How Does The Casino Edge Do the job?
There are many ways that a casino game can be programmed. All types of casinos have several formats for each game type, so it’s important to understand just how each type is programmed so as to choose a machine that may offer the best odds of winning. The chances of a casino game derive from a mathematical equation and so are also dependent on the skill of the players participating. It might take a casino’s expert staff a lot of effort and time to manually analyze all the details of every game. This means that the gambling house cannot guarantee that any provided machine will give the most favorable odds.
Most casinos provide a variety of casino game types that people can choose from. You can find basically three categories of casino games available: table games, gaming machines, and slots. Table games available to individuals include blackjack, baccarat, craps, etc. Many table game titles are progressive, and therefore the jackpot grows as more folks participate in the overall game. Most progressive slots equipment are integrated with electronic devices, which track the results of each spin and update the outcomes on the display.
Countless casinos also offer progressive slots and baccarat, which are famous because they are known for his or her excitement and fast payoff. Progressive slot machines are integrated with gadgets that allow the spins that occurs automatically, which ensures that the result of each spin is certainly unpredictable. In baccarat, the ball player pays a specific sum (either coins or credits) to spin a wheel and receive a specific amount of cards, then has the option to place her or his bet. Each card will probably be worth a certain value, based on the combination that arises.
Another common game offered by casinos is the binomial distribution. A binomial distribution comes after the same basic principle found in figures – random variables are assumed to possess independent values. The binomial distribution comes in two forms: the standard and exponential. The standard binomial distribution is a simple game like roulette where the outcome is pre-determined, while the exponential binomial distribution is a more difficult game like lottery or gambling establishment game where the outcomes depend on the values of real quantities, rather than being pre-determined. For the exponential distribution, the casino game email address details are based on the arithmetic mean of the random variables.
In many casino games, the home advantage means that a person player comes with an edge over other players because he or she has a greater amount of playing successes or less losses. It’s called the gambling establishment game edge, since it gives the one who has it an advantage over others. The size of the edge can be by means of an adjusted rate of come back or perhaps a percent advantage. People tend to view the edges in casino games as a way to an end – for instance, an individual participant might feel a one-percent border because she or he plays lots of games; however, this one-percent edge doesn’t mean anything if all the competitors play lotteries or if there are no restrictions.
Another exemplory case 얀 카지노 of the casino game edge originates from the standard deviation. The typical deviation is the deviation of the random variables. For example, take the jackpot sum in a simple game like roulette. Typical deviation may be the number of points which are distributed randomly between your winning number and the total amount that will be paid out. The standard deviation can be thought of as a “what if” type of scenario in which the actual jackpot may be larger or smaller than the expected jackpot.
One of the most well-known examples of an advantage is the baccarat video game. In a baccarat game, persons place their bets and wait for them to pay off. You can find typically a high roller and a minimal roller. A higher roller bets huge amounts of money, and a minimal roller bets small amounts. When the time comes for the low roller to pay off, then it is time for the great roller to gain because his bet was higher than the low roller’s.
If you have an excellent grasp of statistics, then you know that casino games with high odds of success are known as blackjack or roulette. The key reason why they’re called casino games is due to the casinos. Blackjack, roulette, baccarat, poker, etc. are variations of exactly the same game. They are all about chance, so the chances of your winning on each of the variations above are just as likely.


In the post, someone has written WLC asking him to justify the following assertion he has made in another argument
A collection formed by successive addition cannot be actually infinite.The writer basically says that he intuitively feels that this assertion by WLC makes sense, and yet he was wondering if WLC can unpack the intuition and get into the the details. WLC says yes and starts his argument by thinking about adding finite things together. He says
In the case of beginning with some finite quantity and adding finite quantities to it we can pinpoint the problem clearly: since any finite quantity plus another finite quantity is always a finite quantity, we shall never arrive at infinity even if we keep on adding forever. Infinity in this case serves merely as a limit which we never attain.What he has written here is wrong, but the reason why it is wrong is a little bit subtle, and it depends on what he means by 'never' and 'forever'. To make his example a little bit more concrete, let's assume the things I am adding up are baseball cards. I have a pile of baseball cards in front of me and I keep adding more baseball cards to the pile, every time I add new cards to the pile, I am adding a finite number of them. You might ask at what point in the future will I have a pile with infinite cards, and the answer I would give is never. There is no point in time in the future at which I will have an infinite number of cards in front of me. This is the only way I can think to justify his statement that we will never arrive at infinity.
However, in this situation, I did not "keep on adding forever", I added for an arbitrarily large finite amount of time. These two things are very different. By talking about adding forever, you are talking about taking a limit (which he mentions). In the normal way we think about time, it is true that we can never arrive at that infinity, but we also can't keep adding forever. If we change the setting to where it is possible to add forever, then it is also possible to arrive at infinity. We are adding up all of the baseball cards that are now or ever will be in the pile. That collection is infinite.
There is something very important here that I want to highlight, at any given time on this timeline, how many baseball cards remain that have yet to be put onto the pile? Infinity! No matter how far into the future we go, there is always an infinite number of baseball cards that still need to be put onto the pile.
Back to WLC's argument, he now tackles an infinite past instead of an infinite future. Suppose we have been adding baseball cards to a pile with no beginning, in other words, we have been making this pile forever into the past, and we end at some time and wind up with an infinite pile of baseball cards. he says the following
Now notice that one still hasn’t explained how we are able to form our infinite collection of baseball cards by successive addition. For at any time in the past the collection is already infinite, and yet the total collection has not yet been formed. The total collection will not be formed until the last card is added. From any point in the past one need add only a finite number of cards to complete the collection. But that leaves unsolved the problem of how the entire infinite collection could have been formed by successive addition.Let's pick this apart a bit. He says "For at any time in the past the collection is already infinite", that's true, since we have an infinite past, we have already added up an infinite number of things, no matter how far back in time we go. He also says "From any point in the past one need add only a finite number of cards to complete the collection" which is also true, because every point in the past is a finite distance from the present. But then his last sentence is "But that leaves unsolved the problem of how the entire infinite collection could have been formed by successive addition." But there is actually no problem here. If we return to our other situation where we are counting into the future, at any point in time we saw that there is an infinite number of things remaining to be put onto the pile. Here everything is turned backwards, so at every point there is an infinite number of cards that have already been put on the pile. If at every second we have put a card on the pile, and time goes infinitely into the past, then we will have an infinite number of cards because an infinite number of seconds has past. There is no problem here.
He then writes the following paragraph, which I have seen in various forms at various places by him
in order for the collection to be completed, we must have already enumerated, one at a time, an infinite number of previous cards. But before the final card could be added, the card immediately prior to it would have to be added; and before that card could be added, the card immediately prior to it would have to be added; and so on ad infinitum. So one gets driven back and back into the infinite past, making it impossible for any card to be added to the collection.He seems to think he has a contradiction here, but I think he explains himself quite poorly. However, as far as I can tell, his final example is another version of this same problem, and it is much easier to understand what he is trying to say and identify the mistake he has made:
It gets even worse. Suppose we meet a man who claims to have been counting down from infinity and who is now finishing: . . ., -3, -2, -1, 0. We could ask, why didn’t he finish counting yesterday or the day before or the year before? By then an infinite time had already elapsed, so that he should already have finished. Thus, at no point in the infinite past could we ever find the man finishing his countdown, for by that point he should already be done! In fact, no matter how far back into the past we go, we can never find the man counting at all, for at any point we reach he will already have finished. But if at no point in the past do we find him counting, this contradicts the hypothesis that he has been counting from eternity. This shows again that the formation of an actual infinite by never beginning but reaching an end is as impossible as beginning at a point and trying to reach infinity.I think in this paragraph WLC once again, demonstrates his complete lack of understanding of what he is talking about. This guy has been counting up from negative infinity and he just now reached zero. WLC wants to say this is impossible, and to challenge it he is going to go back in time 10 seconds and confront the guy
WLC "what number are you on?"
WLC "how can you be on -10? you've already counted an infinite number of things!"
guy "be that as it may, I'm at -10 right now"
WLC "but if you've already counted an infinite number of seconds shouldn't you be done?"
guy "nope, you can go as far back in time as you want and talk to me, I will always have counted an infinity of numbers already, it won't mean I'm already at zero."
You see, the problem here is that WLC has conflated two different ideas. One is that the guy has counted down from infinity and finished. The other is that the guy has counted down from infinity, and when he reaches zero it is the first time he has reached infinity. This second idea makes absolutely no sense. You will never have a finite number, add 1 to it, and get infinity. But that is essentially what he is saying. "ah ha! you just got to zero which means you counted an infinite number of things, why didn't you reach infinity 1 number before that?" "well, I did".
The sad thing is that WLC is actually touching on a concept which is actually kind of neat, but of course since he is trying to say this stuff is impossible there is no way he will see how cool it is. Let's consider time with an infinite past as a number line. Every second is a point on that number line but let's not put any labels on the seconds at first. Now, let's say some guy is walking along our number line putting labels down and it just so happens that when he gets to us he puts down the zero. This is only 1 possible labeling of our timeline. We might ask why he didn't have the zero 10 seconds ago, and he will just say that it is not where it goes. He's been counting forever and this is how the labeling worked out. However, a different person, also counting down from infinity, might just have the zero in the same spot the first guy put -10. Absolutely any second on our timeline could potentially be the zero, but once it is set, all other numbers are determined.
Another of WLC's examples demonstrates his lack of understanding. He is searching for "absurdities" with infinities, and describes the following
Consider the scenario imagined by al-Ghazali of our solar system’s existing from eternity past, with the orbital periods of the planets being so co-ordinated that for every one orbit which Saturn completes Jupiter completes 2.5 times as many. If they have been orbiting from eternity, which planet has completed the most orbits? The correct mathematical answer is that they have completed precisely the same number of orbits. But this seems absurd. Think about it: the longer Jupiter and Saturn revolve, the greater becomes the disparity between them, so that they progressively approach a limit at which Jupiter has fallen infinitely far behind Saturn. Yet, being now actually infinite, the number of their respective completed orbits is somehow magically identical. Indeed, they will have “attained” infinity from eternity past: the number of completed orbits is always the same. So Jupiter and Saturn have each completed an infinite number of orbits, and that number has remained equal and unchanged from all eternity, despite their ongoing revolutions and the growing disparity between them over any finite interval of time. This strikes me as nuts.This paragraph demonstrates his complete lack of mathematical understanding. He is right though, it does seem absurd at first, and yet it's true. How can the 2 planets have taken the same number of orbits when one has taken 2.5 times as many orbits? Well let me ask you this? What is 2.5 times infinity? Still infinity, right?
Let's change this question slightly: Which set has more numbers in it, all integers or all even integers? Your first instinct is probably to say all integers, but that is wrong, the correct answer is they are the same size. This is hard to understand at first, but without too much trouble we can get most people to see why it is true. That is why it is a standard problem that most math majors will see in their "introduction to proof" class by the time they are half way through their sophomore year. If William Lane Craig was in that class while I was teaching it, he would have failed.


Hi, everyone. Yesterday I wrote that it was 1-AU day (the day when the earth is exactly the average distance from the sun). Thanks for reading. Later in the day, my friend Marc, an actual astronomer, not a guy like me, pointed out that I had the wrong day: “the JPL Solar System ephemeris (which is as accurate as it gets in this world) has the Earth-Sun distance at 0.99999999989877 AU at 2017-Apr-03 16:39:33 UTC (to within one second — one second later and one second sooner the distance being further from 1 AU that the value at 19:39:33). BTW, (1.0-0.99999999989877) AU is 15 meters.
“This is unsatisfying, so I took the 3600 values (spaced at 1-second intervals) around it and fit a fifth-order spline to it and interpolated to the time corresponding to 1.0 AU. Turns out that’s 2017-Apr-03 16:39:33.031 UTC, the 1-millisecond precision corresponding to a distance error of about 7 cm.”
Much of this is beyond my math reach. Rather than be a peddler of fake news, I’ve corrected the post so that it’s correct for tomorrow, April 3. Sorry about that. I hope I didn’t ruin too many conversations.


Monika Kapoor Math 10 Class Book 17
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Monika Kapoor Math 10 Class Book 17: A Comprehensive Guide for CBSE Students
Monika Kapoor Math 10 Class Book 17 is a textbook of mathematics for class 10 students following the CBSE curriculum. It is written by Monica Capoor, a dedicated mathematics teacher with more than 30 years of professional experience in her chosen field[^2^]. The book covers all the topics prescribed by the CBSE syllabus, such as real numbers, polynomials, linear equations, quadratic equations, arithmetic progressions, triangles, coordinate geometry, trigonometry, circles, constructions, areas related to circles, surface areas and volumes, statistics and probability. The book also includes sample papers, previous year papers, NCERT solutions and additional exercises for practice and revision.
The book aims to help students develop a clear understanding of the concepts and methods of mathematics. It also encourages them to apply their knowledge to solve problems in various situations. The book uses a student-friendly approach that involves activity-based learning, examples, illustrations, diagrams, tables and graphs. The book also provides tips and tricks to enhance the speed and accuracy of calculations. The book is designed to cater to the needs of both average and meritorious students.
Monika Kapoor Math 10 Class Book 17 is one of the most popular and trusted books for class 10 mathematics among CBSE students. It has received positive feedback from teachers and students alike for its comprehensive coverage, clarity of explanations, accuracy of solutions and relevance of questions. The book is available in paperback format and can be purchased online or from any book store. The book is published by Ratna Sagar Private Limited, a leading educational publisher in India.
The book has received rave reviews from both students and teachers who have used it for their class 10 mathematics preparation. Some of the reviews are as follows:
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Dr. Carol Flannagan is a Research Associate Professor and Director of the data center at the University of Michigan Transportation Research Institute. She earned an M.A. in Statistics and a Ph.D. in Mathematical Psychology from the University of Michigan. Dr. Flannagan develops analytical methods for and analyzes a wide variety of transportation-related data, including travel, driving, crash, and injury outcome data. Her work has been implemented both in state planning processes and in in-vehicle algorithms for automatic crash and injury notification. In recent years, she has been thinking about applications of Bayesian statistics and cognitive psychology to improve computer image processing.


TAKE THE NATIONAL NUMERACY CHALLENGE TODAY
‘Don’t let numbers hold you back – join the national numeracy challenge today!’ Chris Humphries CBE, Chairman, National Numeracy
Are you the type of person that says ‘I can’t do maths’? Well, why not try the National Numeracy Challenge and start saying ‘I can do maths’.
Improving your everyday maths can help you make your money go further, choose the best energy deals, plan journeys, check your wage slips, and help with homework.
Lots of people have bad memories of maths at school and often people think that they can’t do it. The National Numeracy Challenge offers a friendly, interactive website designed to help you assess your everyday maths and improve skills in bite-sized steps – building confidence all the way.
Take the challenge today via the links below - you might surprise yourself!
Access code: CELL1
Access code: CELS1
If you feel you need to brush up your skills, CEL offers free qualification courses in Maths to support you in the future. Courses are designed at different levels to suit your needs.
To enroll you will need to book a pre-course assessment interview. Your skills level will be assessed and you will be placed on an appropriate course at a convenient centre. Book a place by phoning 0208 314 3300 or by visiting a centre in person.


This book introduces the commutative property of multiplication to your child. The book will enable your child to learn/recognize the first property used in Algebra. It also enables the child to relate to the skills of basic addition and division. This book is guaranteed to teach the multiplication tables rapidly. This book is originally published last 2016 and has been revised in 2020.
|Publication date||March 11,2021|
|Interior Color||Full Color|
|Book Size||8.500" x 11.000" (280mm x 216mm)|
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To improve the grades or your child in the math subjects, you need to look for a math tutoring program. The math tutoring program will also help in strengthening the learning skills of your child. Thus, when you are considering the math tutoring program for your child, you should ensure that you make the right choice. You need to select the program that will give your child the best. The following are tips to help you to select the right math tutoring program.
The first thing should be determining the needs of your child. In this, you need to understand which area that your child is struggling. When you know on the area that your child needs to improve, you are able to determine the specific type of math tutoring program that will have a positive impact on your child. This site will give you good idea regarding math tutoring program.
Check on how the classes will properly fit into the schedule of your kid when looking for the perfect math tutoring program for your child. The best thing is to select the math tutoring program that is more flexible to allow you to create a class style which is more convenient for the child.
You need to talk to the tutor to make the right selection of the math tutoring program. Through talking with the tutors, you are able to discuss any questions that you have. Also, conversing with the tutors gives an opportunity to understand more about how the program works. You are able to check on which classes which are offered by the program. As a result, you will know of the math tutoring program will be right for your kid. Besides, you are able to check on how the tutor will be available and check on the progress reported of your child. Find more here: https://hellothinkster.com/blog/thinkster-math-vs-kumon-math-a-detailed-review
You need to follow up on the progress of your child when picking the right math tutoring program. When you have enrolled your child in a certain program that you think it fits your child best, it is best to frequently follow up on what your kid is doing. Check whether your child is getting as long with the math tutor. Also, you need to check if your child is making any progresses ad you expect. Talk to your kid to inquire of he is finding fun with the tutoring program. Ensure that you select the math tutoring program that will make your child be more comfortable and happy. This will help your child to focus more and thus can improve well.


- There are none, suggest some!
When working with Revit areas, either through an area scheme or architectural rooms, it’s common to create a schedule of those areas, and utilize the auto-calculation aspect of Revit to get the totals of the floor/department/etc.
If you interrogate your list of rooms, occasionally you will notice that the rounded values presented in the schedule end up not adding up to the totals presented in the schedule. There ends up being a discrepancy between some values, which can cause issues for clients or in planning.
To illustrate and discuss what is happening, we will use this example. Presented here are 4 rooms that were created in Revit, and the values associated with the rounding created in a schedule from Revit. Each column B through F was set to the Area of the space/room, with different rounding precision for each (0 decimals, 1 decimal, etc)
Maximum Precision – Column F
While we have read elsewhere that Revit stores 15 decimals for numbers, it only lets you schedule up to 12 decimals for the Area in a Room or Area schedule (done by “custom” with .000000000001 precision).
Therefore, we are stating that 12 decimals is the “maximum precision” you can get in an Area/Space; and when we discuss the “most precise” area measurement, this is the precision we are talking about (column F).
How Revit Rounds Spaces & Rooms
The main issue here is that Revit is rounding the most precise value it has for the room/space. It rounds this precise value to the exact specified rounding (1,2,3 decimals, etc).
The presented rounded values (Columns B, C, D, E) are simply rounded values, and not stored within the program. Therefore, they are not the values that get added together for the total.
Instead, to maintain accuracy, Revit is adding together the precise room values (Column F) and then rounding this value. This presents the most precise value it can for the Total Area because it is waiting until the last moment to round. This keeps the significant digits to 12 decimal places in the equation, until the final rounding to the decimal precision requested. This means that the value is more precise, but may not match the values of your rounded totals.
We chose this dataset because it can illustrate this issue. If you add up Column B rooms, you will get a value of 283 SF, while the actual rounded value is 282 SF. While minimal, this is still a discrepancy of 1 SF across 4 rooms.
The discrepancy is not limited to 1SF, and can increase; as an example, if we were to double the rooms to the exact same size, Revit’s tabulated total becomes 564 SF across 8 rooms, while the calculated total from the rounded values is 566 SF. This is a difference of 2 SF (double the earlier value).
While the decimals in our example for .1, .01 and .001 precision all are accurate to the totals, this is not always the case. Obviously, by its nature, the more precise values will be closer to the actual totals regardless, but there is the possibility that these are all off as well.
What’s actually happening
It generally depends on how many areas you have, and the distance to “5” (as this is the rounding cutoff) for the value of the decimal you want.
For example 4 spaces
- Values of .949 for 4 values will equal a total of 3.796. This rounds to 3.80 for 2 decimals.
- The 2 decimal value of .95 * 4 is equal to 3.80. This matches.
However, if we look at 5 spaces with this value.
- Values of .949 for 6 values will equal a total of 5.694. This rounds to 5.69 for 2 decimals.
- The 2 decimal value of .95 * 6 = 5.70
Typical Values Balanced by Probability
This is an extreme example. Typically, while .49 values are low, they are normally balanced out by various .51+ values that are high. Therefore, while its possible, its unlikely that your values are off by ‘that’ much ever.
Let’s say that you do want the total to represent an addition of the values that have been rounded. This is possible using a calculated parameter in your schedule, and hiding the ‘actual’ area value. Keep in mind that if you do a workaround, you are actually presenting a ‘less precise’ value as your total.
- Create a new calculated parameter. The type needs to be “Area”.
- Decide on the precision you want, and copy/paste the corresponding precision formula in (below).
- Make sure to set the Area parameter to be a ‘hidden value’
- Make sure to set the “Field Format” of the new calculated parameter to match your precision you set. It will default to your ‘projects’ precision.
Follow the formulas below for each precision level you want. For more precision, follow the pattern below.
Essentially what you are doing is multiplying the most precise value by a certain amount, to get the precision you want to be at the integer level. Then you are rounding that value to the nearest integer. Once you get that rounded value, you are then dividing by the amount you multiplied by in order to get back to a decimal amount. All the dividing and multiplying of SF are to resolve the units.
|0 decimal places||((round((Area * 1) / 1 SF)) * 1 SF) / 1|
|1 decimal places||((round((Area * 10) / 1 SF)) * 1 SF) / 10|
|2 decimal places||((round((Area * 100) / 1 SF)) * 1 SF) / 100|
|3 decimal places||((round((Area * 1000) / 1 SF)) * 1 SF) / 1000|


There is always a lot of discussion around how much rode to let out for a given depth of water. The consensus is always “more is better”. But how do you actually determine how much rode you have let out?
To help determine the correct amount of rode to let out to attain the optimal scope, you can use a rode counter. You can also mark your rode using coloured whipping line at regular intervals, cable ties, plastic ribbons, chain markers or even paint on a chain rode.
Alternatively you can employ a little trick. All you need to know is approximately how high off the water your bow is (B). Multiply this by 5 for a 5:1 scope, and you should see that much rode, when it is fully stretched out and taut, between your bow and where it enters the water (A).
The basis for this is some simple school geometry (Intercept Theorem): The ratio between two sides of a triangle stays the same regardless of the size of the triangle as long as the angles do not change.
Viewing the vertical depth and the outstretched rode (stretched taut) as two sides of a triangle, all you need to know is how high off the surface of the water your bow is. Say it is 5 feet. Multiply this by 5 if you want 5:1 scope, and you should see the resultant length of 25 feet of rode stretched out between your bow and where it enters the water. This means that you will actually have 25 feet visible above water and enough scope underwater regardless of the actual depth.
The ratio of the amount of rode you see above the water (A) to the bow height (B) is equal to the ratio of the length of rode you have let out (R) to the total height of your bow above the bottom (H) and either of these are equal to the Scope. So, if A/B = 25’/5’ = 5/1, then you indeed have a 5:1 scope!
So, finally you have a practical application for the math you were required to learn in school. Of course, this works best for a rope rode. An all chain rode with lots of weight tends sag unless conditions stretch it taut.
Note: this model supposes that the sea bed is horizontal. Naturally, the depth of the water under your boat will vary to a lesser or greater degree as you move away from the anchor’s location. However, this does not affect the scope, as it relates to the height of your bow over the sea bed at the point where your anchor is. Hence the validity of using a horizontal sea bed in this model! –
Thanks to Mike Pool for pointing out this possible source of confusion.
By Alex & Daria Blackwell
Alex and Daria Blackwell are the authors of “Happy Hooking – The Art of Anchoring.” It covers every aspect of anchors and anchoring in a fun and easy to read format with lots of photos and illustrations. It is available from good chandleries, Amazon & CoastalBoating.net


The Tower of Hanoi problem is a classic algorithmic puzzle that can be solved using a simple recursive algorithm. Given a stack of n disks, the algorithm moves the disks from the first stack to the last stack, using only three stacks and moving only one disk at a time. The disks must be moved so that the smallest disk is on top of the largest disk.
The candidate's solution is correct and demonstrates a level of completeness. The approach is clear and easy to follow.
Evaluated at: 2022-11-12 02:16:03


In the year 1961 the meteorologist Edward Lorenz examined a system of a few differential equations, a system, where not concrete points are known, from which one can calculate another. Instead, only the derivation of each point is known, defined by a function, so that approximations for successing points can be calculated. (Please, forgive me for this explanation of differential equations...)
Well, he made experiments and found out, that his result heavily depend on the used numerical precision. An incredibly small change of the starting value caused a totally different result. So the title of one of his publications was: Kann das Flattern eines Schmetterlings in Brasilien einen Orkan in Texas verursachen? (in english: Can the fluttering of a butterfly in Brasil cause a hurricane in Texas?). The answer / his answer was yes. Since then on, this effect is called Schmetterlingseffekt (in english perhaps effect of a bufferfly). Edward Lorenz then greatly simplified his model in order to study it and to experiment with it. It contained only 3 differential equations, thus we can experiment with it, too:
Mr. Lorenz experimented mainly with a=10, b=8/3 and different values for c. The resulting object can be interpreted as a 3-dimensional curve and, if an initial point is given, it's from a mathematical point of view (theoretically) definite, but not in practice.
Mr. Lorenz then chose c=28 and calculated the curve for various initial points. But although the curve started totally different, he found, that after a few seconds always the same figure appeared. It had a very complicated structure, it was built from an infinite number of loops, and the whole thing was very strange... It looked like the 3 differential equations stamped a very complicated structure into the 3-dimensional world in his computer, a fractal attractor, which attracts every point in this world. Because the structure was so complicated, such an attractor is called strange attractor.


Thus we use combinations to compute the possible number of 5-card hands, 52 C 5. This number will go in the denominator of our probability formula, since it is the number of possible outcomes. For the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck. Since there are four Aces and we.
The formula tends to influence vehicle design by prescribing the number of axles, inter-axle spacing and axle weights for a given maximum gross vehicle weight. As the gross vehicle weight increases the bridge formula tends to lengthen the vehicle thereby increasing the cubic cargo capacity which is a desirable productivity attribute. These size and weight changes prescribed by the bridge.
What is the Combination Formula? The number of combinations of n things taken r at a time is written as C(n, r). The formula is given by: If you are not familiar with the n! (n factorial notation) then have a look the factorial lesson. How to use the Combination Formula to solve word problems? Example: In how many ways can a coach choose three swimmers from among five swimmers? Solution: There.
Permutations and Combinations in mathematics both refer to different ways of arranging a given set of variables. Permutations are not strict when it comes to the order of things while Combinations are. For example; given the letters abc. The Permutations are listed as follows Combinations on the other hand are considered different, all the above are considered the same since they have the.
Find all combinations that equal a given sum with an amazing feature. Maybe all of the above methods are somewhat difficult for you, here, I will introduce a powerful-tool, Kutools for Excel, with its Make Up A Number feature, you can quickly get all combinations that equal to a given sum. Tips: To apply this Make Up A Number feature, firstly, you should download the Kutools for Excel, and.
Regression analysis is a set of statistical methods used for the estimation of relationships between a dependent variable and one or more independent variables Independent Variable An independent variable is an input, assumption, or driver that is changed in order to assess its impact on a dependent variable (the outcome). It can be utilized to assess the strength of the relationship between.
A typical combination lock for example, should technically be called a permutation lock by mathematical standards, since the order of the numbers entered is important; 1-2-9 is not the same as 2-9-1, whereas for a combination, any order of those three numbers would suffice. There are different types of permutations and combinations, but the calculator above only considers the case without.
Populating Sheet With Combinations Of Variables Jul 11, 2013. I'm attempting to build a large (let's be honest for a second this is going to be enormous) spreadsheet for a project I'm working on and TBH I'd rather not do it by hand. With the scale of numbers I'm working with I'll miss something and it'll take forever. Here's the deal: I have 5 variables. They must add up to 100. I need to.


Commercial architecture is an intricate fusion of creativity and precision, where mathematical principles lay the foundation for structural integrity and aesthetic harmony. The extent of math’s involvement is profound, impacting various design, planning, and construction stages.
1. Structural Engineering: Math is a cornerstone in calculating commercial structure load distribution, material strength, and stability. Engineers use complex equations to determine column spacing, beam dimensions, and foundational support, ensuring buildings can withstand environmental forces.
2. Space Optimization: Architects use mathematical concepts to maximize space efficiency while adhering to building codes. Measurements, ratios, and spatial arrangements guide floor layouts, ensuring optimal functionality and aesthetics.
3. Proportions and Aesthetics: The Golden Ratio and Fibonacci sequence are mathematical principles often applied to create visually pleasing proportions and compositions in commercial architecture.
4. Building Codes and Regulations: Meeting safety and accessibility regulations require precise measurements and calculations. Compliance with occupancy loads, fire codes, and egress requirements relies on accurate math.
5. Quantity Estimation: Math is pivotal in estimating material quantities for cost calculations. Accurate measurements ensure budgets align with project requirements.
6. Sustainability and Energy Efficiency: Calculations of thermal performance, solar angles, and insulation requirements enable architects to design energy-efficient buildings.
7. Construction Sequencing: Project timelines rely on accurate scheduling based on construction tasks and dependencies, requiring intricate mathematical coordination to ensure timely completion.
8. Digital Modeling: Advanced architectural software employs mathematical algorithms for 3D modeling, simulating real-world physics and structural behavior for informed decision-making.


Why should you care about expected value (EV)? Once you have a basic understanding of how expected value is applied to esports betting, you're one step closer to becoming a profitable bettor.
If we're going all Merriam-Webster on you, Expected Value (EV) is defined as the predicted average return of a variable. Okay now how about for a more helpful definition? It's the average outcome (or value) you can expect when dealing with something that has an element of luck or randomness involved, such as esports betting. We’ll show you how to calculate EV and all the ways it can be used to help you make more profitable decisions!
In any type of wagering, the EV of a bet would be the amount (on average) you expect to win/lose in the long-run by placing the same bet at the same odds many times over. The calculation for this would be:
EV = (Probability of winning x payout if you win) - (Probability of losing x the stake of the wager)
For example, say you and a friend are wagering $1 on many consecutive flips of a normal coin (no tricky business):
The calculation to find our EV of each coin flip would be:
Therefore, you nor your friend can expect to win or lose money, no matter how many times you flip the coin. If your friend decides to change the game and offers you $1.20 every time you win, yet you still only pay $1 when you lose, the EV becomes:
Your average return on a single flip is now $0.10. The game is now more profitable for you in the long run. Could you lose a few in a row and have your friend be winning for a bit? Sure. But as you flip more and more, you'll pull ahead and then pull away.
This same equation can be used when betting on esports. Let’s use this match as an example:
Using a $10 bet the steps to calculate the EV of betting on Luminosity would be:
1. Using the decimal odds, calculate the implied probability of each team winning, dividing 1 by their odds:
2. Calculate the payout of each side of the bet if it were to win:
(*Note: we need our profit, not the total payout!)
3. Substitute these numbers into the EV formula and calculate:
Therefore the EV of betting on Luminosity is -0.483, which means you will on average lose $0.48 for every $10 you bet.
If you calculate EV using the odds from the site, every bet will be -EV since the site needs their side to be +EV in order for it to be worthwhile for them to offer it (IE: Rivalry is a business and needs to make money to exist!). But don’t be deterred by this. Since it’s impossible to know the true win rates of the teams in a match, if you are able to better estimate them than the bookmaker, you will be able to show a profit.
If in the example above, Luminosity’s true win rate is higher than the 41.86% the site is implying, and it’s actually 49%, betting on them will become +EV.
Expected value calculations are a great tool to be used when making bets, EV is the building block in determining any profitable bet before you make it. Do your research and if you get better information and estimate true win rates better than the bookmaker, you can go make some +EV bets and win some money! GL HF!


AVR Single Digit Digital Dice "Guess the number"
Our digital dice will operate similar to that of a normal dice. When the button is pressed in the circuit below a roll will be simulated by displaying a sequence of number from 1- 6 then a number randomly obtained in the sequence would then be displayed on the seven segment display. This operation would repeat whenever the button is pressed.
- Atmel ATmega 16 Chip
- 7 Segment Display (CA or CC)
- Reset Button + 10k Resistor
- Power Source (5V)


STEAM and SEND tutoring
Developing confidence in numeracy and literacy skills from KS2 to KS3 and helping student to achieve to their potential for GCSE, AS and A Level sciences (KS4 and KS5 Science).
At ehcpservices we believe in helping your child to overcome their barriers to learning so they are able to achieve to their potential. Our approach is to deliver bespoke support focusing on the strengths of your child to build the skills they need and the confidence to exceed their academic potential in Cambridge.
Bachelors degrees in the subjects being taught
Postgraduate Teaching Qualification (or equivalent)
KS2 student comment
A letter written by one of our year 3 students telling his mum how his confidence in numeracy skills has developed.
KS4 Numeracy Intervention
Stretching able student beyond their key stage
KS4 maths skills being developed by looking at shapes found in nature.
The student was able to use her maths skills to decide how this shape should be drawn using a pencil, ruler and protractor.


By Brian Burke
The new Bayesian draft model is nearly ready for prime time. Before I launch the full tool publicly, I need to finish describing how it works. Previously, I described its purpose and general approach. And my most recent post described the theoretical underpinnings of Bayesian inference as applied to draft projections. This post will provide more detail on the model's empirical basis.
To review, the purpose of the model is to provide support for decisions. Teams considering trades need the best estimates possible about the likelihood of specific player availability at each pick number. Knowing player availability also plays an important role in deciding which positions to focus on in each round. Plus, it's fun for fans who follow the draft to see which prospects will likely be available to their teams. Hopefully, this tool sits at the intersection of Things helpful to teams and Things interesting to fans.
Since I went over the math in the previous post, I'll dig right into how the probability distributions that comprise the 'priors' and 'likelihoods' were derived.
I collected three sets of data from the last four drafts--best player rankings, expert draft projections (mock drafts), and actual draft selections. In a nutshell, to produce the prior distribution, I compared how close each player's consensus 'best-player' ranking was to his actual selection. And to produce the likelihood distributions I compared how close each player's actual selection was to the experts' mock projections.
Consensus Best Player Rankings
The prior distribution only needs to get us in the ballpark, and be approximately of the right shape. In this model, the prior is based on a consensus of best player available rankings. Best player rankings are overall rankings without regard to exactly where they would fit in the draft based on things like position or team need. I averaged rankings from ESPN's top 300, CBS Sports top 300, Gil Brandt's top 125, Kiper's top 100, and Scouts Inc.'s top 150.
Over the past four years, the consensus rankings are very good starting points for predicting when a player will be taken, but are rarely spot-on accurate. I assessed the accuracy of the rankings as predictors by looking at the distribution of each selection's error. In other words, how often was the ranking spot-on? What proportion of the time was it off by 1 pick late? 1 pick early? 2 picks late? etc...
This produced the plot below. The black line is the observed proportion of each degree of error. (It should actually be a probability mass rather than a density because the picks are discrete numbers, not a continuous variable. But for the sake of clarity, I plotted the distributions as densities. The results are unchanged.) An error of -1 means the player was actually chosen 1 pick earlier than his overall rank. An error of +1 means the players was actually chosen 1 pick later than his overall rank, and so on.
It's much easier to predict the pick # of the very top prospects than the third-day prospects of the 4th round and later. Not only is there far more attention and analysis for the top players, but there is a bound to the error of any prediction--You can't have a player chosen before the 1st pick. Because accuracy is largely dependent on where players are generally ranked, I created separate distributions for different regions of the draft. The one depicted above happens to be from the second half of the 1st round. I divided the prospects into groups of 1-5, 6-14, 15-32, 2nd round, 3rd round, and later.
You might have noticed that if the 15th best ranked player has an error of -20, that would mean he was actually 5 picks before the draft began. That's a result of aggregating the players into groups. Ideally, I'd prefer unique distributions for every individual pick #, but that's impossible given the amount of data available. To correct for the "negative pick #" flaw, I reallocated any probability mass that fell before the #1 pick into the positive region, by setting the probability of being picked before #1 to zero, and adjusting the rest of the distribution so it summed to 1.
The Bayesian 'likelihoods' were computed based on expert mock drafts using the same general techniques, with one important difference. As discussed in the previous post, Bayesian inference relies on inductive reasoning.
The prior distribution is the probability a player was taken with each selection number given his overall ranking. But the likelihood looks at the question from the other direction. The likelihood distribution is the probability a player was projected to be taken at each selection given when he was actually chosen. For instance, 'what's the probability the player who was actually chosen 10th was projected to be taken 10th? Projected to be taken 9th? 11th? Etc...
I treated each expert as a clone, using the assumption no one expert is significantly better than another, assuming the experts considered are among the most serious and most widely respected. While I would grant that one guy might be 'truly' better than another, it's probably very close, and we might need 100 years worth of data to tell with any degree of confidence. I'd prefer to have unique distributions for each expert, but that creates distributions that are too sparse. To establish the likelihood distributions, I used Kiper, McShay, and Mayock, but also included a couple additional mocks from Sports Illustrated and CBS Sports to round out the data.
Again, using the bkde function in R's Kern Smooth package, I created smoothed estimates of the likelihood distribution.
The complete results of the model applied to all of this year's prospects will be published soon.


Consider the following riddle:
You are robbing a bank and can only carry 100 pounds (45 kilograms) of currency out of the vault. Once inside, you see that the bank has organized its money into large piles of Dimes, Quarters, and Half Dollars. How do you pack your bag, taking from these piles, so that you maximize the amount of money you get?
(answer is below)
The answer is that it does not matter how you pack your bag, as all three of the listed coins have the same "value density" (i.e: 100 pounds of dimes is worth exactly as much as 100 pounds of quarters and 100 pounds of half dollars).
|Coin Name||Coin Value in Cents||Coin Weight in Grams||Density (cents/gram)|
Consult the United States Mint's currency specifications for proof.
By the way, the bag you take from the bank will have just shy of two grand.
100 (lb) * 453.59 (g/lb) * 4.409 (cents/g) * 0.01 (dollar/cent) = 1999.88 dollars


[tennisbiz] Looking for the proper magazine
I wrote an article "Ivanisevic's model" but can't publish it. Because in our
countries (Russia, Ukraine, ...) there are not tennis science magazines. My
article - it's mathematics analysis of Goran Ivanisevic play and synthesis
of formula to predict future develops of the power tennis.
Help me to find the proper magazine.
All the best.
Received on Sun Aug 26 2001 - 08:34:34 CDT


5 Statistical summaries
5.1 Revealing uncertainty
If you have information about the uncertainty present in your data, whether it be from a model or from distributional assumptions, it’s a good idea to display it. There are four basic families of geoms that can be used for this job, depending on whether the x values are discrete or continuous, and whether or not you want to display the middle of the interval, or just the extent:
- Discrete x, range:
- Discrete x, range & center:
- Continuous x, range:
- Continuous x, range & center:
geom_smooth(stat = "identity")
These geoms assume that you are interested in the distribution of y conditional on x and use the aesthetics
ymax to determine the range of the y values. If you want the opposite, see Section 16.1.2.
y <- c(18, 11, 16) df <- data.frame(x = 1:3, y = y, se = c(1.2, 0.5, 1.0)) base <- ggplot(df, aes(x, y, ymin = y - se, ymax = y + se)) base + geom_crossbar() base + geom_pointrange() base + geom_smooth(stat = "identity")
base + geom_errorbar() base + geom_linerange() base + geom_ribbon()
Because there are so many different ways to calculate standard errors, the calculation is up to you. For very simple cases, ggplot2 provides some tools in the form of summary functions described below, otherwise you will have to do it yourself. R for Data Science (https://r4ds.had.co.nz) contains more advice on working with more sophisticated models.
5.2 Weighted data
When you have aggregated data where each row in the dataset represents multiple observations, you need some way to take into account the weighting variable. We will use some data collected on Midwest states in the 2000 US census in the built-in
midwest data frame. The data consists mainly of percentages (e.g., percent white, percent below poverty line, percent with college degree) and some information for each county (area, total population, population density).
There are a few different things we might want to weight by:
- Nothing, to look at numbers of counties.
- Total population, to work with absolute numbers.
- Area, to investigate geographic effects. (This isn’t useful for
midwest, but would be if we had variables like percentage of farmland.)
The choice of a weighting variable profoundly affects what we are looking at in the plot and the conclusions that we will draw. There are two aesthetic attributes that can be used to adjust for weights. Firstly, for simple geoms like lines and points, use the size aesthetic:
# Unweighted ggplot(midwest, aes(percwhite, percbelowpoverty)) + geom_point() # Weight by population ggplot(midwest, aes(percwhite, percbelowpoverty)) + geom_point(aes(size = poptotal / 1e6)) + scale_size_area("Population\n(millions)", breaks = c(0.5, 1, 2, 4))
For more complicated geoms which involve some statistical transformation, we specify weights with the
weight aesthetic. These weights will be passed on to the statistical summary function. Weights are supported for every case where it makes sense: smoothers, quantile regressions, boxplots, histograms, and density plots. You can’t see this weighting variable directly, and it doesn’t produce a legend, but it will change the results of the statistical summary. The following code shows how weighting by population density affects the relationship between percent white and percent below the poverty line.
# Unweighted ggplot(midwest, aes(percwhite, percbelowpoverty)) + geom_point() + geom_smooth(method = lm, linewidth = 1) #> `geom_smooth()` using formula = 'y ~ x' # Weighted by population ggplot(midwest, aes(percwhite, percbelowpoverty)) + geom_point(aes(size = poptotal / 1e6)) + geom_smooth(aes(weight = poptotal), method = lm, linewidth = 1) + scale_size_area(guide = "none") #> `geom_smooth()` using formula = 'y ~ x'
When we weight a histogram or density plot by total population, we change from looking at the distribution of the number of counties, to the distribution of the number of people. The following code shows the difference this makes for a histogram of the percentage below the poverty line:
ggplot(midwest, aes(percbelowpoverty)) + geom_histogram(binwidth = 1) + ylab("Counties") ggplot(midwest, aes(percbelowpoverty)) + geom_histogram(aes(weight = poptotal), binwidth = 1) + ylab("Population (1000s)")
5.3 Diamonds data
To demonstrate tools for large datasets, we’ll use the built in
diamonds dataset, which consists of price and quality information for ~54,000 diamonds:
diamonds #> # A tibble: 53,940 × 10 #> carat cut color clarity depth table price x y z #> <dbl> <ord> <ord> <ord> <dbl> <dbl> <int> <dbl> <dbl> <dbl> #> 1 0.23 Ideal E SI2 61.5 55 326 3.95 3.98 2.43 #> 2 0.21 Premium E SI1 59.8 61 326 3.89 3.84 2.31 #> 3 0.23 Good E VS1 56.9 65 327 4.05 4.07 2.31 #> 4 0.29 Premium I VS2 62.4 58 334 4.2 4.23 2.63 #> 5 0.31 Good J SI2 63.3 58 335 4.34 4.35 2.75 #> 6 0.24 Very Good J VVS2 62.8 57 336 3.94 3.96 2.48 #> # … with 53,934 more rows
The data contains the four C’s of diamond quality: carat, cut, colour and clarity; and five physical measurements: depth, table, x, y and z, as described in Figure 5.1.
The dataset has not been well cleaned, so as well as demonstrating interesting facts about diamonds, it also shows some data quality problems.
5.4 Displaying distributions
There are a number of geoms that can be used to display distributions, depending on the dimensionality of the distribution, whether it is continuous or discrete, and whether you are interested in the conditional or joint distribution.
For 1d continuous distributions the most important geom is the histogram,
ggplot(diamonds, aes(depth)) + geom_histogram() #> `stat_bin()` using `bins = 30`. Pick better value with `binwidth`. ggplot(diamonds, aes(depth)) + geom_histogram(binwidth = 0.1) + xlim(55, 70) #> Warning: Removed 45 rows containing non-finite values (stat_bin). #> Warning: Removed 2 rows containing missing values (geom_bar).
It is important to experiment with binning to find a revealing view. You can change the
binwidth, specify the number of
bins, or specify the exact location of the
breaks. Never rely on the default parameters to get a revealing view of the distribution. Zooming in on the x axis,
xlim(55, 70), and selecting a smaller bin width,
binwidth = 0.1, reveals far more detail.
When publishing figures, don’t forget to include information about important parameters (like bin width) in the caption.
If you want to compare the distribution between groups, you have a few options:
- Show small multiples of the histogram,
- Use colour and a frequency polygon,
- Use a “conditional density plot”,
geom_histogram(position = "fill").
The frequency polygon and conditional density plots are shown below. The conditional density plot uses
position_fill() to stack each bin, scaling it to the same height. This plot is perceptually challenging because you need to compare bar heights, not positions, but you can see the strongest patterns.
ggplot(diamonds, aes(depth)) + geom_freqpoly(aes(colour = cut), binwidth = 0.1, na.rm = TRUE) + xlim(58, 68) + theme(legend.position = "none") ggplot(diamonds, aes(depth)) + geom_histogram(aes(fill = cut), binwidth = 0.1, position = "fill", na.rm = TRUE) + xlim(58, 68) + theme(legend.position = "none")
(We’ve suppressed the legends to focus on the display of the data.)
Both the histogram and frequency polygon geom use the same underlying statistical transformation:
stat = "bin". This statistic produces two output variables:
density. By default, count is mapped to y-position, because it’s most interpretable. The density is the count divided by the total count multiplied by the bin width, and is useful when you want to compare the shape of the distributions, not the overall size.
An alternative to a bin-based visualisation is a density estimate.
geom_density() places a little normal distribution at each data point and sums up all the curves. It has desirable theoretical properties, but is more difficult to relate back to the data. Use a density plot when you know that the underlying density is smooth, continuous and unbounded. You can use the
adjust parameter to make the density more or less smooth.
ggplot(diamonds, aes(depth)) + geom_density(na.rm = TRUE) + xlim(58, 68) + theme(legend.position = "none") ggplot(diamonds, aes(depth, fill = cut, colour = cut)) + geom_density(alpha = 0.2, na.rm = TRUE) + xlim(58, 68) + theme(legend.position = "none")
Note that the area of each density estimate is standardised to one so that you lose information about the relative size of each group.
The histogram, frequency polygon and density display a detailed view of the distribution. However, sometimes you want to compare many distributions, and it’s useful to have alternative options that sacrifice quality for quantity. Here are three options:
geom_boxplot(): the box-and-whisker plot shows five summary statistics along with individual “outliers”. It displays far less information than a histogram, but also takes up much less space.
You can use boxplot with both categorical and continuous x. For continuous x, you’ll also need to set the group aesthetic to define how the x variable is broken up into bins. A useful helper function is
ggplot(diamonds, aes(clarity, depth)) + geom_boxplot() ggplot(diamonds, aes(carat, depth)) + geom_boxplot(aes(group = cut_width(carat, 0.1))) + xlim(NA, 2.05) #> Warning: Removed 997 rows containing missing values (stat_boxplot).
geom_violin(): the violin plot is a compact version of the density plot. The underlying computation is the same, but the results are displayed in a similar fashion to the boxplot:
ggplot(diamonds, aes(clarity, depth)) + geom_violin() ggplot(diamonds, aes(carat, depth)) + geom_violin(aes(group = cut_width(carat, 0.1))) + xlim(NA, 2.05) #> Warning: Removed 997 rows containing non-finite values (stat_ydensity).
geom_dotplot(): draws one point for each observation, carefully adjusted in space to avoid overlaps and show the distribution. It is useful for smaller datasets.
What binwidth tells you the most interesting story about the distribution of
Draw a histogram of
price. What interesting patterns do you see?
How does the distribution of
Overlay a frequency polygon and density plot of
depth. What computed variable do you need to map to
yto make the two plots comparable? (You can either modify
5.5 Dealing with overplotting
The scatterplot is a very important tool for assessing the relationship between two continuous variables. However, when the data is large, points will be often plotted on top of each other, obscuring the true relationship. In extreme cases, you will only be able to see the extent of the data, and any conclusions drawn from the graphic will be suspect. This problem is called overplotting.
There are a number of ways to deal with it depending on the size of the data and severity of the overplotting. The first set of techniques involves tweaking aesthetic properties. These tend to be most effective for smaller datasets:
Very small amounts of overplotting can sometimes be alleviated by making the points smaller, or using hollow glyphs. The following code shows some options for 2000 points sampled from a bivariate normal distribution.
df <- data.frame(x = rnorm(2000), y = rnorm(2000)) norm <- ggplot(df, aes(x, y)) + xlab(NULL) + ylab(NULL) norm + geom_point() norm + geom_point(shape = 1) # Hollow circles norm + geom_point(shape = ".") # Pixel sized
For larger datasets with more overplotting, you can use alpha blending (transparency) to make the points transparent. If you specify
alphaas a ratio, the denominator gives the number of points that must be overplotted to give a solid colour. Values smaller than ~\(1/500\) are rounded down to zero, giving completely transparent points.
norm + geom_point(alpha = 1 / 3) norm + geom_point(alpha = 1 / 5) norm + geom_point(alpha = 1 / 10)
If there is some discreteness in the data, you can randomly jitter the points to alleviate some overlaps with
geom_jitter(). This can be particularly useful in conjunction with transparency. By default, the amount of jitter added is 40% of the resolution of the data, which leaves a small gap between adjacent regions. You can override the default with
Alternatively, we can think of overplotting as a 2d density estimation problem, which gives rise to two more approaches:
Bin the points and count the number in each bin, then visualise that count (the 2d generalisation of the histogram),
geom_bin2d(). Breaking the plot into many small squares can produce distracting visual artefacts.18 suggests using hexagons instead, and this is implemented in
geom_hex(), using the hexbin package.19
The code below compares square and hexagonal bins, using parameters
binwidthto control the number and size of the bins.
norm + geom_bin2d() norm + geom_bin2d(bins = 10)
Estimate the 2d density with
stat_density2d(), and then display using one of the techniques for showing 3d surfaces in Section 5.7.
If you are interested in the conditional distribution of y given x, then the techniques of Section 2.6.3 will also be useful.
Another approach to dealing with overplotting is to add data summaries to help guide the eye to the true shape of the pattern within the data. For example, you could add a smooth line showing the centre of the data with
geom_smooth() or use one of the summaries below.
5.6 Statistical summaries
geom_bin2d() use a familiar geom,
geom_raster(), combined with a new statistical transformation,
stat_bin2d() combine the data into bins and count the number of observations in each bin. But what if we want a summary other than count? So far, we’ve just used the default statistical transformation associated with each geom. Now we’re going to explore how to use
stat_summary_2d() to compute different summaries.
Let’s start with a couple of examples with the diamonds data. The first example in each pair shows how we can count the number of diamonds in each bin; the second shows how we can compute the average price.
ggplot(diamonds, aes(color)) + geom_bar() ggplot(diamonds, aes(color, price)) + geom_bar(stat = "summary_bin", fun = mean)
ggplot(diamonds, aes(table, depth)) + geom_bin2d(binwidth = 1, na.rm = TRUE) + xlim(50, 70) + ylim(50, 70) ggplot(diamonds, aes(table, depth, z = price)) + geom_raster(binwidth = 1, stat = "summary_2d", fun = mean, na.rm = TRUE) + xlim(50, 70) + ylim(50, 70) #> Warning: Raster pixels are placed at uneven horizontal intervals and will be #> shifted. Consider using geom_tile() instead. #> Warning: Raster pixels are placed at uneven vertical intervals and will be #> shifted. Consider using geom_tile() instead.
To get more help on the arguments associated with the two transformations, look at the help for
stat_summary_2d(). You can control the size of the bins and the summary functions.
stat_summary_bin() can produce
ymax aesthetics, also making it useful for displaying measures of spread. See the docs for more details. You’ll learn more about how geoms and stats interact in Section 14.6.
These summary functions are quite constrained but are often useful for a quick first pass at a problem. If you find them restraining, you’ll need to do the summaries yourself (see R for Data Science https://r4ds.had.co.nz for details)
So far we’ve considered two classes of geoms:
Simple geoms where there’s a one-on-one correspondence between rows in the data frame and physical elements of the geom
Statistical geoms where introduce a layer of statistical summaries in between the raw data and the result
Now we’ll consider cases where a visualisation of a three dimensional surface is required. The ggplot2 package does not support true 3d surfaces, but it does support many common tools for summarising 3d surfaces in 2d: contours, coloured tiles and bubble plots. These all work similarly, differing only in the aesthetic used for the third dimension. Here is an example of a contour plot:
ggplot(faithfuld, aes(eruptions, waiting)) + geom_contour(aes(z = density, colour = ..level..))
The reference to the
..level.. variable in this code may seem confusing, because there is no variable called
..level.. in the
faithfuld data. In this context the
.. notation refers to a variable computed internally (see Section 14.6.1). To display the same density as a heat map, you can use
ggplot(faithfuld, aes(eruptions, waiting)) + geom_raster(aes(fill = density))
# Bubble plots work better with fewer observations small <- faithfuld[seq(1, nrow(faithfuld), by = 10), ] ggplot(small, aes(eruptions, waiting)) + geom_point(aes(size = density), alpha = 1/3) + scale_size_area()
For interactive 3d plots, including true 3d surfaces, see RGL, http://rgl.neoscientists.org/about.shtml.


Yankton math teacher named South Dakota Teacher of the Year
Yankton Middle School math instructor Beth Kaltsulas has been honored as South Dakota's Teacher of the Year.
State Education Secretary Melody Schopp announced the award late Thursday. Kaltsulas will now be a candidate for National Teacher of the Year, an award given out at the White House.
Kaltsulas has taught in the Yankton School District for 30 years. She has mentored 13 student teachers, runs a daily math academy and piloted a math class for special education students.
(Copyright 2016 The Associated Press. All rights reserved. This material may not be published, broadcast, rewritten or redistributed.)


This fascinating pattern is created by the following steps:
- Pick 3 vertices for a triangle
- Assign a color for each vertex (In this case, red, green and blue)
- Pick an arbitrary point "p" that is inside the triangle, with an arbitrary color
- Pick, at random, one of the vertices.
- Find the midpoint, "m", between "p" and the randomly picked vertex, along with the mid-point color
- Draw "m", and replace "p" with "m"
- Repeat from step 4


An Introduction to Topological Data Analysis for the Digital HumanitiesLECTURE at 2-3pm on 28 Apr 2017 ⌖ Butler Studio (208b)
Persistent homology is a powerful tool from the field of topological data analysis that can help you understand the shape of and extract features from your high-dimensional datasets. In this workshop, Dr. Rachel Levanger will offer a friendly introduction to the subject at the conceptual level and then work through some examples. No background in mathematics is assumed.
Dr. Levanger (PhD, Rutgers U) is a is a postdoctoral researcher in the Department of Electrical and Systems Engineering at the University of Pennsylvania, specializing in topological data analysis.


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My Million Pound Penny Stock Punt Experiment V2.0:
Aim: to go from £1,000 to £1,000,00 in 10 trades.
Method: To invest £1,000 into a penny stock with 100% upside potential. As soon as I’ve doubled my money I will sell that stock and look for the next penny stock with 100% upside potential.
You can subscribe to this blog by clicking HERE
This is how the maths works:
Trade 1: £1,000 turns into £2,000 – you can see my 1st stock choice by clicking here
Trade 2: £2,000 turns into £4,000
Trade 3: £4,000 turns into £8,000
Trade 4: £8,000 turns into £16,000
Trade 5: £16,000 turns into £32,000
Trade 6: £32,000 turns into £64,000
Trade 7: £64,000 turns into £128,000
Trades 8: £128,000 turns into £256,000
Trade 9: £256,000 turns into £512, 000
Trade 10: £512,000 turns into £1,024,000
I have a shortlist of penny stocks I’ve screened to fit certain criteria that I believe is needed in order for them to achieve 100% upside.
In the last year or so I’ve also become a lot more familiar with the mechanics of penny stocks and how the junior market works.
So I’ve decided to go for, “My One Million Pound Penny Stock Punt Experiment v2.0”
To find out the first stock I’m going with please click here to sign up to my weekly blog


Calculus videos created by Mike McGarry, BA in Physics (Harvard), MA in Religion (Harvard), content creator at Magoosh (http://magoosh.com).
This lesson shows how to perform the Right-end, Midpoint, and Trapezoid method in estimating the area between curves using geometric figures. In this ...
Need more help? Set up a tutoring session with me over skype. Email me at email@example.com for more information. You can evaluate expressions and ...
From Graphing Calculator Review
This video demonstrates how to find the zeros of a function using any of the TI-84 Series graphing calculators. The Wabbitemu TI Calculator Emulator was used ...
From Cole's World of Mathematics
http://mcstutoring.com/ Private math tutoring and test preparation in Huntington Beach, CA. Subjects include ACT, SAT 1, algebra, geometry, and calculus.