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Thread: Chris Carter on I.R./Marshall McFadden now on Active Roster/Jamie McCoy back on P.S.

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    Legend RuthlessBurgher's Avatar
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    Chris Carter on I.R./Marshall McFadden now on Active Roster/Jamie McCoy back on P.S.

    Steelers Make Transactions
    Posted 14 minutes ago

    The Steelers made a number of roster moves today that included promoting linebacker Marshall McFadden to the active roster from the teamís practice squad. McFadden will wear No. 40.


    McFadden spent the first 10 weeks of the 2012 season on the teamís practice squad. He originally was signed to the teamís offseason roster on January 18, 2012.


    To make room for McFadden on the active roster, the Steelers placed linebacker Chris Carter on the teamís Reserve/Injured List with an abdominal injury. Carter saw action in eight games this season, making his first three career starts.


    The Steelers also signed tight end Jamie McCoy to the teamís practice squad. McCoy will wear No. 80.
    http://www.steelers.com/news/article...d-e9b628d52575

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    It's looking more and more like OLB is becoming a high draft need next year as well.

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    Legend hawaiiansteel's Avatar
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    Quote Originally Posted by NW Steeler View Post
    It's looking more and more like OLB is becoming a high draft need next year as well.
    don't we have Jason Worilds waiting in the wings?

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    Quote Originally Posted by hawaiiansteel View Post
    don't we have Jason Worilds waiting in the wings?
    Maybe he shouldn't be in the wings. Doesn't he have more QB pressures and sacks than Harrison? Was Harrison even on the field against the Chiefs? Can't recall hearing his name called once during the game.
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    Can anyone tell me what the heck they see in McCoy that they keep him around? Really hard for me to believe he is the best talent on the street available with PS eligiability.
    Playing Fantasy Football does not qualify you to be the in the front office or on the coaching staff of the Pittsburgh Steelers. They are professionals and you are not!

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    Quote Originally Posted by Oviedo View Post
    Can anyone tell me what the heck they see in McCoy that they keep him around? Really hard for me to believe he is the best talent on the street available with PS eligiability.
    McCoy is the backup H-back/FB for Will Johnson along with being able to play TE. we all know how Tomlin loves position flexibility...

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    Quote Originally Posted by Oviedo View Post
    Maybe he shouldn't be in the wings. Doesn't he have more QB pressures and sacks than Harrison? Was Harrison even on the field against the Chiefs? Can't recall hearing his name called once during the game.
    well, then we have our future starting OLBs in place with Woodley and Worilds and Chris Carter as the main backup.

    in that case, why would we waste a high draft choice on an OLB when we have so many other pressing needs at ILB, S, WR etc. that need to also be filled?

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    Quote Originally Posted by hawaiiansteel View Post
    well, then we have our future starting OLBs in place with Woodley and Worilds and Chris Carter as the main backup.

    in that case, why would we waste a high draft choice on an OLB when we have so many other pressing needs at ILB, S, WR etc. that need to also be filled?
    You can't tell me you still think Adjepong and particularly Carter are our future. Carter was handed the starting OLB job, and flopped as badly as you can flop. I agree with a previous poster, who said Harrison has been largely invisible in his return. He has.

    But Carter was invisible before that.

    To think, we've basically played 10 man football this whole season... and Adjepong still can't get on the field, makes me have even less faith in him.

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    Quote Originally Posted by NW Steeler View Post
    It's looking more and more like OLB is becoming a high draft need next year as well.

    Coin flip between OLB, ILB and Safety. Can you have a three sided coin??????
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    Quote Originally Posted by Oviedo View Post
    Coin flip between OLB, ILB and Safety. Can you have a three sided coin??????
    http://www.statisticool.com/3sided.htm
    Constructing a fair 3 sided coin 3/21/05
    For a 2 sided coin, the kind we are all familiar with, being "fair" means that P(Heads) = P(Tails) = .5. What would a 3 sided coin look like? Can a 3 sided coin be made "fair" so P(Heads) = P(Tails) = P(Sides) = 1/3? And how do we know fairness can even be obtained? This informal paper explores these questions.


    A 3 sided "coin" can take several shapes. I put "coin" in quotes because it won't be a coin as we commonly use the term. Here are several possible ways a 3 sided coin look, with their outcomes, Heads, Tails, and Sides, labeled.


    The first 3 sided coin I thought of was a long triangular prism. In the first version, each long face is an outcome, and the triangular edges could be rounded to prevent the coin from landing on the ends. In the second version, each triangular end is an outcome, and the long faces collectively count as one outcome.




    Some real life examples of this are the delicious Toblerone chocolate bars, and a ruler (which is not delicious).



    One could also make a 3 sided dreidel (please forgive my Paint skills!).




    Another example is a die with opposite sides painted the same colors to count as the same outcome.


    As interesting as these shapes are, I was more interested in maintaining the coin shape as much as possible, so I chose to focus on a right circular cylinder.


    Here is a diagram of a right circular cylinder, our 3 sided coin




    David Boll made some postings on this topic on internet newsgroups, and I contacted him around 1998, saying that I'd be interested in flipping the coins he made.


    Here are the coins, or varying heights, that he made, with a quarter for reference




    The coins, going from left to right, not including the quarter: h = .5r, h = .75r, h = r, and h = 1.334r. David sent me the coins, and I ended up flipping three of these coins over 10,000 times total! From the data, it was determined that the coin where h = r was the most fair, with P(Sides) = 1240/3800 = .326. In David's own experiments, he obtained P(Sides) = 319/1000 = .319. Intuitively this makes sense, because having h = r makes the coin more die-like. However, the results could have been flukes.


    It is important to note that it is not too clear how to fairly flip such a coin because we now have more than one axis of rotation to think about. When I flipped them, I tossed them, one at a time, as haphazardly as I could into a wall, and then the coin would fall on to a short carpet, sort of like rolling dice in a game of craps.


    I set out to explore the mathematics behind these 3 sided coins. But first, what reason do we have in thinking fairness, that is, P(Heads) = P(Tails) = P(Sides) = 1/3, can be obtained?


    For a 2 sided coin, h is small (see the quarter above), and therefore P(Sides) is very small. Conversely, think of a Pringles can, where h is very large, thus making P(Sides) very large. Therefore, it stands to reason that there is high probability that some h exists between the extremes where P(Sides) is "just right".


    In Frederick Mosteller's Fifty Challenging Problems in Probability, Problem 38 reads (paraphrasing using my terminology of "height" for "thick" and "Sides" for "edge")


    What height should a coin be to have a 1/3 chance of landing on Sides?


    The story goes, the great mathematician John von Neumann solved this problem in his head and provided an answer to three decimal places, all in less than half a minute! Going by von Neumann's remarkable work, I tend to believe that story.


    The solution: put the coin in a sphere, with the center of the coin being the center of the sphere, then select a random point on the surface of the sphere. Draw a straight line from this point to the center, and if that line hits the side of the coin, the coin can be considered to have landed on Sides.


    Mosteller writes that a theorem from solid geometry simplifies this problem. Apparently when parallel planes cut a sphere, they produce an "orange-peel-like band" between the planes. This band is called a "zone". The surface area of a zone is proportional to the distance between the planes, and therefore our coin should have a height 1/3 of the sphere's height.




    Let R be the radius of the sphere and r the radius of the coin. The Pythagorean Theorem gives


    R2 = r2+(1/9)R2
    Solving for (2/3)R, which is h, gives


    (2/3)R = (sqrt(2)/2)r
    or h ~ .707r


    Therefore, to obtain P(Heads) = P(Tails) = P(Sides) = 1/3, construct the coin so h = .707r. This theoretical result is surprising because it does not coincide with the experimental result of the best performing coin being where h = r. In fact, when I originally flipped the coins I obtained P(Sides) = 138/3800 = .046 using a coin where h = .75r, and David obtained P(Sides) = 88/800 = .11.


    The discrepancies could be due to statistical flukes, the material of the coin, the material of where the coin was landing, the flipping method, the mathematical theory being too simple, or some combination of these.


    For a detailed discussion of 3-sided coins, I strongly recommend the article Teaching Bayesian Model Comparison With the Three-Sided Coin by Scott Kuindersma and Brian Blais in The American Statistician, August

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