Wikipedia:Reference desk/Archives/Mathematics/2007 July 19

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July 19

mathematics of poker (no limit texas hold 'em)

hello, I enjoy playing this game online and live. It seems difficult to find some nice down to earth nuts and bolts mathematics........from first principles. The wiki article is great but a bit over my head.

Newsgroups and forums seem to say....."hey idiot.....you're an idiot..LOL... etc".

What I'm after is how to play successful solid play that constantly gets results (over the long term).

Do any of you think this is possible?

People program bots to play online (so is there a formule?) and there are software programs, do they help?

sorry to ramble, please post any links or opinions.

many thanks. —Preceding unsigned comment added by 91.105.116.186 (talk • contribs) 13:01, 19 July 2007

Well, live and online are different beasts (different factors bluffing being the main one, also card counting). I imagine the bots have a certain threshold for play/fold/bet based on probability. As for a way to win consistently, well, people have been trying to do that since throwing bones was invented. :) --Cronholm¹⁴⁴ 13:16, 19 July 2007 (UTC)

I'll go into some more details. If you play better than your opponents, then you will usually win, and if you're betting money you will probably make a profit in the long run. However, everyone tries to win and play the best they can, so this will be very difficult. In online play, I suppose the key factor is evaluating probabilities of different events, something that a computer program can be very good at (much better than a human). However, using them probably counts as cheating. I don't think there is any easy solution to improve your unassisted play, but learning a table of probabilities for each hand can help you evaluate what your opponents are likely to have. -- Meni Rosenfeld (talk) 13:25, 19 July 2007 (UTC)

I should also point out that poker isn't solely a mathematical game. A great deal of the game is the psychological subjective analysis of gauging how your opponents will respond to your betting actions based on your potential hands and their potential hands. Even in online play there is going to be a good deal of psychology involved. It's not too surprising that the winner of the 2007 World Series of Poker, Jerry Yang (poker player), has a psychology degree. Dugwiki 15:01, 19 July 2007 (UTC)

If your actually planning to make a bot, or to play for a large profit, I see no reason it would be better then stealing. Capitalism is supposed to make it so you need to do something good with your life in order to survive. If you make money gambling or stealing, rather then having a job, or having a full-time as opposed to a part-time job, you take more from society then you give back. That, being said, you never actually said whether you where doing it for that, you just are curious (probably pretty harmless), or you want to have more fun playing with small amounts of money (ethically questionable, along with gambling in general). — Daniel 16:51, 19 July 2007 (UTC)

Just a suggestion, but the philosophical opinions on whether or not gambling is stealing and the stated purpose of capitalism are better left for philosophical discussion boards.

Dugwiki 18:01, 19 July 2007 (UTC)

(from OP) I am not building "a bot"... just using it as an example of mathematical poker processing. I guess my original question is "Is it possible to play poker and win consistantly over the long term?". As for gambling=stealing.......I hope your pension scheme busts!

Practise and become skilled at the game. There is no mathematical shortcut.--Cronholm¹⁴⁴ 18:48, 19 July 2007 (UTC)

"Is it possible to play poker and win consistantly over the long term?" Well, think of it this way - assume that there is a perfect poker strategy, that will give you an advantage over all other players (not necessarily win every hand, but certainly over the long term). Now, assume that all players in a single game know this strategy. Because order of play rotates around the table, even if the strategy involves order of play all players still get an equal shot. Follow the possibilities through to the conclusion, and suddenly everyone has the same advantage, which means that no-one has an advantage. So therefore the "perfect strategy" cannot exist. Confusing Manifestation 22:36, 19 July 2007 (UTC)

This is assuming everybody else follows the Nash equilibrium. They virtually never do. If at least one person on the table is imperfect, everyone else can consistently make money. I think there is a strategy that would allow you to make money when possible, but this likely wouldn't be the best because you aren't taking full advantage of your opponents mistakes. Someone could take advantage of you trying to take full advantage of them, but this is unlikely. About my comment earlier, I figured the annoyance I gave everyone would be worth the small possibility that anon was planning on trying to make a significant amount of money and my comment would stop him. This is apparently not the case. — Daniel 23:33, 19 July 2007 (UTC)

Wikipedia cannot make you a winner, because your opponents also read wikipedia. Bo Jacoby 19:35, 22 July 2007 (UTC).

One factor that hasn't been mentioned is that in a "friendly game of poker", if every player played equally well, each would be expected to go home with exactly the money they brought to the game, on average. In a casino or online game, however, "the house always wins". That overstates it a bit, as it is possible to beat the house, on occasion, but the odds are in the house's favor so it wins more often, on average. This is accomplished via subtle rule changes such as "the dealer wins on ties, and the house always deals". Thus, any system you develop, like card counting, would need to sufficiently shift the odds in your favor to overcome this rules bias, or you would still lose, on average. StuRat 07:48, 24 July 2007 (UTC)

In poker, there's nothing subtle about the "house advantage". It's usually just a rake. J Elliot 17:44, 25 July 2007 (UTC)

Mathematics at the investment bank

Knowing mathematics, I am thinking about seeking employment in one of the following areas:

• quantitative analysis
• algorithmic trading
• risk management

I would very much appreciate your comments about this. Are they as mathematical as I think? Do they differ in terms of working hours and salary? Are they perhaps not separated from eachother at all, but handled by the very same employees? Stochastic calculus is one thing to read up on I guess, anything else? Oh, and about quantitative analysis, is there such a thing as qualitative analysis? Thanks! —Bromskloss 14:07, 19 July 2007 (UTC)

There are possibilities of employment in mathematical finance inside the academia; if you are really interested in the math that's what you should look into. Many great financial firms will ask for PhD's or high qualifications in math, but those requirements are more of a guarantee for them that you are serious than actual knowledge/proficiency requirements. In the end you are just plugging in numbers into calculators and doing the same kind of analysis day after day. On the other hand, of course, the salaries in the academia simply do not compare. As for your question about qualitative analysis, it really depends on what meaning you want to give quantitative or qualitative in this context. People in mathematical finance sometimes rely on methods that are quite soft (although not as much as mathematical economists), but I'm not sure that's what you mean. Even then, I personally know people who would dismiss most of probability, and especially measure theory methods, as soft, and that's what they need the most in finance. Phils 16:42, 19 July 2007 (UTC)

When I hear qualitative, I think for example about finding the zero-crossings and extreme points of a curve, whereas quantitative would be computing actual values of it (perhaps numerically). I'm looking forward to more answers from everyone! —Bromskloss 07:18, 20 July 2007 (UTC)

Here is my input ...

• Algorithmic trading has the least mathematical content of these three areas. It is mostly programming and data mining, looking for correlations between different prices, interest rates etc. Of course, if you look hard enough at enough data, you are bound to find some patterns ...
• The theory of risk management is very mathematical, but the practice is much less so. One of the practical problems faced by risk management departments is that they have to explain their results to senior management, regulators and traders, who want simple, clear cut messages backed up by a few numbers. So the theory is usually simplified a lot before it is applied.
• Quantitative anaylsis (in a trading context) has the potential for the most mathematical content. There is less pressure to have simple, easily understandable models - no one worries to much about whether they understand a quant model, as long as it makes money. May involve early starts and long hours, depending on which markets you are working in.

This is based on my experience in UK/European investment banks and fund managers - but I have never actually done any of these jobs myself, so I am not an expert. Gandalf61

That's very valuable information, thank you! I would be prepared to get up early and work very late (not forever, perhaps) if the pay is great. Is it? How about the other areas? Are there any overlaps between the areas? —Bromskloss 09:08, 20 July 2007 (UTC)

If you have a practical bent, you could consider applied maths. Engineering has some cool applied maths problems in Image Processing, for example. Challenging, infinitely variable, creative. What more could a geek ask for? Oh, yeah, money. Probably not as financially rewarding as being a good quant, but you do want to keep your soul, don't you? :) 196.2.111.133 14:50, 20 July 2007 (UTC)

Ah, right, my soul. Just the other day, someone told me, only half jokingly, that investment bankers must be the most callous people there are. Is there anything to this and how does it show? When you say that applied maths in engineering is "probably not as financially rewarding", do you meant that it is even close? —Bromskloss 19:32, 20 July 2007 (UTC)

An actuary gets paid very well, I believe. 80.2.205.183 22:26, 24 July 2007 (UTC)

Mental calculation

Could you, please, cite as many books about mental calculation as possible? Other sources of information such as web pages (printer-friendly if possible) are also accepted. Thanks. --Taraborn 17:05, 19 July 2007 (UTC)

The article Mental calculation looks like a good starting point. Dugwiki 17:44, 19 July 2007 (UTC)

This man's a genius (irony mode off). Now, anyone? --Taraborn 13:27, 20 July 2007 (UTC)

Well, that article has all of the obvious internet links, as for listing books, well, google book search... What can we do that it can't?--Cronholm¹⁴⁴ 13:41, 20 July 2007 (UTC)

(After edit conflict) Taraborn - Dugwiki gave you good advice - there are more than a dozen links to other web resources at the mental calculation page. Or you could look at the articles on specific systems, such as Vedic mathematics, the Trachtenberg system or Chisanbop - these have links to relevant web sites and books. Or you could go to your favourite on-line bookshop and search for books with "mental arithmetic" in the title - Amazon lists over 200 books. Gandalf61 13:56, 20 July 2007 (UTC)

Well, that article has all of the obvious internet links, as for listing books, well, google book search... What can we do that it can't? Of course, when I ask something, it's because it's not obvious. As for the latter part of your statement, maybe you had already read (something I currently seriously doubt) a book about the topic which you found to be particularly useful, something Google Book search can't do, but apparently that's not among your capabilities. Google book searching isn't useful: just try either mental arithmetic or mental calculation and you get loads of garbage or material directed towards children. The last reply by Gandalf61 was a better one: unfortunately, I'm already aware of those articles (but they're too short to significantly speed up my calculations) and the links provided in the mental calculation article don't fit my needs, they are too obvious. I was looking for something more advanced, such as the book title provided in the Trachtenberg system article, I was wondering if there were more of this kind. --Taraborn 15:45, 20 July 2007 (UTC)

I have read such a book, thank you very much. Its name escapes me now(it was directed a accountants and the like), but I will look through my bookshelves and find it. The reason that you were getting vague replies was the ambiguous nature of your question. "Cite as many books about mental calculation as possible" --Cronholm¹⁴⁴ 15:52, 20 July 2007 (UTC)

OK, sorry then. English is not my first language (nor my second nor third, for instance), therefore don't expect absolute precision in my statements. Instead of poking fun at my poorly written question by providing vague answers, I guess it would be better for everyone just to try to answer my question as if it was better written. It may not be perfect English, but it isn't written in hieroglyphics either. --Taraborn 08:21, 21 July 2007 (UTC)

Taraborn - your original question is not poorly written - its grammar and spelling are fine - and no-one poked fun at it. It is clear and easily understood. It is, however, imprecise because you did not explain what routes you had already tried or what constraints you would place around your search. If you had explained at the beginning that you had already read the relevant Wikipedia articles, had already tried various search engines, and wanted to find something similar to the Trachtenberg system, you may have got more useful responses. As things stand now, any editors who could give you more targetted information may decide not to offer their suggestions because of the uncivil tone of your responses above. You may have, as we say, "shot yourself in the foot". Gandalf61 10:26, 21 July 2007 (UTC)

Indeed. "Don't bite the hand that feeds you" is still good advice. Communication involves transmission, channel, and reception. Question your transmission before criticizing the reception. Read How To Ask Questions The Smart Way, and take it to heart. --KSmrq^T 10:40, 21 July 2007 (UTC)

Maybe my expectations of reference desk users are too high since I guess (or hope) everyone that asks here doesn't ask trivialities or things than can be answered with a 5 second Wikipedia/Google search, but apparently this doesn't seem to be something uncommon. Even more, there is a list of tips at the top of the page, and one of them begins: "Search first...". So, I assume every question asker has already read, at the very least, the Wikipedia article about the main topic of their question. If I ask a question about mental calculation and I'm told to go read the Wikipedia article, I am told, in other words, that I haven't done the _least_ research on my own, that I'm unable to use the simplest Internet search engine and, summing up, that I want my job to be done by others, which is far from being the case. Maybe I'll have to introduce myself every time I ask a question, saying that I've already read at least some directly-related Wikipedia articles and I've already done basic research. Replying just with a "go read the Wikipedia article about your question" when that is assumed is an insult. Aha?

For the lazy reader: If you had explained at the beginning that you had already read the relevant Wikipedia articles, had already tried various search engines, and wanted to find something similar to the Trachtenberg system, you may have got more useful responses. That's something all question askers have, supposedly, already done.

As things stand now, any editors who could give you more targetted information may decide not to offer their suggestions because of the uncivil tone of your responses above. If somebody is not going to answer a question for that reason, it's highly unlikely they can come up with a good response. --Taraborn 15:35, 21 July 2007 (UTC)

Unfortunately, we do get too many questions "that can be answered with a 5 second Wikipedia/Google search", and it is only natural for us to start with the most obvious answers. Even if Dugwiki's original response was not useful for you, it is certainly more productive to politely say so (and explain more clearly what it is that you are after) than to bite. -- Meni Rosenfeld (talk) 11:07, 22 July 2007 (UTC)

Okay, I didn't know that. I'll moderate my tone next time. --Taraborn 16:29, 22 July 2007 (UTC)

Ugh

Remove the brackets from:

$(2-x)(1-x)$

So:

$=2-x-2x+x^{2}$

$=2-3x+x^{2}$

However, my book gives $x^{2}-3x+2$ as a result. What have I done wrong? 86.138.36.33 17:25, 19 July 2007 (UTC)

Nothing, you just need to rearange the equation at the end. Addition is commutative so (a+b) = (b+a). --Salix alba (talk) 17:37, 19 July 2007 (UTC)

(edit conflict) Indeed, nothing. Your answer has constant term 2, linear term −3x, and quadratic term</math> x²; so does the book answer. Does the word commutativity mean anything to you? --KSmrq^T 17:41, 19 July 2007 (UTC)

It does, but consider x = 3. Then in my result we have 2 - 9 + 9 = -16 whereas the book result we have 9 - 9 + 2 = 2, right? Unless there is a priority to go from left to right in the first equation. I can't see where it shows that. 86.138.36.33 17:43, 19 July 2007 (UTC)

2 - 9 + 9 = 2, not 16. –King Bee ^{(τ • γ)} 17:58, 19 July 2007 (UTC)

Can you show why that is? Why is there a priority over the subtraction? 86.138.36.33 18:03, 19 July 2007 (UTC)

Order of operations, the good ol' days :). Anyway, there isn't 2 - 9 = -7 and -7 + 9 = 2 oila! --Cronholm¹⁴⁴ 18:16, 19 July 2007 (UTC)

The secret is to treat all subtractions as adding the opposite, so 2 - 9 = 2+ (-9) and then rearrange as much as you like. Likewise division is really just multiplying by the reciprocal. Donald Hosek 18:27, 19 July 2007 (UTC)

Off-topic, but an irresistible old joke:

• Q. Expand (a+b)⁴.
• A.
  (a + b)⁴
  (a  +  b)⁴
  (a    +    b)⁴
  …

Sorry, I couldn't help myself. ;-) --KSmrq^T 07:41, 20 July 2007 (UTC)

As to why 2 - 9 + 9 is 2 not 16, its due to the fact that + and - are Left-associative (see Common operator notation), 2 - 9 + 9 is always interpreted as (2-9)+9 not 2-(9+9). --Salix alba (talk) 11:51, 20 July 2007 (UTC)

The minus sign is used in two ways, and that is confusing to the beginner but it is too late to change. The minus sign in the expression −9 means 'change of sign', and the minus sign in the expression 2−9 means subtraction. Beginners make sign errors due to this convention. If you want to make fever sign errors, then substitute '+m ' for each minus sign. Then 2−9+9 = 2+m9+9 . The latter expression contains no minus signs. The symbol m, shorthand for (−1), satisfies the equation m+1 = 0. By the way it also satisfies the equation mm = 1, [because m+1 = 0 = 0m = (1+m)m = m+mm, and m+1 = m+mm implies 1 = mm]. So 2+m9+9 = 2+(m)9+(1)9 = 2+(m+1)9 = 2+(0)9 = 2+0 = 2. This way of killing minus signs makes calculations safer for the pupil even if the teacher dislikes it because she has long ago grown accustomed to the minus sign. Your original problem is, step by step

(2−x)(1−x)

= (2+mx)(1+mx) [substituting '+m ' for '−']

= (2+mx)(1)+(2+mx)(mx) [using the rule a(b+c)=ab+ac]

= (2)(1)+(mx)(1)+(2)(mx)+(mx)(mx) [using (b+c)a=ba+ca twice]

= 2+mx+2mx+mxmx

= 2+m3x+xx [because mx+2mx = 1mx+2mx = (1+2)mx = 3mx = m3x and mm = 1]

= 2−3x+x² [reinstalling the minus sign in order to make the teacher happy].

Bo Jacoby 17:20, 20 July 2007 (UTC).

What a neat method, once I realized that subtraction was the equivalent of adding a negative number I was good to go(age=?), but we all learn differently.--Cronholm¹⁴⁴ 19:40, 20 July 2007 (UTC)

This elaborate computation is both confusing and irrelevant: it only reproduces the answer we were presented in the question!

From an abstract algebra perspective, we have the group of (real) numbers under addition, with identity (zero) and additive inverse (negation). As a distinctly separate, and more difficult, algebraic structure we have the field with a second binary operator (multiplication) distributing over addition, and its identity (one) and multiplicative inverse (reciprocal). The field causes enormous headaches for students, namely fractions, division by zero, and multiplying negative numbers.

Yet here we have an anomaly. How is it that the person posting the question could handle the more complicated algebra of multiplying two linear polynomials (in fact, properly multiplying negatives), yet could still be confused about the elementary group of addition? Peculiar! --KSmrq^T 20:25, 20 July 2007 (UTC)

I was a bit bewildered the first time I was asked to divide by a negative, ${\frac {a}{-b}}$; I think I wrote two possible answers, equivalent to ${\frac {-a}{b}}$ and $-ab$, because when you add or subtract a negative you do the opposite with a positive, so might not division with a negative be the same as multiplication? My teacher didn't understand my perplexity. —Tamfang 18:05, 24 July 2007 (UTC)

The basic rules of algebra for positive integers: (a+b)+c = a+(b+c), (ab)c = a(bc), a+b = b+a, ab = ba, a(b+c) = ab+ac, and a+b = a+c implies b = c, also applies for polynomials involving positive integers. So one can define 0 as the solution to x+1 = 1, so that 0+1 = 1, and m = −1 as the solution to x+1 = 0, so that m+1 = 0, and i as a solution to xx = m, so that ii = m, and the difference a−b as the solution to the equation x+b = a, which is x = a+mb, and the fraction a/b as the solution to bx = a, and the algebraic numbers as solutions to f(x) = g(x) where f and g are polynomials with positive integer coefficients. To me this is the easier way to algebra. All we need is plus and times and solving polynomial equations. We don't need special rules for subtraction and division and roots. Real numbers belong to topology, not algebra. The stated anomaly is due to the fact that each positive integer has a standard representation, such that equality of integers are reflected by equality of representations, (in the equality 9+9 = 18, the question is 9+9, and the answer is 18), but the equality of polynomials are not reflected by equality of representations, $2-3x+x^{2}$ and $x^{2}-3x+2$ are equivalent answers to the question $(2-x)(1-x)$. Bo Jacoby 08:53, 21 July 2007 (UTC).

Now you're just being silly.

• If I want a solution to x²−2 = 0, or if I want to know the area inside a unit circle, I should take a topology class?!! (That's a rhetorical question; please do not try to answer it.)
• Unless you can read minds, you cannot explain the anomaly I refer to, which concerns the peculiar state of the poster's understanding. It is as if someone can pole vault, but not walk.

At this point, I would prefer to hear about the poster's thinking, not yours. --KSmrq^T 09:53, 21 July 2007 (UTC)

You are free to disagree, but not to be rude. I cannot read minds but I can read the text saying that the poster assumed that his result should match exactly that of the book: "$2-3x+x^{2}$. However, my book gives $x^{2}-3x+2$ as a result. What have I done wrong? ". Bo Jacoby 22:05, 21 July 2007 (UTC).

But you can also read KSmrq's text, where the anomaly he has referred to (which is not related to multiple representations of polynomials) is quite clearly explained. -- Meni Rosenfeld (talk) 23:11, 21 July 2007 (UTC)

Exactly. My interest, as from the beginning, is to answer the poster's question. What mental model is in that mind? The multiplication was done correctly, but the comparison with the book was not. Unless we better understand the thinking behind that mistake — which only the poster can tell us — our attempts to correct it are blind. --KSmrq^T 23:34, 21 July 2007 (UTC)

I think this has been said above, but when you list terms, I'm pretty sure you do them in decreasing power, so x⁷ would come before x⁶ and anything less, etc. 68.39.174.238 18:36, 22 July 2007 (UTC)

The problem was put $(2-x)(1-x)$ rather than $(-x+2)(-x+1)$, thus violating that rule. A power series is always written with the constant term first like this

$a_{0}+a_{1}x+a_{2}x^{2}+\cdots$

because there is no term with maximum exponent. Bo Jacoby 19:16, 22 July 2007 (UTC).

There is no fixed convention about term order. Two examples: With a monic polynomial, where the coefficient of the term of highest degree is stipulated to be 1, the term degrees would typically be highest on the left, decreasing to the right. When describing a generic polynomial of flexible degree, typically the terms would begin with the constant term and increase in degree to the right.

When a polynomial has more than one indeterminate, as in x²−2xy+y², it is obviously impossible to choose increasing order for all indeterminates. In fact, the question of term order has interesting consequences in the construction of a Gröbner basis for a collection of polynomials in multiple indeterminates ("multinomials"); options include

• lexicographic
• degree lexicographic
• degree reverse lexicographic
• and more

We're not done yet. The sum-of-monomials form is only one of many useful ways to write a polynomial. For example, another choice of great practical interest is the Bernstein-Bézier form.

Incidentally, convention conflicts in all areas of mathematics create an unending challenge for Wikipedia editors.

• Do the natural numbers begin 0, 1, 2, … or do they begin 1, 2, 3, …?
• Ever heard of Archimedes' constant?
• Should we write [0,1) or [0,1[ for the interval 0 ≤ x < 1?
• Is a "rng" different from a "ring"?

In an encyclopedia that anyone in the world can edit, many readers and editors are sure to assume that their convention is the correct (or only!) one. As if that isn't bad enough, we also get the occasional editor who invents a new notation, and insists Wikipedia ought to use it because it is so much better than the ones in professional use. --KSmrq^T 14:39, 23 July 2007 (UTC)