Wikipedia:Reference desk/Archives/Mathematics/2014 November 17

= November 17 =

Largest constant
It has always boggled my mind how our mathematical reality is structured, considering that some important constants, such as e and π|pi are irrational. I just learned that even natural numbers can amaze me in the same way. To my surprise, a natural classification of simple groups gives birth to a monster, whose order is 246 · 320 · 59 · 76 · 112 · 133 · 17 · 19 · 23 · 29 · 31 · 41 · 47 · 59 · 71. There's a large number, not just because someone is trying to make one, like the googolplex or Graham's number. To me, the monster number feels like a true constant. Are there any more such large constants? Skewes' number may be considered another one, although it doesn't feel as constant to me. &mdash; Sebastian 05:46, 17 November 2014 (UTC)


 * You're conflating human interest with mathematical structure. Constants are constants in so far as they are meaningful to us; any specific number that someone cares enough about is a "constant". So, unless you're asking for large named numbers (which doesn't appear to be the case), or you have a more rigid definition for what fits as an answer, I'm not sure you are actually asking a mathematical question; it seems more like: "What large numbers are there that occur naturally in mathematical works that are not bounds?". Is that an accurate rendering?Phoenixia1177 (talk) 07:06, 17 November 2014 (UTC)
 * Graham's number wasn't constructed "just because someone was trying to make a large number". It was presented as an upper bound to a specific problem in Ramsey Theory. -- Meni Rosenfeld (talk) 08:37, 17 November 2014 (UTC)


 * I think I get your point: The size of the monster is a "natural" fact that (after much labor) falls out of the definition of a finite simple group, that is, the size of the monster only depends on the axioms of group theory, while Graham's number was just a bound, that could in principle have achieved the same function if it were a decent bit larger or smaller, or in any case not that exact number. I personally was always amazed at how we get several infinite families of FSG, but also the strangely finite number of sporadic groups (26). Anyway, I would agree that the size of the Monster is somehow more innate/natural than Graham's number. However, a constant is a constant, so they are both equally constant, but it seems that the monster is more on par with e and pi as being constants that emerge from fundamental properties, not from human choices. Unfortunately, I can't think of any other similar large numbers at present. SemanticMantis (talk) 15:54, 17 November 2014 (UTC)


 * Thanks all for your answers. I feel that SemanticMantis really understands where I'm coming from, and therewith also answers Phoenixia's question and corrects my inexactness, which is pointed out by Meni. I didn't really expect there to be any other similar large numbers, at least not by the time I posted this. During the process of reading up to it, I came across the other articles mentioned, which made that less likely. &mdash; Sebastian 20:13, 18 November 2014 (UTC)

"n1 = n2 = n3"
How do I express "n1 = n2 = n3" in set builder notation, set notation, or mathematical notation?174.3.125.23 (talk) 07:34, 17 November 2014 (UTC)


 * What are your n's? As it stands, there is no set to express, you are simply saying three things are equal. On an aside: mathematical notation and set notation are not the same, one is a superset of the other, are you asking about expressing sets or are you asking about something general and thinking sets might be involved? More context is needed to answer this.Phoenixia1177 (talk) 08:14, 17 November 2014 (UTC)
 * I trying to express "s", where "s = the number of the variable that is equal". I guess I am trying to delineate that "n = 3", but there are 3 "n"s.  How to express?174.3.125.23 (talk) 09:30, 17 November 2014 (UTC)
 * I'm sorry, that doesn't appear to mean anything - or if it does, it can mean a multitude of things. What are you trying to do in plain English?Phoenixia1177 (talk) 09:33, 17 November 2014 (UTC)
 * Example: "n1 = 4, n2 = 4, n3 = 4", but "s = 3", because there are 3 "n"s. How to express?174.3.125.23 (talk) 09:37, 17 November 2014 (UTC)
 * There is no specific notation for that, you can define some, but by the time you explain the notation, you could have just said it. Not everything needs, nor should be, symbolic. If you truly need some form of symbols, "Ni = |{j : nj = ni}|, s = sup Ni", meaning that s is equal to the largest collection of n that are equal to each other. You can write this sans the big N, but there is no reason to do that.Phoenixia1177 (talk) 09:42, 17 November 2014 (UTC)
 * Yes! Exactly!, I believe that is what I want.  Can you explain the notation?  For example, is that the absolute sign?174.3.125.23 (talk) 09:54, 17 November 2014 (UTC)
 * I would strongly caution against using the notation in this case, but I'll explain it. || around a set denotes the number of elements, so |{1, a, 66, cow}| = 4; sup is the supremum, the largest value assumed. Thus, Ni is the number of nj equal to ni, the supremum of the Ni is the largest elements of that sequence. In other words, it is defining s to be the maximal number of equal ni's. So, if you had n0 = n2 = n3 = 8 and n1 = n4 = 7, then the sequence of Ni would be 3, 2, 3, 3, 2, this, obviously, has largest element 3, so s = 3.Phoenixia1177 (talk) 10:02, 17 November 2014 (UTC)
 * What is the name of the absolute looking sign? For example, is it the same sign as used in "|" such as in Bayes theorem?  And would you provide a link?
 * Is there a notation that can replace "sup"?174.3.125.23 (talk) 10:18, 17 November 2014 (UTC)
 * Also, correct me if I am wrong, confirm if I am right: In your example you mean:  "n0 = n2 = n3 = 8 and n1 = n4 = 7, then the sequence of Ni would be 8, 7, 8, 8, 7, this, obviously, has largest element 8, so s = 8.".174.3.125.23 (talk) 10:23, 17 November 2014 (UTC)
 * Or do you mean: "n0 = n2 = n3 = 7 and n1 = n4 = 8, then the sequence of Ni would be 7, 8, 7, 7, 8, this, obviously, has largest element 8, so s = 8."?174.3.125.23 (talk) 10:25, 17 November 2014 (UTC)
 * I don't know the name of the bars, they indicate the size of the set, they're extremely common in set notation. You could write max for sup. No, Ni is the number of j's so nj = ni; in my example, for example, nj = n0 exactly when j = 0, 2, 3, thus, there are 3 such j so that N0 = 3. I return to my point that you ought not use such notation, that this is not immediate is a strong indicator that it's only going to obfuscate meaning, not clarify it (which is the point of notation). I suggest, instead, grabbing a book that covers basic set theory stuff - or, to mature your mathematical sense, a nice intro to abstract algebra or groups, and go from there. I'm not trying to sound condescending, it's just the best route to go instead of learning things piece meal, trust me.Phoenixia1177 (talk) 10:38, 17 November 2014 (UTC)
 * I understand that this is quite complex notation. Long story short, I'd like still to understand this.  So what I don't understand is why nj = ni:  They shouldn't be equal because the values that they indicate are not equal.174.3.125.23 (talk) 11:00, 17 November 2014 (UTC)
 * Essentially, Ni counts how many n's equal ni. So, using my example, n0 = 8, so N0 will be how many n's are equal to 8; in this case, there are 3, n0, n2, n3. The notation after "s =" means to pick the largest value of the N's, in this case it will be 3. For another example(I'm going to write ni for ni because it's easier) if you had n0 = 0, n1 = 2, n2 = 0, n3 = 5, n4 = 2, n5 = 0, n6 = 0. Then, N0 is how many equal 0, which is 4; N1 is how many equal 2, which is 2; N2 is, again, how many equal 0, so 4; N3 is how many equal 5, so 1; etc. The whole sequence of N's is 4, 2, 4, 1, 2, 4, 4. The maximum of this sequence will be 4, so s = 4.Phoenixia1177 (talk) 11:39, 17 November 2014 (UTC)


 * The notation for size of a set is described at Set_(mathematics). Vertical bar is the descriptive name for the character. It is also used in a similar (but different!) way for absolute value, sometimes norm_(mathematics) -- in often carries a vague connotation of "size" for different mathematical objects. SemanticMantis (talk) 17:58, 17 November 2014 (UTC)

Ball-drawing problem
I've been trying to calculate the expected value of how many random draws it would take to get three different colours when taking successive balls without replacement from a bag containing 5 red, 5 blue and 5 yellow, without success. So I wrote a program to generate all permutations of the order in which the 15 balls could appear and to determine the critical draw for each. Averaging over all permutations gave the fraction 917281/252252. Firstly, is this correct and secondly, how could it (or the correct value) be calculated directly?→86.171.209.142 (talk) 13:13, 17 November 2014 (UTC)
 * I guess you could do it by calculating the probability that the critical draw is $$n$$ for every $$n$$ from 3 to 11, and make the appropriate sum. That probability would be $$\frac{5!5!5!}{15!} \cdot \sum_{k = \max(n-6, 1)}^{\min(5,n-2)} 3 * \frac{(n-1)!(15-n)!}{k!(n-k-1)!4!(5-k)!(6+k-n)!}$$. The 3 represents the choice of which color is drawn with the critical draw.  Fixing one of the other colors, $$k$$ is the number of balls drawn of that color before the critical draw.  $$\frac{(n-1)!}{k!(n-k-1)!}$$ is the number of ways to permute the balls before the critical draw, while $$\frac{(15-n)!}{4!(5-k)!(6+k-n)!}$$ is the number of ways to permute the balls after the critical draw.  So $$\sum_{n = 3}^{11} \frac{n\cdot 5!5!5!}{15!}\cdot \sum_{k = \max(n-6, 1)}^{\min(5,n-2)} 3 * \frac{(n-1)!(15-n)!}{k!(n-k-1)!4!(5-k)!(6+k-n)!}$$ should be the expected draw.--80.109.80.31 (talk) 14:06, 17 November 2014 (UTC)
 * I plugged

Sum[n/(5!)^3 Sum[  3*((n - 1)! (15 - n)!)/(k! (n - k - 1)! 4! (5 - k)! (6 + k - n)!), {k, Max[n - 6, 1], Min[5, n - 2]}], {n, 3, 11}]
 * Into Mathematica and got 32487/16000 ~ 2.03, which is obviously wrong. So there's probably an error somewhere. -- Meni Rosenfeld (talk) 14:25, 17 November 2014 (UTC)
 * Furthermore, dropping the n doesn't give you probabilities that sum to 1.
 * I think you should replace $$\frac{1}{5!5!5!}$$ with $$\frac{5!5!5!}{15!}$$ (one over the number of different permutations). This gives you probabilities that sum to 1; and a different result (51/11) from the OP's, which actually seems more plausible to me (the OP's result is 3.64, which represents a very high chance to get it right the first time). -- Meni Rosenfeld (talk) 14:45, 17 November 2014 (UTC)
 * You're absolutely right. I've edited my first answer.--80.109.80.31 (talk) 14:58, 17 November 2014 (UTC)
 * Update: The OP's fraction is very close to 51/11 - 1. Yet the denominator is close to 756756, the number of permutations. Also, my own numerical testing confirms the result of roughly 51/11. So I expect that:
 * The OP had a few off-by-one errors
 * Both answers given (up to the corrections above) are correct.
 * Meanwhile I'm trying to run an exhaustive search (I did it in a really inefficient way so it might take some time), which might further confirm the results. -- Meni Rosenfeld (talk) 15:00, 17 November 2014 (UTC)
 * My Brute-force search has completed and confirmed the result of 51/11. -- Meni Rosenfeld (talk) 17:34, 17 November 2014 (UTC)
 * Being told that my answer was wrong made me re-examine the logic of how the permutating-generating program counted the draws until the third colour was seen, with the result that the ratio should have been 1169532/252252, which does indeed reduce to 51/11. Thanks for the answers.→86.171.209.142 (talk) 18:21, 17 November 2014 (UTC)
 * Not exactly what the OP wanted, but coupon collector's problem is close,. Robinh (talk) 19:28, 17 November 2014 (UTC)

What is the percentage of 5 out of 6?
? Thanks.  Uhlan  talk  23:39, 17 November 2014 (UTC)
 * 83.33....%. This has no finite decimal representation, but you can round it wherever appropriate for your application. If it's really 5 out of 6, I'd call it 83%. --Stephan Schulz (talk) 00:17, 18 November 2014 (UTC)
 * Thanks for that! Sorry had a complete mindblank.  Uhlan  talk  01:54, 18 November 2014 (UTC)


 * Have you memorised the percentage value of all other fractions (quite some feat), or have you forgotten how to work them out (not quite so impressive)? Because just being told the answer would only help the former case, not the latter.  --   Jack of Oz   [pleasantries]  03:42, 18 November 2014 (UTC)


 * Either he's trolling (odd, given his contribs) or thinks this is easier that typing 5/6= into google. Either way, asked, answered, and no ref relevant. μηδείς (talk) 05:53, 18 November 2014 (UTC)


 * I just forgot how to work it out.  Uhlan  talk  08:06, 18 November 2014 (UTC)
 * The easiest way to approximate it is to divide 6 into 500, then shift the decimal sign (long division). So, 6 x 83 = 498, 500 / 6 is around 83. Since we added two 0's, we need to shift the decimal by two 0's; so 83 becomes 0.83. Of course, since we are converting to % we have to shift it back by two places, and get around 83%. In general, going from decimal to % you shift the decimal two places to the right; so 0.546 becomes 54.6%. When estimating with small numbers, adding two or three zeroes to the top, then finding a close division (like with 83, above) will give you a close estimate for most situations. If you don't need an estimate by hand, just use a calculator to do the division, shift the decimal two to the right to get the %.Phoenixia1177 (talk) 08:12, 18 November 2014 (UTC)
 * I was trying to do 5 divided 1 times 100, rather than 5 divided by 6 times 100, but thanks. I didn't just google it because I was on the Reference desk already so just quickly posted it.  Uhlan  talk  08:20, 18 November 2014 (UTC)
 * No problem; that's why we're here: math questions:-)Phoenixia1177 (talk) 08:28, 18 November 2014 (UTC)