Wikipedia:Reference desk/Archives/Mathematics/2011 October 3

= October 3 =

Scrabble probabilities
I was playing a game of regulation rules Scrabble earlier today, and was led to contemplate the question: What is the likelihood that throughout the game I will possess every letter of the alphabet at least once? Clearly this depends upon my play strategy, my knowledge of the English language, and the English language itself. For simplicity, assume I play an average of L (1 ≤ L ≤ 7) letters per turn, and that I'm a reasonably good player (at least half of the letters I play each turn score above the average of the points available to me from the sum of the tiles). Also assume my opponent plays exactly as I do. There's a lot of variables here, and I'm not sure if my stipulations are reasonable or complete; please fill in the blanks if I've missed something. But the essential question is what intrigues me, and perhaps it's actually too complex for meaningful computation. Thanks for the input. — Anonymous Dissident  Talk 10:37, 3 October 2011 (UTC)
 * By symmetry, I suppose it stands to reason that I'd have 50 tiles throughout the course of the game. That probably makes things significantly simpler. — Anonymous Dissident  Talk 10:49, 3 October 2011 (UTC)
 * It looks like a form of Hypergeometric distribution. But I'm having problems wrapping my enfeebled brain around the maths :( --Tagishsimon (talk) 11:12, 3 October 2011 (UTC)


 * If you assume both players are equal, then it is natural to make the approximation that you both have an equal chance of encountering any tile, and that the odds of doing so are solely determined by the letter frequency. You each get 50 draws.  Based on Scrabble letter distribution and a quick Monte Carlo I get that probability of encountering every letter (not including blanks) is about 1 occurrence for every 1700 games.  What really hurts your odds are the five singleton tiles.  You'll miss at least one of them more than 96% of the time.  Dragons flight (talk) 11:17, 3 October 2011 (UTC)


 * Incidentally, if you let blanks count towards missing letters, you can "encounter" every letter in one game out of every 75. Dragons flight (talk) 11:25, 3 October 2011 (UTC)
 * Umm. Having found a hypergeometric calculator, I get the probability that you'll draw all five singletons as 0.028. IINAM, &c. --Tagishsimon (talk) 11:26, 3 October 2011 (UTC)

Hypergeometric distribution wtf
As a follow-up to the scrabble question above, would someone be good enough to explain how this formula from Hypergeometric distribution works - specifically, the step from the bracketed numbers to the 5.18... / 1027... thanks --Tagishsimon (talk) 11:32, 3 October 2011 (UTC)


 * $$ P(X=4) = f(4;50,5,10) = {{{5 \choose 4} {{45} \choose {6}}}\over {50 \choose 10}} = {5\cdot 8145060\over 10272278170} = 0.003964583\dots. $$
 * the dot between 5 and 8145060 indicates they are multiplied, it's not a decimal point. Apart from that it's just substituting in the values for the binomial coefficients and doing the sum.-- JohnBlackburne wordsdeeds 12:07, 3 October 2011 (UTC)
 * Umm, sorry, I'm still not getting it. How does the 50 10 bracketed number equal 1027... it's the sum that I cannot do :( --Tagishsimon (talk) 12:18, 3 October 2011 (UTC)


 * $${50\choose 10}=\frac{50!}{40!\,10!}=\frac{50 \times 49 \times \dots \times 42 \times 41}{10 \times 9 \times \dots \times 2 \times 1}=10272278170$$ Gandalf61 (talk) 12:34, 3 October 2011 (UTC)


 * See Binomial coefficient. —Bkell (talk) 14:43, 3 October 2011 (UTC)

You might consider doing obvious cancellations before multiplying. You would find that
 * $$ {5\cdot 8145060 \over 10272278170} = {5\cdot 6 \over 7567}. $$

Michael Hardy (talk) 01:21, 4 October 2011 (UTC)

If you order the calculation of the binomial coefficient $${50\choose 10}$$ like this
 * $$(((((((((41/1)42/2)43/3)44/4)45/5)46/6)47/7)48/8)49/9)50/10$$

all divisions produce integer results
 * $$(((((((((41))43)11))23)47)2)49)5$$

The multiplication can be done by hand. The J (programming language) provides extended precision when you postfix an x to the last number */41 43 11 23 47 2 49 5 1.02723e10 */41 43 11 23 47 2 49 5x 10272278170 and you may also use the build-in binomial coefficient verb (!) 10!50x 10272278170 Bo Jacoby (talk) 12:47, 4 October 2011 (UTC)

Thanks both. The information I was missing was Binomial coefficient, something I'm not at all familiar with. Am now. (well, a little). --Tagishsimon (talk) 13:07, 4 October 2011 (UTC)

Why do we have odd and even numbers
It seems obvious to me that some properties of numbers in maths must exist for the system to work. 10 must be larger than 6 for our number system to have meaning, and I've heard somewhere that certain things are impossible without the concept of the number 0. But why were odd and even numbers...invented? And at the very least, why are they still taught? I don't see what the concept gains us, or what it does at all. Thanks. Prokhorovka (talk) 14:39, 3 October 2011 (UTC)


 * Odd and even properties are used a lot. For example, I was working on a method in a paper that had a step in which if width of the bast was even, it had to be divided by two. If it is was odd, it shouldn't be divided. It could have stated "if the width of the base if evenly divisible by two", but stating "if the width of the base is even" is shorter. -- k a i n a w &trade; 14:45, 3 October 2011 (UTC)


 * (edit conflict) It's because lots of things work either one way or another, depending on whether some number is even or odd. For example, $$(-1)^n$$ is 1 if n is even, or −1 if n is odd. Negative numbers have real odd roots (cube roots, fifth roots, seventh roots, and so on), but do not have real even roots (square roots, fourth roots, sixth roots, and so on). A polynomial always has a real zero if its degree is odd, but might not if its degree is even. The graph of a power function is symmetric about the y-axis if the power is an even integer, or symmetric about the origin if the power is an odd integer. A few examples from graph theory: A cycle graph has a perfect matching if it has an even number of vertices, but not if it has an odd number of vertices. A graph is bipartite if and only if it contains no cycles of odd length (even-length cycles don't matter). A connected graph has an Eulerian circuit if and only if the degree of every vertex is even. And so on. This dichotomy between even and odd is ubiquitous in mathematics. See Parity (mathematics) for more. —Bkell (talk) 14:58, 3 October 2011 (UTC)


 * For a more "real-world" example, in many parts of the United States at least, houses with odd numbers are on one side of the street, and houses with even numbers are on the other side, so if you understand the difference between odd and even numbers you know which side of the street to look at when searching for an address. A similar scheme is often used for office numbers in large buildings, with even-numbered offices on one side of a corridor and odd-numbered offices on the other side. —Bkell (talk) 15:07, 3 October 2011 (UTC)

How to calculate Deming regression
Can I have the expressions used to determine the "a", "b", and the correlation coefficient of a data set using Deming regression? --Melab±1 &#9742; 19:17, 3 October 2011 (UTC)
 * The formulas, except for the correlation coefficient, can be found in our article on Deming regression. Looie496 (talk) 21:52, 3 October 2011 (UTC)
 * It is not clear to me which ones are which. --Melab±1 &#9742; 22:22, 3 October 2011 (UTC)
 * It's not clear what you mean by "a" and "b". If you mean the coefficients of y = a + bx, they are given by the values β0 and β1.  And also, I should have realized before, the correlation coefficient is not altered by Deming regression -- you just use the standard formula. Looie496 (talk) 21:57, 4 October 2011 (UTC)

Given 2 lines that cross in a multidimensional space, how to find the angle?
Hi, I have 2 points in a 50-dimensional space. If I have the straight lines that join those points with the origin, is there a way to find the angle between them? — Preceding unsigned comment added by 190.226.26.168 (talk) 20:58, 3 October 2011 (UTC)
 * The dot product, which works in all dimensions not just two or three. See in particular the definition and geometric interpretation.-- JohnBlackburne wordsdeeds 21:15, 3 October 2011 (UTC)


 * ... and this defines the angle between the lines even if they don't cross.   D b f i r s   08:10, 4 October 2011 (UTC)
 * Well, they do cross, because the lines you're considering when you take the dot product of the coordinates of two points both pass through the origin. —Bkell (talk) 14:18, 5 October 2011 (UTC)
 * Apologies for not reading the question carefully -- the lines in the 50-D space clearly pass through the origin. I was thinking that, in general, the dot product of the direction vectors ignores the positions of the lines, so the method also defines the angle between skew lines.   D b f i r s   21:07, 5 October 2011 (UTC)

Lebesgue integrals
Hello math buddies! I am interested in learning how to do Lebesgue integrals as an independent study project to extend my (fairly strong and rigorous) calculus knowledge. Unfortunately I have a very woeful set theory background so I need something that builds up to measure theory (from which Lebesgue is constructed) assuming little or no prior knowledge of set theory, ie by covering the set theory needed. I am not looking for something that will take me really deep into set theory because I anticipate remediating my lack of set theory in the near future, I'm looking for just enough so I can do Lebesgue. Can anyone recommend a work or work(s), again preferably not unnecessarily heavy on the other aspects of set theory or measure theory, and that has a nice amount of rigour but is still pretty relaxed so I can get through a lot of material quickly? Thanks a bunch :) 187.115.202.178 (talk) 21:20, 3 October 2011 (UTC)
 * An introduction to the Lebesgue integral and measure theory, at least at my school, is covered in a second-semester undergraduate real analysis course. For the most part, it is rather self-contained. If you're that uncomfortable in working with sets, perhaps Halmos' "Naive Set Theory" could be a sufficient introduction to set theory; it can give you a fairly good idea of what is going on behind the curtain without getting bogged down in a lot of details, and fairly quickly at that. Nm420 (talk) 14:09, 4 October 2011 (UTC)
 * Lebesgue integration on Euclidean spaces, by Frank Jones. Buy it.  You'll thank me later.  :-)  Sławomir Biały  (talk) 17:39, 4 October 2011 (UTC)
 * This is the problem with the mathematics curriculum. Set theory is never taught properly, yet they dump a whole load of crap on us without letting us understand the very foundations. Students typically end up memorizing definitions, formulas etc. What a great way to waste time. My advice is to spend the time to obtain a strong background in set theory if you're going to do research. Money is tight (talk) 10:00, 9 October 2011 (UTC)