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Posts posted by oldmanhawkins


10 minutes ago, RBM said:
Sounds like he isn't playing this week.
Didn't see that coming  got hurt in the first quarter but returned in the second and finished the game. Same ankle as preseason?

5 minutes ago, shadyridr said:
Maybe if he was healthy but what has he done this year to put him in that tier?
That's a fair question. After the guys I mentioned you could really make a case for a dozen other guys.
This is my personal check list for TEs
1. Talent. He's got tons
2. Targets. This is more important than snap %. If a Te isn't getting 100+/year they won't amount to much. Ebron is right at that threshold
3. Offense. Good QB and system. I think it's a yes to both. I was more confident before last week.

Wanted to add this link to the bad penalty on his TD catch.
To me, Ebron is in the convo right after healthy gronk, Jordan reed, and Olsen. Same tier as kelce, ertz, jimmy.

Any thoughts on e squared after 4 games? It's a boring year for TEs but it looks to me that Ebron could end the year top 6ish. He's getting about 7 targets per game and his yards per target is solid. He's lacking in the TD department but there is a good argument that he could see a healthy dose of red zone opportunities going forward with their current situation at RB and WR.

Could Rod Streater play the Jordan Matthews role of a size mismatch in the slot?
 1

Out of curiosity, how do you think his value would change if dwill is already rostered in a keeper? Half a round lower?

Traded a 4th for victor Cruz.

This write up: http://web.mit.edu/neboat/Public/6.042/randomwalks.pdf
does a pretty good job of deriving the answer but I couldn't follow a lot of it. If anyone knows where I can learn more about this I'm all ears.

I did what I could working out basic examples. I had some fun figuring out ways to sum a few of the series, but unfortunately had to start brute forcing the expectation at X=2 and Y= 3. Thankfully, by the time I did X=3, Y=3 I saw the answer:
it's just XY

looks like a good opportunity for me to learn a bit about random walks.

7 hours ago, ZWK said:
With a circle, there are infinitely many possible cuts and they all go through the same point at the center. Every diameter is a possible cut, and those are the only possible cuts.
I think that is also true with a pentagon, although I'm more confident in the "infinitely many" part than the "same point at the center" part. There should be infinitely many possible cuts, because if you pick any one point on the perimeter as your starting point there must be some point across the pentagon which you can cut to in order to make it exactly a 5050 split. I'm less sure about them going through the same point, but it does happen with circles and squares and my intuition is that it works with all regular polygons.
I fiddled around enough to convince myself that these cuts (which are indeed infinite in number) don't all go through the same point. But there is some star like shape that they all pass through. So if your equipment were sitting on this shape it would definitely get destroyed.
I will try to recreate the sketch on Desmos tonight so I can share.

3 hours ago, ZWK said:
Think in terms of object with some area, not dimensionless points. A given object takes up a smaller fraction of the area of its ring if it is more towards the outside, and there is the same amount of laser cuts in each ring, so the odds of getting cut must be dropping as you move outward.
For example, imagine that you divide the circle into "rings" which are each one foot wide. Say that a ring near the center is 50 feet around its circumference, and a ring near the outside is 5,000 feet around. Now, take an object that is one square foot in size. Whichever ring you put it in, it takes up the entire width and 1 foot of the circumference. If you put it in the ring near the center then there is a 1/50 chance that the laser will go through it. If you put it in the ring near the outside then there is a 1/5000 chance that the laser will go through it.
(In general, the chance that the laser goes through it is proportional to 1/r, where r is its distance from the center of the circle.)
Thanks, I understand what you are saying now! It helped to (correctly) imagine the equipment as having some fixed area.
Still, isn't there some shape in the center that every equal areas dividing line goes through? Some collection of points with the property that, if you left the equipment there, there would be 100% chance of it getting cut?
I tried to sketch it on geometer's sketch pad and came up with a tiny pentacle or 5 point star. The radius of the circumscribed circle about the pentacle was almost 10% of the radius of the circumscribed circle of the pentagon. I had to make a false assumption to make the sketch easier which might account for the difference.

On 7/8/2016 at 5:48 PM, ZWK said:
As a simpler version, we can imagine that the shape
of the building is a circle. Then the laser has to cut along a diameter to cut it in half (I assume "planar ray" means that it cuts in a straight line). We can assume that all diameters are equally likely. So the mostly likely spots to get hit are near the center of the building (with the point at the center guaranteed to get hit), and the least likely spots to get hit are near the edge.
For example, you can look at a ring of the circle near the edge, between 99% of the way from the center to the edge and 100% of the way to the edge. Compare that to a ring near the center of the circle, between 1% of the way from the center to the edge and 2% of the way to the edge. The outer ring is 66x as big in area as the inner ring, but each of them will have the same amount of laser cut within them. So a given object is 1/66 as likely to get hit if you put it near the outer edge than if you put it in that inner ring.
The answer should be pretty similar given that the building is pentagonshaped  you want to move the important objects close to the perimeter (and presumably close to the corners, which are farther from the center).
I'm having some trouble understanding this (the problem itself and your explanation). If you don't mind, could you explain how (with the exception of the center point) how any point on the circle was more likely to be hit than any other? I agree that the "rings" on the outside have larger area than those on the inside, but that doesn't mean they have more points. Or am I missing something?
I interpreted the problem to mean that given an arbitrary point on the edge of the pentagon, a straight line would pass through the shape in such a way that the shape was divided perfectly in half. For instance, if you start on a vertex, the line would obviously go through the midpoint of the opposite side. Of course, you need not start on the vertex. So, if you consider the set of all lines with this property, they will sweep out some shape. I think the shape is a 5 pointed star made by connecting the vertices of the pentagon. I'll play around a bit more tonight.

2 hours ago, maf005 said:
You can't possibly think you overpaid
It's only my second year playing dynasty  so I'm not really sure what is fair and what is an overpay.
It wouldn't surprise me if a lot of people value Benjamin and Evans similarly. I traded reed for hill a few months back but lost some interest in him when gio signed. So it felt like I was giving hill away too easily.

Jeremy Hill, kelvin Benjamin and Flacco
for
mike Evans and Carson Palmer
i may may have overpaid (sold low on hill), but thought my window to acquire Evans was closing

I did the same as ZWK but got
5 and 1/6

6 hours ago, ZWK said:
I think that the shape you get for any polygonshaped bread is made of
intersecting parabolas. One definition of a parabola is the set of points which are equally distant from a particular point and a particular line, and that is basically what we're doing here (with each edge playing the role of the line, and the center as the point).
I believe that you are right. My initial attempt was four intersecting circles. It made a pretty picture, but I didn't have a logical reason for choosing circles. It worked for the few points I tested so I went with it. When you switch to parabolas it changes the answer to:
(1/3)[4sqrt2  5] or about 21.9% of the original
Not sure I want to put the hours in needed to generalize it though...

On 6/18/2016 at 8:39 AM, bostonfred said:
1/4
It doesn't matter
Say you have a square with area = 1, and center O. If you surround it with another square with area = 4 and center O, the smaller square doesn't satisfy the criteria in the problem. To see this, consider any of the vertices of the smaller square. The distance to O, measured radially, is sqrt(2)/2. The distance to the edge of the larger square, measured perpendicularly, is 1/2. Thus, the vertices are outside the necessary cut.

I'm getting around 22% of the original square. Is there any easy way to post a picture?

Haven't proved it yet, but my initial guess for the square is:
pi/16
ETA: that's wrong.
I know the shape that is created though!

The difference between the puzzles seems like it might matter  each dragon not only sees 9 red eyed dragons, each also knows that every one of the others sees at least 8 red eyed dragons.
Meanwhile each mathematician knows that the other 9 have made errors but cannot know for certain if the others know at least 8 made errors.
Unless "informing others" is a rule they agreed to collectively follow, rather than individually follow.
Or not.

24 minutes ago, Ignoratio Elenchi said:
Yeah, this one seems pretty straightforward. Feel free to post your answer in spoiler tags and if you're interested there are links to all the previous puzzle discussions in the OP.
I agree that it sounds like a rewrite of the green eyed dragon problem. The thing I'm not completely clear on is whether or not the mathematicians know their group behavior when someone makes mistakes. I'm having trouble deciding if it even matters.

18 minutes ago, Ignoratio Elenchi said:
I think I agree with this. Here's the reasoning:
If it gets down to the final 2 pirates (pirates P9 and P10), P9 will propose giving himself all the gold. He only needs his own vote for this to pass, so that's sufficient.
If it gets down to the final 3 pirates (P8, P9 and P10), P10 will vote for any plan in which he gets more than 0 gold, since that's what he'll end up with if P8's proposal doesn't pass. So P8 proposes [9 0 1] as the distribution.
If it gets down to the final 4 pirates, P9 will vote for any plan in which he gets any gold at all, since he'll get none if it proceeds to P8. So P7 proposes [9 0 1 0]
It carries on this this down the line. The proposals would be:
P6: [8 0 1 0 1]
P5: [8 0 1 0 1 0]
P4: [7 0 1 0 1 0 1]
P3: [7 0 1 0 1 0 1 0]
P2 [6 0 1 0 1 0 1 0 1]
So finally, P1 proposes P1 [6 0 1 0 1 0 1 0 1 0], and all the pirates with nonzero gold amounts vote for it to pass.
This also illustrates the answer to the generalized problem with P pirates and G gold. The plan that will pass is as follows: The first pirate gets [G  floor ((P1)/2)] pieces of gold and the rest get 0 and 1 pieces in alternating fashion.
unless (p1)/2 exceeds g

The last two weeks were a little beyond me with the math involved, but I think this week is easy:
6 0 1 0 1 0 1 0 1 0
Jordy Nelson
in The Shark Pool (NFL Talk)
Posted
Traded Crowell and Shephard two weeks ago for Jordy. Thought I got a steal...now not so sure.