The Importance of Steals

I read an interesting article about the importance of steals in the NBA last night. The writer, Benjamin Morris, measured the performance of teams with and without certain players. He compared the team's points per game with and without the player to the player's average stats and a replacement player's average stats; the chart below shows his results. For example, if a team is playing without a player that gets 12 rebounds per game who is being replaced by a player with just 2 rebounds per game, the team will probably average 17 fewer points.

As you can see in this chart, a steal is 9 times more valuable than a point! It's high value is understandable, as a steal will almost always result in two easy points in the resulting fast-break scoring opportunity and also shifts the momentum, but it's amazing that such an overlooked stat is so valuable. It's easier to glorify high-scoring players because the numbers are so much grander; Kevin Durant is leading the league with 32.2 points per game, while Chris Paul leads the league in steals with 2.53 steals per game. But let's look at the stats of these two players with the logic of this chart:

Durant:

32.2 PPG

1.3 STLPG

0.8 BLKPG

5.6 APG

7.6 RPG

3.6 TOPG

32.2(1) + 1.3(9.1) + 0.8(6.1) + 5.6(2.2) + 7.6(1.7) - 3.6(5.4) = 54.71

Chris Paul:

18.5 PPG

2.5 STLPG

0.1 BLKPG

11.0 APG

4.3 RPG

2.4 TOPG

18.5(1) + 2.5(9.1) + 0.1(6.1) + 11(2.2) + 4.3(1.7) - 2.4(5.4) = 60.41

With the logic of a player's contributions compared to a replacement player, it seems like our own Chris Paul of the Los Angeles Clippers is an even better player than Kevin Durant, the leading MVP candidate!

Works Cited:

http://espn.go.com/nba/player/_/id/2779/chris-paul

http://espn.go.com/nba/player/_/id/3202/kevin-durant

http://fivethirtyeight.com/features/the-hidden-value-of-the-nba-steal/

10.8 Polar Equations of Conics

In this section, we are finding the polar equations of conics and graphing them on polar graphs. In this example, we are given the type of conic (parabola), the eccentricity (1), and the directrix (y=-2). With a horizontal directrix below the origin, we can tell that the equation's denominator will be (1-esinØ). The directrix is also 2 units away from the focus [which is always at (0,0) for these problems], so we know that p is 2. The bottom row of this picture displays how to insert those numbers and find the final equation (boxed on the right).

10.6 Polar Coordinates

This section introduces us to graphing using polar coordinates rather than the usual rectangular (x,y) coordinates. Polar coordinates are (r,Ø) [that Ø is the closest thing to theta I could get]. The four equations in the picture below are used to transfer polar equations into rectangular equations and vice versa. 

This next picture is an example of changing a polar equation into a rectangular one. Knowing the angle, you can plug it into the top-left equation from the above picture. The final answer is boxed below:

This printable paper is helpful to use for graphing polar equations:

http://www.embeddedmath.com/downloads/files/polargraph/polargraph-letter.pdf

Wilt Chamberlain vs. The Modern NBA

In my free time, I like to look at stats of NBA legends to understand why they were so famous and successful. Today, I was reading about Wilt Chamberlain and I saw that in his 1961-1962 season on the Philadelphia Warriors, he had an average of 48.5 minutes per game in a game where there are only 48 minutes per game. I researched the history of the length of NBA games to see if they were longer back then or something, but I found that they've been 48 minutes since the beginning of the NBA. The only reason Wilt averaged more than 48 minutes per game is because of the few games that went into overtime.

On basketball-reference.com, I found that 10 overtime periods were played by the Warriors that season (5 single-overtime games, 1 double-overtime game, and 1 triple-overtime game). Each overtime period is 5 minutes long, so since there were 10 OTs, 50 extra minutes were played.

There were only 80 games in the season at the time, so the number of regulation minutes that season (48X80) was 3,840. With the 50 OT minutes, the total was 3890. Wilt played 3882 minutes that season. 3882/3890 = 99.79% of the total minutes played. Interestingly, I found on Wikipedia that the only reason he didn't play those 8 minutes was that he was ejected from one game with 8 minutes left after getting his second technical foul.

His 3882 minutes over 80 games in a season (3882/80) means he averaged exactly 48.525 minutes per game.

Wilt Chamberlain was 25 years old that season. To compare to a modern NBA superstar, Kevin Durant is 25 years old this season and averages 38.4 minutes per game this season. This number is already ten less than Wilt's average at the same age, and is probably higher than it would have been because Oklahoma City's other superstar Russell Westbrook missed 31 games this season. But even at this pace (in today's 82-game season), Durant would play 3148.8 minutes, over 700 minutes fewer than Chamberlain! This is certainly partially why Chamberlain retired after 14 seasons, while current NBA stars can last much longer. For example, Tim Duncan is in his seventeenth season on the San Antonio Spurs and still plays at an elite level. Because of increasing awareness about injuries and the fact that athletes usually want their career to last as long as possible, I don't think we'll ever see a player play as many minutes as Wilt again.


Works Cited:
http://www.basketball-reference.com/teams/PHW/1962_games.html
http://en.wikipedia.org/wiki/Wilt_Chamberlain
http://espn.go.com/nba/player/_/id/3202/kevin-durant

10.4 Rotation of Conics

In this section, we learned how to find the equation and sketch the graph of a conic section that is not perfectly vertical or horizontal, but on tilted axes. The following picture is of three equations that are necessary to start every one of these problems:

First, you must use the top equation to find the angle of rotation of the axes. You then use that angle to simplify the bottom two equations. The ultimate goal is to replace x and y in the original conic's equation with x' and y', or the new x and y that correspond with the new axes. The next picture shows the classifications of each type of conic section using its determinant. From the value of the determinant, we can tell which type of conic section an equation represents before completing the problem. This will be helpful to double-check the result.

10.3 Hyperbolas

Here are the equations for a hyperbola:

The first equation is used for horizontal hyperbolas, while the second is used for vertical hyperbolas.

Below is an example of a problem that asks you to sketch a hyperbola. Unless the coefficients of x^2 and y^2 are already one, the first step to these types of problems is completing the square. This is done on the second line of this problem. (Don't forget to add equal amounts to the other side of the equation!) to sketch the graph, mark the center (h,k), then mark points that are "a" distance away in the direction "a" corresponds to (in this case, x). Do the same for "b" (in this case, in the y direction) and make a box; the asymptotes, or lines the graph will never touch but approach infinitely, will run diagonally out of the corners of the box. Then draw the graph as shown, leaving from the vertex, or the "a" points you marked, and approaching the asymptotes.

10.2 Ellipses

This section focuses on graphing and finding equations for ellipses. The basic equations for ellipses are:

[(y-k)^2]/a^2 + [(x-h)^2]/b^2              AND          [(x-h)^2]/a^2 + [(y-k)^2]/b^2            (h,k) are the coordinates of the center. "a" and "b" are the distance between vertices and the center; a is the longer of the two. Although "c" is not in the equation, it is important to know that c is the distance between the center and each focus. c2 = a2 – b2 Some problems will give you the equation of an ellipse and you have to find out information from it (center, foci, vertices, and eccentricity) and sketch it using that information. Other problems will give you hints and ask you to find the equation. This problem (#33) gives you a sketch and the vertices, and asks for the equation:

Works Cited:
http://www.purplemath.com/modules/ellipse.htm

The REAL True Shooting Percentage

So there's this stat that ESPN NBA analysts use all the time called true shooting percentage. It is intended to measure a player's shooting ability more accurately, as it accounts for free throws and three-pointers. However, when I saw the equation used to calculate it, I didn't understand how it accomplished its goal at all. Here's the equation:

True Shooting Percentage = Total points / [(FGA + (0.44 x FTA)]

In my research, I learned that the equation was created to account for the number of possessions a shot consumes. This helps explain the (.44 x FTA) part of the equation, as usually, the other team gets the ball after a pair of free throws, and some free throws come after a made shot on a three-point play, so such a free throw is a chance for points that doesn't even take up a possession. 0.44 was reasonable because it was slightly less than one-half, or .5, and therefore accounted for the possessions a free throw consumes fairly well. 

However, this means that "True Shooting Percentage" is a misleading name for this statistic; it is more a measure of efficiency per possession than per shot. So, I made my own, more reasonable True Shooting Percentage equation:

Laski True Shooting Percentage = Total Points / [FTA + (2 x FGA) + 3PA]

Or, more simply: LTS% = Points Scored / Total Possible Points 

This equation encompasses all types of shots accurately, as each three-point shot is worth three points,each regular field goal is worth two, and each free throw is worth one. The resulting percentage represents the percentage of points a player scored out of the possible number of points they could have scored if they never missed a shot of any kind.

Stephen Curry's Laski True Shooting Percentage so far this season:

1459/[276 + (2 x 1108) + 492] = 0.4889 --> 48.9%

Works Cited:

http://theoldnorthking.blogspot.com/2013/01/true-shooting-percentage.html

http://espn.go.com/nba/statistics/player/_/stat/scoring

10.1 Parabolas

The basic equations for a parabola are (x-h)^2 = 4p(y-k) and (y-k)^2 = 4p(x-h). As you can see, these two equations are very similar, but the variables are just rearranged. The first equation is used to represent parabolas that go up-and-down, while the latter represents parabolas that go side-to-side. The p is equal to the distance between the vertex of the parabola and it's focus or it's directrix. (The focus is a point in the same direction of the parabola, and the directrix is a line that the parabola goes away from). Finally, (h,k) is the vertex pf the parabola.

Passer Rating: Russell Wilson

These are the steps to finding Passer Rating. You must add together the answers of four equations, then multiply by 100 and divide by 6.

Total 1:

1. Divide a quarterback's completed passes by pass attempts.
2. Subtract 0.3.
3. Divide by 0.2 and record the total. The sum cannot be greater than 2.375 or less than zero.

Total 2:

1. Divide passing yards by pass attempts.
2. Subtract 3.
3. Divide by 4 and record the total. The sum cannot be greater than 2.375 or less than zero.

Total 3:

1. Divide touchdown passes by pass attempts.
2. Divide by 0.05 and record the total. The sum cannot be greater than 2.375 or less than zero.

Total 4:

1. Divide interceptions by pass attempts.
2. Subtract that number from 0.095.
3. Divide that product by 0.04 and record the total. The sum cannot be greater than 2.375 or less than zero.

Final Steps:

1. Add the four totals you recorded.
2. Multiply that total by 100.
3. Divide by 6.
4. The final number is the passer rating.

I used this to calculate Russell Wilson's Quarterback Rating in the Super Bowl:

(2.1+1.31+1.6+2.375)X(100/6)=123.08

Works Cited:

http://football.about.com/od/frequentlyaskedquestions/ht/How-To-Calculate-A-Quarterback-Rating.htm

9.8 and 9.9 - Exploring Data

These sections expand on the main ideas of mean, median, and mode that we've known for a long time. But now, Variance and Standard Deviation are added. The equations for these are given in the chapter, but most of the problems involve too many numbers to do in a reasonable amount of time, so it's a good decision to just use standard deviation and variance calculators online. The following example finds the mean, median, and mode of the set 3, 7, 14, 15, 21, 21.

Mode: 21

Median: 14.5

Mean: 13.5

9.7 Probability

Probability problems can seem really complicated, but are often a lot simpler than they appear. At the base, every problem is asking you the ratio of favorable outcomes to total circumstances 

(# of successes/# of possibilities). For example, if you want to get a coin flip to be heads, there is one successful outcome, but two total outcomes (heads and tails), so the probability of getting heads is 1/2. There are twelve face cards in a deck of 52 cards, so the probability of drawing a face card is 12/52 (3/13). Let's try an example:

14.) In a 52-card deck, one card is drawn. What are the chances the number is six or lower?

• Ace, 2, 3, 4, 5, and 6 are six or lower, and there are four of each of those, so 6X4=24.
• There are 52 total cards, so the probability is 24/52 (Simplified: 6/13).