is a bayesian approach. You need to pre-specify an impression of what a human generated sequence is like, and what a good rng is like, and set some prior probabilities. Then just update your beliefs. Specifically, there isn't a good way to just say a probability - your view of whether the sequence is random or not can only be stated relative to a well (and in advance, especially not based on perceived specialness of the sequence) defined alternative hypothesis and/or a set of priors.--Fangz (talk) 00:48, 19 June 2008 (UTC)[reply]

You have raised the psychological warfare issue - "the student knows I will suspect 77777, and thus will avoid it if he cheats". The problem with this is that it can always be taken one step further - "The student knows I know he knows, and thus will choose 77777 if he cheats in order to trick me". It's thus not very reliable reasoning.

It is not reliable reasoning, but it does serve to point out a flaw in your argument. Implicit in any attempt at discriminating one hypothesis from another is a model - and indeed, a psychological model of how the student is likely to react. It is impossible to argue broadly that out of *all* possible algorithms the student may use instead of a random number generator, certain sequences are objectively more likely than according to random chance.

You say that trying to find projections from the space of sequences to a lower-dimensional space in which 77777 is projected to a salient outcome is dangerous. However, it is only dangerous if the projection is complicated and arbitrary. For example, we could define " $p(X)$ is 0 if $X=38026$ , otherwise $p(X)=1$ ". Then $p(X)=0$ will have a very low probability, so we will reject an answer of 38026. However, this projection is very arbitrary and silly. If we stick to something simple and natural, like " $p(X)$ is the number of distinct digits in $X$ ", we should be good.

But the issue is that there is no rigorous mathematical framework around that definition of 'simple and natural'. I stress, the difference is that you need to pre-define your specialness criteria. Not choose it based on the number itself. (For example, regarding 38026, 38026 is 2 * 19013, which is a very rare 5 digit centered square prime number. 77777 is not a multiple of two primes like that. Are prime numbers more special and non-arbitary than repeating digits? Who can tell?) --Fangz (talk) 15:57, 21 June 2008 (UTC)[reply]

In your Bayesian framework, you say we need "an impression of what a... sequence is like". But this can only be formalized in terms of some projection like above. The essence of your suggestion is that instead of arbitrarily choosing the probability of a student cheating and the distribution of $p(X)$ given that he is, we should calculate them based on observations. However, such observations are hard to come by. Cheating students will generally not admit to doing so, and asking random students to invent their own sequences may not give you a lot of insight into the mind of the cheater.

-- Meni Rosenfeld (talk) 11:45, 19 June 2008 (UTC)[reply]

If you want to get psychological you also need to account for the fact that if a student genuinely gets 7,7,7,7,7 out of a RNG, they are likely to think something has gone wrong and do it again. --Tango (talk) 15:04, 19 June 2008 (UTC)[reply]

Basically, to give a sensible assignment like that, you should ask for enough numbers that a sensible randomness test will give a clear answer. Say you ask for 200 numbers and you get the following output from a student:

1549248135779511892489513549165798

7691249876216984798416498765194327

9165498732498490987951979164984691

9498496519496195498795195746249832

4981951976249519762495462498746579

5162195762169795194957264916

Did he do this by tapping on a keyboard or by using a RNG? (Hint - I am lazy!) So maybe you can do the chi-square test and the runs test and get a pretty good answer. --Slashme (talk) 13:52, 19 June 2008 (UTC)[reply]

Maybe the runs test isn't very sensitive! For example, for the numbers I gave (monkey style), I got a runs test value of -1.2535104, giving a P-value of 0.10601 and for 200 random numbers from Excel, I got a runs test result of -1.5420651, giving a P-value of 0.06258. (done using a javascript applet found on the web, so maybe there's a bug? I'm too lazy to check the code for correctness) --Slashme (talk) 14:40, 19 June 2008 (UTC)[reply]

Follow up

Thanks for all the responses above. They were all very helpful. What makes the most sense to me lies within Grom's reply, which essentially states that some data will "look" randomly generated and some data will not "look" randomly generated. (Thus, arousing the teacher's suspicions.) So, let me re-phrase my original question. Let me use "39710" as a value that does look randomly generated, and let me use "77777" as a value that does not look randomly generated. I understand that the definition of "look" is fuzzy, but I assume that people generally know what I mean. (For example, a value such as "12345" does not look randomly generated, but a value such as "83016" does.) So, to re-phrase my original question … in terms of mathematics and random number generation and probability … is a value that does look randomly generated (e.g., 39710) equally like / probable as a value that does not look randomly generated (e.g., 77777)? If you want to clarify the fuzzy definition of "look", I can ask it this way: is a random number which has no repeating digits (e.g., 39710) equally likely / probable as a random number which has all 5 digits repeated (e.g., 77777)? In a true random generator, will both of those values appear with equal likelihood / equal probability? Thanks. (Joseph A. Spadaro (talk) 16:24, 19 June 2008 (UTC))[reply]

The chance of a specific sequence coming up is the same for any sequence (assume a truly uniform and independent random number generator). So 83016 is just as likely as 77777. The real question is about how many "random looking" sequences there are and how many "non-random looking" sequences there are, because those are what determine the chance of the sequence you generate being random looking or not, because you don't actually care which (non-)random sequence you get, you just care if it looks random or not (ie. you could replace 77777 with 55555 and it wouldn't make any difference). There are 10⁵=100,000 possible sequences of 5 digits. How many of those don't look random depends on your definitions (for example, does 12121 look random or not?), but you even with quite restrictive definitions you can find quite a few. Also, it's important to note, as someone pointed out above, that even quite unlikely occurrences end up being quite likely to occur if you do the experiment enough times. --Tango (talk) 16:37, 19 June 2008 (UTC)[reply]

The problem isn't in the randomness, but in the prejudice of the reader. When selecting randomly from the set of integers from 1 to 100,000, the numbers 77777, 12345, 39710, and 83016 all have the same probability of occurring (1 out of 100,000). It is only the reader's misplaced "feeling" that 77777 is less likely than 83016 that is responsible for the suspicion that 77777 was not randomly selected. (In fact, one might exploit that prejudice as follows: A (large) list of numbers that contains no numbers such as 77777, 12345, 54321, etc. is likely "faked.") Wikiant (talk) 16:35, 19 June 2008 (UTC)[reply]

If I was the teacher then 77777 would make me suspicious and 83016 would not. Not because 77777 seems less likely than 83016 to be the result of a random process, but because 77777 seems more likely than 83016 to be the result of a non-random process. It could for example be cheating to avoid home work, or a cheeky student testing how the teacher or other students would react, or the result of using a poor method to generate the sequence, or simply a stuttering student. But without more information I don't think there is a meaningful way to assign percentages to these and other possibilities. PrimeHunter (talk) 17:21, 19 June 2008 (UTC)[reply]

I'm not sure that your reasoning is clear. Let F be the event "student has fabricated the data". Let 7 be the event that the student reports 77777, and let 8 be the event that the student reports 83016. What I heard you say is, "Pr(7|F) > Pr(8|F)." I believe that the pertinent question is, "is Pr(F|7) > Pr(F|8)?" Wikiant (talk) 21:13, 19 June 2008 (UTC)[reply]

I estimate Pr(7|not F) = Pr(8|not F), and Pr(7|F) > Pr(8|F). Based on this (and 0 < F < 1), I conclude Pr(F|7) > Pr(F|8). So I think 77777 gives rational reason to suspicion even though a random number would have the same chance of being 83016. PrimeHunter (talk) 22:19, 19 June 2008 (UTC)[reply]

P(F|7) > P(F|8) with my assumptions is intuitively clear to me and can probably be shown easier than the following.

P(8) = P(F)P(8|F) + (1-P(F))P(8|not F).
P(7) = P(F)P(7|F) + (1-P(F))P(7|not F).

I assume 0 < P(F) < 1, and 0 < P(7|not F) = P(8|not F).
Let a = (1-P(F))P(7|not F) = (1-P(F))P(8|not F). Then 0 < a < 1.
P(8) / P(7) = [P(F)P(8|F) + a] / [P(F)P(7|F) + a].

From Bayes' theorem:
P(F|7) = P(7|F)P(F)/P(7), and P(F|8) = P(8|F)P(F)/P(8).
Then P(F|7) / P(F|8) = P(7|F)/P(8|F) * P(8)/P(7).
Substituting the earlier expression for P(8)/P(7) gives:
P(F|7) / P(F|8) = P(7|F)/P(8|F) * [P(F)P(8|F) + a] / [P(F)P(7|F) + a]
                = [P(F) + a/P(8|F)] / [P(F) + a/P(7|F)].
I estimated P(8|F) < P(7|F), so (remembering a > 0):
P(F) + a/P(8|F) > P(F) + a/P(7|F), and then P(F|7) > P(F|8).

PrimeHunter (talk) 01:02, 20 June 2008 (UTC)[reply]

follow on (from another person)

(please excuse my bad manners)
Seeing as we've now had the discussion I would like to point out that "the teacher asks for five random numbers from his students, and then proceeds to mark them." does represent some sort of 'false hypothesis' - only the teacher is wrong here... Clearly though the sequence "1","7","banana","3.4","8","0","4" is wrong in at least two ways... Can you spot them?

Here's a 'spin' on the question - you ask me to generate 5 random numbers ignoring the original "only 0-9" qualification)- I present "pi","e","sqrt(2)","1" and "0" .. is this sequence any less random than eg "2","6","5","7","1" ? Do I gain or lose marks for 2i ?87.102.86.73 (talk) 17:16, 19 June 2008 (UTC)[reply]

Well, say the teacher asks for 2000 random digits and proceeds to mark them, and one set of numbers, according to one of the standard randomness tests has a probability of 0.0000001% of having been generated by a random process (because the student just banged on the numeric keypad 100 times and cut and pasted the resulting numbers 20 times) should he fail the student?. --Slashme (talk) 13:13, 20 June 2008 (UTC)[reply]

My opinion

I think the answer depends on whether or not the order of the numbers matters. If order does matter then I say yes something like "77777" is more unlikely than say "12345". If order does not matter then it is just as likely as anything else.

When order does matter, then the second you get something thats not a 7 you're screwed so to speak. So the chance of "77777" being the result is (1/10)^5 because you have to get a 7 each time. But with "12345" you could, for example, get any of the five digits, then any of the four remaining, then three etc so the probability can be worked out with combinations and it will be more likely than "77777"

When order does not matter, the probability of any number sequence (of 5 digits in our case) is always (1/10)^5 --212.120.247.132 (talk) 17:43, 20 June 2008 (UTC)[reply]

Yes, that sounds right, 5 different specific digits should be 120 (5!) times more likely than one specific digit 5 times. --Tango (talk) 18:15, 20 June 2008 (UTC)[reply]

Thanks to all for the above discussion. You have all been helpful. And your mathematical insights have been very elucidating for me, on this topic. Thanks. I really appreciate the input and the discussion. (Joseph A. Spadaro (talk) 01:13, 20 June 2008 (UTC))[reply]

Seems to me that the fault lies with the teacher for not specifying clearly what they meant by "5 random numbers". If the teacher wanted 5 different random numbers then they should have said so. Without this qualification, it is unfair to penalise the 7 7 7 7 7 response unless you also penalise every other response that might have been generated some non-random process (or not even pseudo-random) process. What about 1 2 4 8 6, or 1 3 9 7 1, or 1 4 9 6 5, or 1 1 2 3 5, or 3 1 4 1 5, or 2 7 1 8 2, or 5 7 7 2 1, or 1 6 4 4 9 ? Each of those sequences has a pattern, and so should also be penalised. Of course, the teacher's ambiguity could be deliberate, and the responses could be a springboard for a discussion of how assigning a meaning to the term "random number" is not as straightforward as it appears at first sight. Gandalf61 (talk) 12:07, 21 June 2008 (UTC)[reply]

Exactly - a single sample from any distribution tells us precisely nothing..87.102.86.73 (talk) 14:12, 21 June 2008 (UTC)[reply]

To clarify for Gandalf61 ... the teacher's instructions would have been something like: "Generate a random number from 0 to 9 inclusive. Then do so a second time. Then do so a third time. Then do so a fourth time. Then do so a fifth time." Or such. Thanks. (Joseph A. Spadaro (talk) 00:28, 22 June 2008 (UTC))[reply]

In that case 7 7 7 7 7 is a valid response - at any rate, it is just as valid as 1 2 4 8 6 or 1 3 9 7 1 or 1 4 9 6 5 etc. etc. Gandalf61 (talk) 17:29, 22 June 2008 (UTC)[reply]

Agreed. The question (and ensuing discussion above) was ... is the teacher's suspicion justified? (Joseph A. Spadaro (talk) 19:13, 22 June 2008 (UTC))[reply]

We're drifting into the realms of education psychology rather than maths here ... is the student who gives 7 7 7 7 7 as an answer simply correct, stupid, naively right, trying to prove a point? etc etc .. hypothetical teacher, hypothetical students - difficult to give answers - too many unknowns.87.102.86.73 (talk) 19:42, 22 June 2008 (UTC)[reply]

Not difficult at all ... and the answer was, in fact, given ... namely, that "77777" is equally likely to occur as any other 5-digit string of numbers such as "13974". (Joseph A. Spadaro (talk) 21:57, 22 June 2008 (UTC))[reply]