Wikipedia:Reference desk/Archives/Mathematics/2012 November 4

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November 4

"Random article" chances of landing on a football-related page.

What are the chances of clicking Wikipedia "Random article" and recieving an article related to football? Nicholasprado (talk) 06:23, 4 November 2012 (UTC)

I guess one could use a Category-searching tool to estimate what fraction of articles are related to football. But not all articles have an equal chance of being chosen ... —Tamfang (talk) 06:39, 4 November 2012 (UTC)

Random article is not really random in the sense of having uniform probability. There are tweaks in it, to increase the likelihood of showing the reader something interesting. I suggest just clicking it 20 times or so, counting how many times you click it and how many football articles you get. 67.119.3.105 (talk) 08:43, 4 November 2012 (UTC)

Can you provide a link or any more information about those "tweaks"? —Steve Summit (talk) 13:46, 4 November 2012 (UTC)

The explanation at Wikipedia:TFAQ#random suggests that the selection is not truly random due to limits in the methodology. But beyond that, it does not mention any tweaks, and looking at other related pages, I can find no evidence of the existence of any such tweaks. A rough estimate of the proportion of football articles can be gained from the table at Wikipedia:Version 1.0 Editorial Team/Football articles by quality statistics, which lists all pages tagged as falling under the scope of WikiProject Football. While not every football-related article is tagged, in my experience the proportion is over 90%. After stripping out non-article categories, at the time of writing Wikipedia:Version 1.0 Editorial Team/Football articles by quality statistics lists 141,850 tagged football articles, and Special:Statistics gives the total number of articles as 4,090,691. This gives a lower bound for the proportion of football articles of 3.47%. Oldelpaso (talk) 22:42, 4 November 2012 (UTC)

more odd than even natural numbers?

with the formula "for all even natural numbers n, n<->n+1" (goes with), we have drawn a 1:1 relationship between every even number and an odd number. e.g. 2 goes with 3, 4 goes with 5, 6 goes with 7 and so on. Every even number is thus matched. There are an equal number of them as odd numbers, since you can bring them into 1:1 correspondence.

But wait, I trickily left out one odd number, namely "1". So after we have matched every even number with an odd number (without exception - the formula gives the match for every even number) we still have an odd number left over that no even number is matched with! So, in this pairing, there isn't it fair to say there is one more odd natural number than even natural number? Is this reasoning correct? Thanks. 87.97.96.159 (talk) 16:50, 4 November 2012 (UTC)

No, because you could just as well do it the other way around, and leave out an even. StuRat (talk) 17:07, 4 November 2012 (UTC)

That doesn't address the "flaw" in the proof. All you've shown is that counting one way, there is one more odd number than even number, with all the even numbers being paired 1:1 with an odd number, but one odd number left out; and counting another way, there is one more even number than odd number, with all odd numbers being paired 1:1 with an even number, but one even number left out. So, are both these statements true? 87.97.96.159 (talk) 17:16, 4 November 2012 (UTC)

The reason is that one number, or indeed any finite set of numbers, is infinitely small when compared to infinity. In other words, the limit of 1/x, or c/x, goes to zero, as x approaches infinity. StuRat (talk) 17:53, 4 November 2012 (UTC)

But that does not really address my argument. Look, if every boy is matched with a girl, and no girl is unmatched but there is one boy unmatched, then there is 1 more boy than girl, is that not fair to say? So, with infinite boys and girls, perhaps it is possible to match every girl to a boy but leave one boy alone, and in a different arrangemnt it's possible to match every girl to a boy but leave one girl alone? 87.97.96.159 (talk) 18:07, 4 November 2012 (UTC)

With infinite numbers (cardinalities), there is always the possibility of mismatching exactly matched sets by pretty much any number you choose, e.g., one could say that the set of natural numbers is ten elements larger than itself. The sensible (and standard) way of matching set sizes, then, is to say they have the same cardinality if a one-to-one mapping between them exists, or one is larger than the other if it has unmatched elements with every possible matching. — Quondum 18:31, 4 November 2012 (UTC)

Thank you, this makes a lot of sense and resolves my question. 87.97.96.159 (talk) 20:02, 4 November 2012 (UTC)

Read the definition of equipollence carefully — it says 'there exists a bijection from the set A to the set B', but not 'any function from the set A to the set B is a bijection'. You can present many functions that leave some items not paired in either set, but that doesn't prove or disprove anything. On the the other hand if you can show a single function which matches two sets with a one-to-one correspondence, this implies immediately the two sets have equal cardinality.
We can match even and odd integers in multiple ways, one of which is 2n ↔ 2n+1 — and that's enough to say they are equinumeros. --CiaPan (talk) 12:06, 5 November 2012 (UTC)

Your pairing would show that there is the same number of evens as there is of odds except 1, the flaw in your reasoning is tacitly assuming that there are more odds than there are odds except 1, there aren't, they're the same size. A bijective pairing tells you that the domain and range are the same size, that is all; you're basically just restating that an infinite subset can be the same size as a proper subset in a way that looks problematic, but isn't. Sorry to be redundant with the posts above:-) Phoenixia1177 (talk) 05:08, 6 November 2012 (UTC)

You will enjoy Hilbert's paradox of the Grand Hotel. Bo Jacoby (talk) 08:49, 6 November 2012 (UTC).

Let's count the numbers.

$\sum _{n=1}^{\infty }1=\zeta (0)=-{\frac {1}{2}}$

$\sum _{n=1}^{\infty }{\frac {1}{(2n)^{s}}}=2^{-s}\zeta (s)$

which is -1/2 for s = 0. So, while the total number of even number is -1/2, the total number of odd numbers is zero. :) Count Iblis (talk) 20:32, 6 November 2012 (UTC)

Aside from equipollence as mentioned above, check out the rather longer article on Cardinal number Gzuckier (talk) 08:21, 10 November 2012 (UTC)