What is the best shape for a soccer goal post?

[kshitij]

Yes I get OB=$S\sqrt2\cos(\frac{\pi}{4}-\theta)$, next step?

Are we just considering the cases where $d \gt S\sqrt2\cos(\frac{\pi}{4}-\theta)$ but this will not give the cases when the ball goes in after deflection from the post, this was not what I was asking

 

[kshitij]

Are we just considering the cases where $d \gt S\sqrt2\cos(\frac{\pi}{4}-\theta)$ but this will not give the cases when the ball goes in after deflection from the post, this was not what I was asking

I thought you calculated the probability of a shot going in after hitting the post, this was what I was originally asking, "of all the shots that hit the post which of the two posts (square or round) has a greater probability for those shots going in"

 

[haruspex]

this will not give the cases when the ball goes in after deflection from the post

It will. Draw the diagram for $\theta>\pi/2$ and d just sufficient to pass to the right of A.

 

[kshitij]

It will. Draw the diagram for $\theta>\pi/2$ and d just sufficient to pass to the right of A.

The blue circle is of radius $D$ and if the path of the ball is tangent to that circle (because $d$ is the perpendicular distance of the path from origin and this distance is equal to $D$ in the limiting case) it will not hit the post

 

[kshitij]

View attachment 287709
The blue circle is of radius $D$ and if the path of the ball is tangent to that circle (because $d$ is the perpendicular distance of the path from origin and this distance is equal to $D$ in the limiting case) it will not hit the post

I realized you said that the limiting case is $d=D\cos(\frac{\pi}{4}-\theta)$ not $d=D$ but now I cannot understand why?
What I do understand is that $\left|\frac{\pi}{4}-\theta\right|$ is the angle between the vectors $\vec {OA}$ and $\vec {OB}$ where $\vec {OB}$ is the perpendicular distance from origin $O$ to the path of the ball and $D\cos(\frac{\pi}{4}-\theta)$ is the projection of $\vec {OA}$ on $\vec {OB}$

 

[kshitij]

I think now I somewhat understand what you are trying to do,
Here in this link the blue line gives all possible paths of the ball (move point $B$ to get different cases) and the red line is the limiting case for a given blue path.

As long as we stay to the left of the red line, we can score a goal thus these are our favourable cases.

 

[kshitij]

A goal is scored if and only if the ball passes to the right of (−D,D). That is, if d>S2cos⁡(π4−θ).

Now finally I understand what you said here!

But now how did you get this?

The probability of a goal is therefore $\frac 1{2D\pi}\int_0^\pi D-S\sqrt 2\cos(\frac{\pi}4-\theta)d\theta$

And also, please assume $D=S\sqrt 2=R\sqrt 2$ in further calculations because the diameter of the round post is equal to side length of square post according to the laws of the game

 

[kshitij]

According to what I understood reading post #91, I similarly created a graphic for the circular post as well!

 

[haruspex]

But now how did you get this?

And also, please assume $D=S\sqrt 2=R\sqrt 2$ in further calculations because the diameter of the round post is equal to side length of square post according to the laws of the game

There's no need to plug in S=R at the start. I prefer to keep things general initially and apply specifics at the end. There are advantages.
The value of D does not matter as long as it encompasses both posts.

We have established a goal is scored if $d>S\sqrt 2\cos(\frac{\pi}4-\theta)$.
d is uniformly distributed over the range (-D,D), a line of length 2D. Of that line, length $D-S\sqrt 2\cos(\frac{\pi}4-\theta)$ satisfies that condition. So the probability of a goal for a given angle is $\frac 1{2D}(D-S\sqrt 2\cos(\frac{\pi}4-\theta))$.
To get the overall probability we need to integrate over the range of the angle and divide by that range, pi.
The probability of a goal is therefore $\frac 1{2D\pi}\int_0^\pi D-S\sqrt 2\cos(\frac{\pi}4-\theta)d\theta$.

(Love the animations!)

 

[kshitij]

There's no need to plug in S=R at the start. I prefer to keep things general initially and apply specifics at the end. There are advantages.
The value of D does not matter as long as it encompasses both posts.

We have established a goal is scored if $d>S\sqrt 2\cos(\frac{\pi}4-\theta)$.
d is uniformly distributed over the range (-D,D), a line of length 2D. Of that line, length $D-S\sqrt 2\cos(\frac{\pi}4-\theta)$ satisfies that condition. So the probability of a goal for a given angle is $\frac 1{2D}(D-S\sqrt 2\cos(\frac{\pi}4-\theta))$.
To get the overall probability we need to integrate over the range of the angle and divide by that range, pi.
The probability of a goal is therefore $\frac 1{2D\pi}\int_0^\pi D-S\sqrt 2\cos(\frac{\pi}4-\theta)d\theta$.

(Love the animations!)

That is so clever!

First we calculated favourable cases for line whose slope is fixed and then we fixed the distance of that line from origin then calculated favourable cases for all possible values of slope!

But if we consider the maximum value of $D$ to be $\infty$ (i.e. the incident path can be anywhere in the 2D plane) then logically we should get the probability in case of square post as 33.33% as all three faces have equal chance of striking and striking only one of these three face leads to goal, but using your formula I get 15.91%, why?

 

[kshitij]

According to what I understood reading post #91, I similarly created a graphic for the circular post as well!

I updated the graphic for the circular post case, now you don't have to move two points separately to get all possible paths you can get it by moving only point B in the link below,
https://www.geogebra.org/calculator/khj7d6hk

 

[kshitij]

I evaluated your integral for $\theta \in (0,\pi)$ and got

$P(square)=\dfrac{D-2S}{2D\pi}$

$P(round)=\dfrac{D-2R}{2D\pi}$

Now for $D \to \infty$, we get $P(square)=P(round)=\dfrac{1}{2\pi}=0.1591%$ regardless of what the dimensions of the square and round posts are as long as they are finite, i.e., we can even have $R=1000S$ and still get the same probability for both posts as 15.91% which seems totally illogical to me, am I missing something?

 

[kshitij]

I evaluated your integral for $\theta \in (0,\pi)$ and got

$P(square)=\dfrac{D-2S}{2D\pi}$

$P(round)=\dfrac{D-2R}{2D\pi}$

Now for $D \to \infty$, we get $P(square)=P(round)=\dfrac{1}{2\pi}=15.91%$ regardless of what the dimensions of the square and round posts are as long as they are finite, i.e., we can even have $R=1000S$ and still get the same probability for both posts as 15.91% which seems totally illogical to me, am I missing something?

And if we take $D=7.32m$ which is equal to the distance between the two posts and $R=S=6cm$ according to the laws of the game, we get,

$P(square)=P(round)=0.1516%$

Which still doesn't make any sense to me, I mean Probability for the square ones should be close to 33% and for the round one should be atleast 25%!

 

[kshitij]

I evaluated your integral for $\theta \in (0,\pi)$ and got

$P(square)=\dfrac{D-2S}{2D\pi}$

$P(round)=\dfrac{D-2R}{2D\pi}$

Now for $D \to \infty$, we get $P(square)=P(round)=\dfrac{1}{2\pi}=0.1591%$ regardless of what the dimensions of the square and round posts are as long as they are finite, i.e., we can even have $R=1000S$ and still get the same probability for both posts as 15.91% which seems totally illogical to me, am I missing something?

And if we take $D=7.32m$ which is equal to the distance between the two posts and $R=S=6cm$ according to the laws of the game, we get,

$P(square)=P(round)=0.1516%$

Which still doesn't make any sense to me, I mean Probability for the square ones should be close to 33% and for the round one should be atleast 25%!

I realized that 33% and 25% where the probabilities of a shot going in after hitting the post whereas in your method the ball doesn't need to hit the post, but now this confuses me even more! that means 15% is the probability for any shot (no matter where it is hit in the 2D plane) to go in!

So now suddenly 15% changes from being a little less than expected to a lot more than expected

 

[kshitij]

Well I'm too confused so let me state what I think I understood then you correct me where I'm wrong.

• $d$ is the distance of a given path of the ball from origin and origin is the center of one of the posts (right post from our point of view and left from the ball's) and the maximum value of $d$ is equal to the maximum value of $D$
• $D$ is all possible values of the distance of the path of the ball from origin starting from the edge corner of one post to the edge corner of other (the other post was at $(-\infty,0)$)
• $\theta$ is the inclination of the path with x-axis or the goal line which ranges from $(0,\pi)$
• A goal is scored if the ball passes to the left (looking at the ball from our point of view its left otherwise w.r.t ball it should be right) of the minimum value of $D$
• for scoring a goal we have, $d>S\sqrt 2\cos(\frac{\pi}4-\theta)$ in case of square post and in case of round one we get $d>R\cos(\frac \theta 2)$ ($R$ is the radius of round post and $2S$ is the side length of square one)

And the points below are the ones that I concluded from the points above,

• the value (here I am only talking about the magnitude) of $D$ can range from $(R,\infty)$ or $(S\sqrt{2},\infty)$ if we consider the other post to be at $\infty$, and if we considered a more realistic system then value of then maximum value of $D$ must be equal to the distance of the left (again its left w.r.t us and right w.r.t ball) touchline from the edge of the right post
• Since a line is bidirectional and not unidirectional like the path of the ball, we can consider goal posts at both ends of the pitch so along path of the ball the ball can travel both ways backwards or forwards and scoring a goal at both ends of the pitch is equivalent for us

 

[haruspex]

D is all possible values of the distance of the path of the ball from origin starting from the edge corner of one post to the edge corner of other (the other post was

No, D is a constant such that if d>D then it is sure to be a goal without touching the post. Likewise, if d<-D it surely will not be a goal. D's value won't matter as long as it is large enough.
We only need to compare the posts for what happens when -D<d<D.

 

[kshitij]

No, D is a constant such that if d>D then it is sure to be a goal without touching the post. Likewise, if d<-D it surely will not be a goal. D's value won't matter as long as it is large enough.
We only need to compare the posts for what happens when -D<d<D.

So $D$ is a large constant and $d$ belongs to $(S\sqrt 2\cos(\frac{\pi}4-\theta),D)$ for square post and for round post it lies between $(R\cos(\frac \theta 2),D)$

So still, when $D \to \infty$ it should surely give us the probability when the path of the ball can lie anywhere in the 2D plane because now $d$ belongs to $(S\sqrt 2\cos(\frac{\pi}4-\theta),\infty)$, so still, 15% is the probability of any shot in the 2D plane going in, right?

 

[haruspex]

So $D$ is a large constant and $d$ belongs to $(S\sqrt 2\cos(\frac{\pi}4-\theta),D)$ for square post and for round post it lies between $(R\cos(\frac \theta 2),D)$

So still, when $D \to \infty$ it should surely give us the probability when the path of the ball can lie anywhere in the 2D plane because now $d$ belongs to $(S\sqrt 2\cos(\frac{\pi}4-\theta),\infty)$, so still, 15% is the probability of any shot in the 2D plane going in, right?

I don't know how you get 15%.
If D goes to infinity and d is still uniformly distributed from somewhere up to that then the probability of going in tends to 1.

 

[kshitij]

I don't know how you get 15%.

I evaluated your integral for $\theta \in (0,\pi)$ and got

$P(square)=\dfrac{D-2S}{2D\pi}$

$P(round)=\dfrac{D-2R}{2D\pi}$

Now for $D \to \infty$, we get $P(square)=P(round)=\dfrac{1}{2\pi}=0.1591%$

 

[haruspex]

$\frac 1{2D\pi}\int_0^\pi D-S\sqrt 2\cos(\frac{\pi}4-\theta)d\theta$
$=\frac 1{2D\pi} [D\theta+S\sqrt 2\sin(\frac{\pi}4-\theta)]_0^\pi$
$=\frac 1{2D\pi} (D\pi+S\sqrt 2\sin(\frac{\pi}4-\pi)-S\sqrt 2\sin(\frac{\pi}4))$

 

[kshitij]

$\frac 1{2D\pi}\int_0^\pi D-S\sqrt 2\cos(\frac{\pi}4-\theta)d\theta$
$=\frac 1{2D\pi} [D\theta+S\sqrt 2\sin(\frac{\pi}4-\theta)]_0^\pi$
$=\frac 1{2D\pi} (D\pi+S\sqrt 2\sin(\frac{\pi}4-\pi)-S\sqrt 2\sin(\frac{\pi}4))$

Oh yes I forgot the $\pi$ in the numerator but now the probability is 50% right?

 

[kshitij]

I evaluated your integral for $\theta \in (0,\pi)$ and got

$P(square)=\dfrac{D-2S}{2D\pi}$

$P(round)=\dfrac{D-2R}{2D\pi}$

Now for $D \to \infty$, we get $P(square)=P(round)=\dfrac{1}{2\pi}=0.1591%$ regardless of what the dimensions of the square and round posts are as long as they are finite, i.e., we can even have $R=1000S$ and still get the same probability for both posts as 15.91% which seems totally illogical to me, am I missing something?

*correction,

$P(square)=\dfrac{D\pi-2S}{2D\pi}$

$P(round)=\dfrac{D\pi-2R}{2D\pi}$

Now for $D \to \infty$, we get $P(square)=P(round)=\dfrac{1}{2}=0.5$

 

[haruspex]

*correction,

$P(square)=\dfrac{D\pi-2S}{2D\pi}$

$P(round)=\dfrac{D\pi-2R}{2D\pi}$

Now for $D \to \infty$, we get $P(square)=P(round)=\dfrac{1}{2}=0.5$

Yes, in post #118 I meant d being from 0 to infinity. That would give 1.

It depends what you are trying to do here. Initially you were asking what the probability of a goal is given that the ball hits the post. You can answer that by making D a variable depending on the angle. But that means a) you are changing the distribution of angles, making shots at 45 degrees relatively more common than ones full on, and b) you cannot use it to compare the two post shapes.
To compare the posts you must set D to a constant finite value large enough to encompass both posts from all angles. This will not give an actual probability of a goal in either case, but it does tell you what the difference between them is.

 

[kshitij]

Yes, in post #118 I meant d being from 0 to infinity. That would give 1.

It depends what you are trying to do here. Initially you were asking what the probability of a goal is given that the ball hits the post. You can answer that by making D a variable depending on the angle. But that means a) you are changing the distribution of angles, making shots at 45 degrees relatively more common than ones full on, and b) you cannot use it to compare the two post shapes.
To compare the posts you must set D to a constant finite value large enough to encompass both posts from all angles. This will not give an actual probability of a goal in either case, but it does tell you what the difference between them is.

Again I can't understand what you mean here, how am I changing the distribution of angles? And why can't this be used to compare the two post shapes?

Let me state the whole method as I understand it.

Now, I need diagrams and figures to understand pretty much everything so I will now talk w.r.t to the diagrams in the links below.

Square post Circular post

We assumed a path of shot whose angle with the goal line is $\theta$ and distance from center of one of the posts is $d$

As we can see that moving point $B$ changes both the distance of path of the ball from the center of the post and its angle with the goal line.

(Note that distance to the left side of the post in the figures are taken -ve)

And we assumed the other post to be at distance $(-\infty)$ from our post in the figure, so our total values of $d$ ranges from $(-D,D)$ where $D$ is a large enough constant and the favourable values of $d$ ranges from $(-D,-S\sqrt 2\cos(\frac{\pi}4-\theta))$ in square post and $(-D,-R\cos(\frac \theta 2))$ in round post.

So now for a given value of $\theta$ our required probabilities are,

$P(square)=\dfrac{d \space \text{(favourable)}}{d \space \text {(total)}}=\dfrac{D-S\sqrt 2\cos(\frac{\pi}4-\theta)}{2D}$

$P(round)=\dfrac{d \space \text{(favourable)}}{d \space \text {(total)}}=\dfrac{D-R\cos(\frac \theta 2)}{2D}$

Now this was for a given $\theta$ but we also have $\theta \in (0,\pi)$ and thus we got,

$P(square)=\frac 1{2D\pi}\int_0^\pi D-S\sqrt 2\cos(\frac{\pi}4-\theta)d\theta$

$P(round)=\frac 1{2D\pi}\int_0^\pi(D-R\cos(\frac\theta 2))d\theta$

Now in this method we calculated the probability assuming uniform distribution of $d$ from $-D$ to $D$ and uniform distribution of $\theta \in (0,\pi)$

Now, if we want to compare the posts we can take the value of $D=S\sqrt2=R$ (as it doesn't matter what value of $D$ we take as long as its larger than or equal to $S\sqrt2=R$)

So finally we get that if $\theta \in (0,\pi)$ then, the probabilities for a shot (whose distance from the center of the post is equal to or less than $|D|$) going in for both posts are equal to $\dfrac{\pi-2}{2\pi}=0.1816$

That is, both posts are equivalent as long as a shot's angle with the goal line is between $(0,\pi)$ and distance from the center of the post is less than or equal to $|D|$

 

[jbriggs444]

Again I can't understand what you mean here, how am I changing the distribution of angles? And why can't this be used to compare the two post shapes?

You are selecting from distributions over sets that include only shots that hit the posts. If the two posts have different shapes, those will be distributions over different sets of shots. So the distributions will certainly be different.

If you select a more inclusive outline that encompasses the entirety of both post outlines then you can find a distribution over a set of shots that intersect with that outline. That distribution can be identical for both posts. Then you can do an apples to apples comparison of which of those shots go in and which do not.

 

[kshitij]

You are selecting from distributions over sets that include only shots that hit the posts.

But our selection depends on what value we choose for $D$ (see post #124) and if we chose $D$ to be large enough then it would also include the cases where the ball didn't hit the post.

If you select a more inclusive outline that encompasses the entirety of both post outlines then you can find a distribution over a set of shots that intersect with that outline. That distribution can be identical for both posts. Then you can do an apples to apples comparison of which of those shots go in and which do not.

Isn't setting the same constant value of $D$ for the probability calculation of both posts a like to like comparison?

 

[kshitij]

Just tell me whether my this conclusion is correct or not?

both posts are equivalent as long as a shot's angle with the goal line is between (0,π) and distance from the center of the post is less than or equal to |D|

 

[haruspex]

Now, if we want to compare the posts we can take the value of D=S$\sqrt 2$=R (as it doesn't matter what value of D we take as long as its larger than or equal to S$\sqrt 2$=R)

No. It doesn’t matter what D is as long as it exceeds both S$\sqrt 2$ and R. This is independent of the relationship between S and R.
You are specifically interested in the case S=R (and it turns out that in that case the probabilities are the same). So choose D$\geq S\sqrt 2=R\sqrt 2$.

But our selection depends on what value we choose for D (see post #124) and if we chose D to be large enough then it would also include the cases where the ball didn't hit the post.

Yes. It is not possible to compare the posts correctly without allowing some cases where one of the posts would be missed. This is because no matter what the relative values of S and R are there will be trajectories that would miss one post and hit the other.

Isn't setting the same constant value of D for the probability calculation of both posts a like to like comparison?

Yes, but if you insist on only considering trajectories that hit a post then you will not be using constant D.
Even just looking at the square post, if you restrict attention to shots that hit the post then you will be considering more trajectories at 45 degrees than at any other angle. That distorts the distribution, making 45 degree angles relatively more likely.

 

[kshitij]

No. It doesn’t matter what D is as long as it exceeds both S2 and R. This is independent of the relationship between S and R.
You are specifically interested in the case S=R (and it turns out that in that case the probabilities are the same). So choose D≥S2=R2.

I need a numerical value of $D$ because that will give us a number, i.e., if $D=R=6cm$ then we get 18% chance as calculated in post #124 instead of getting just that the probability in square and round post are equal, we get that probability in square and round post are equal which is equal to 0.1816 which is better

Yes, but if you insist on only considering trajectories that hit a post then you will not be using constant D.
Even just looking at the square post, if you restrict attention to shots that hit the post then you will be considering more trajectories at 45 degrees than at any other angle. That distorts the distribution, making 45 degree angles relatively more likely.

I now understand that if I restrict the shots to hit the post then that will be an unfair comparison of the two posts, but as I said earlier that if I just fix the value of $D$ then wouldn't that be a fair comparison?

e.g., say that I fix the value of $d$ to range from $-3.66m$ (which is the distance from one post to the midpoint of the goal) to $+41.34m$ (which is the distance from one post to the corner flag) then I get (using $R=S=0.06m$)

$P(square)=\frac 1{45\pi}\int_0^\pi 3.66-0.06\sqrt 2\cos(\frac{\pi}4-\theta)d\theta=0.08048$

$P(round)=\frac 1{45\pi}\int_0^\pi(3.66-0.06\cos(\frac\theta 2))d\theta=0.08048$

Then can I say that the chance of any shot hit within the legal area of play has an 8% chance of going in for both the posts?

 

[kshitij]

Also, I have another doubt,

Say if I set the value of $D=R$ in the round post, then it should mean that now I'm considering only the shots that hit the round post, so now I get the probability of a shot hitting the round post going in as

$P(round)=\dfrac{D\pi-2R}{2D\pi}=0.18$

But that seems logically incorrect as this probability must be atleast 0.25 (as any shot hitting sector AB {see the second diagram in post #1} must go in). What is wrong here?

 

[haruspex]

this probability must be atleast 0.25 (as any shot hitting sector AB {see the second diagram in post #1} must go in).

No, that was incorrect. With D=R, the probability of going in is $1-\cos(\theta/2)$. Sketch 1-cos. It rises quadratically at first.
As regards hitting sector AB, that won't even start to happen until $\theta=\pi/2$, and again it is quite unlikely to begin with. The overall probability of hitting that sector is $\frac 1\pi\int_0^{\pi/2}(1-\cos(\theta)).d\theta=\frac 14-\frac 1{2\pi}$. (Exactly half the total probability of going in.)

 

[kshitij]

As regards hitting sector AB, that won't even start to happen until $\theta=\pi/2$, and again it is quite unlikely to begin with. The overall probability of hitting that sector is $\frac 1\pi\int_0^{\pi/2}(1-\cos(\theta)).d\theta=\frac 14-\frac 1{2\pi}$. (Exactly half the total probability of going in.)

I am thinking with a different model, consider all possible shots that hit the post, now the probability that a shot hits the sector AB is $\frac14$, right? the whole post is divided into 4 quarters and hitting anyone of these quarter is equally likely? correct?

Now back to your model, we again considered all possible paths that hit the post (by stetting the ranges of $d$ from $-R$ to $R$), then still of all these paths the chance that one of them ends up at sector AB has to $\frac14$ right? again the post can be divided into 4 quarters and each path should be equally likely to end up at anyone quarter I cannot understand why won't it?

 

[kshitij]

I am thinking with a different model, consider all possible shots that hit the post, now the probability that a shot hits the sector AB is $\frac14$, right? the whole post is divided into 4 quarters and hitting anyone of these quarter is equally likely? correct?

Now back to your model, we again considered all possible paths that hit the post (by stetting the ranges of $d$ from $-R$ to $R$), then still of all these paths the chance that one of them ends up at sector AB has to $\frac14$ right? again the post can be divided into 4 quarters and each path should be equally likely to end up at anyone quarter I cannot understand why won't it?

Why is the model I'm talking about a subset of your model? for $R \gt d \gt -R$ aren't both the models equivalent?

 

[haruspex]

the post, now the probability that a shot hits the sector AB is 1/4, right?

Wrong.
If the shots could come equally likely around the whole 360 degrees you would be right, but for the 180 degrees allowed the sector AB is more to the rear than to the front, so the probability must be less than 1/4.

 

[kshitij]

Wrong.
If the shots could come equally likely around the whole 360 degrees you would be right, but for the 180 degrees allowed the sector AB is more to the rear than to the front, so the probability must be less than 1/4.

Okay, I think now I understand (but still it doesn't feel right)

Anyway, let me ask one final thing, for a given post round or square, $\dfrac{D\pi-2S}{2D\pi}=\dfrac{D\pi-2R}{2D\pi}$ is the probability of a shot that is hit within the semicircular area of radius $D$ going in goal assuming the other post to be at $-\infty$?

 

[haruspex]

Okay, I think now I understand (but still it doesn't feel right)

Anyway, let me ask one final thing, for a given post round or square, $\dfrac{D\pi-2S}{2D\pi}=\dfrac{D\pi-2R}{2D\pi}$ is the probability of a shot that is hit within the semicircular area of radius $D$ going in goal assuming the other post to be at $-\infty$?
View attachment 287821

No, that won't give a uniform distribution for d, so it will be a different result.
E.g. treating the post as a point, if the ball comes from a point at angle theta around the arc it has probability $\frac \theta{2\pi}$ of scoring, giving a total of 1/4. For the distribution we used before, setting the post to a point makes the probability 1/2.

 

[kshitij]

treating the post as a point, if the ball comes from a point at angle theta around the arc it has probability θ/2π of scoring, giving a total of 1/4.

I didn't get how you did this?

Also if that is not how we can think of this probability then what is the physical meaning of the probability that we calculated?

 

[haruspex]

I didn't get how you did this?

Also if that is not how we can think of this probability then what is the physical meaning of the probability that we calculated?

If the ball comes from $(D,\theta)$ in polar then, of the whole $2\pi$ range of angles it can go in, only a sector of angle width $\theta$ is headed for the goal. The probability of a goal is therefore $\frac \theta{2\pi}$. Integrating, the overall probability is $\frac 1\pi\int_0^\pi\frac \theta{2\pi}.d\theta=\frac 14$.

The probability we calculated before is what I defined it to be: the probability of a goal if the angle of the trajectory is uniformly distributed over its legal range $(\pi)$ and its displacement from the centre of the post is uniformly distributed over (-D,D).
Changing it to the shot being made from an arc of radius D, uniformly distributed along the arc, leads to a different distribution of trajectories and a different probability.

I note that throughout this thread you have not really understood the importance of clearly defining the assumed distribution. This is fundamental to dealing with probabilities.

 

[kshitij]

If the ball comes from $(D,\theta)$ in polar then, of the whole $2\pi$ range of angles it can go in, only a sector of angle width $\theta$ is headed for the goal. The probability of a goal is therefore $\frac \theta{2\pi}$. Integrating, the overall probability is $\frac 1\pi\int_0^\pi\frac \theta{2\pi}.d\theta=\frac 14$.

The probability we calculated before is what I defined it to be: the probability of a goal if the angle of the trajectory is uniformly distributed over its legal range $(\pi)$ and its displacement from the centre of the post is uniformly distributed over (-D,D).
Changing it to the shot being made from an arc of radius D, uniformly distributed along the arc, leads to a different distribution of trajectories and a different probability.

I note that throughout this thread you have not really understood the importance of clearly defining the assumed distribution. This is fundamental to dealing with probabilities.

I think that you assumed that a shot is hit from a point on the arc of a semicircle of radius $D$ but I meant that a shot hit from any point in the area of the semicircle of radius $D$, because now we see that $\theta$ is uniformly distributed in $(0,\pi)$ and $d$ (displacement form the center of the post) is uniformly distributed in $(-D,D)$.

 

[haruspex]

I meant that a shot hit from any point in the area of the semicircle of radius D,

That's different again. And it's messier because now there are three degrees of freedom to integrate over instead of two.

 

[kshitij]

That's different again. And it's messier because now there are three degrees of freedom to integrate over instead of two.

But only $d$ and $\theta$ are variable, what do you mean by 3 degrees of freedom?

 

[kshitij]

I just want to visualise the probability, saying that "this is the probability when the path of the ball is such that $d$ and $\theta$ are uniformly distributed over $(-D,D)$ and $(0,\pi)$ respectively" doesn't give me any hint about what is happening physically. Saying that "this is the probability when the path of the ball is always intersected in a semicircular area of radius $D$" gives a much better understanding of how the ball is hit.

 

[haruspex]

But only $d$ and $\theta$ are variable, what do you mean by 3 degrees of freedom?

If the ball is coming at angle theta from a uniform distribution of points in the semicircle then there are two degrees of freedom just to specify the point.

 

[kshitij]

If the ball is coming at angle theta from a uniform distribution of points in the semicircle then there are two degrees of freedom just to specify the point.

Yes, but isn't it correct that for a given point in that semicircular area, there is only one path possible that passes through that point and is perpendicular to the origin

 

[haruspex]

Yes, but isn't it correct that for a given point in that semicircular area, there is only one path possible that passes through that point and is perpendicular to the origin

Perpendicular to a point? What does that mean?
Why can't I pick any point in the semicircle and any angle from that point that's in the 180 degree range?

 

[kshitij]

Perpendicular to a point? What does that mean?

Sorry I meant, perpendicular to the line joining that point and origin

 

[kshitij]

Why can't I pick any point in the semicircle and any angle from that point that's in the 180 degree range?

For every point you pick we can define a unique line which is perpendicular to the line joining that point and origin

 

[kshitij]

I was trying to define the equation of the line in parametric form with a given slope and given distance (perpendicular distance) from origin

 

[haruspex]

I was trying to define the equation of the line in parametric form with a given slope and given distance (perpendicular distance) from origin

If you pick a point from a uniform distribution over the semicircle and an angle uniformly distributed $(0,\pi)$ then look at the distance d that path is from the origin you will not get a uniform distribution for d in (-D,D), so the probability of a goal will be different from what we previously calculated.

 

[kshitij]

If you pick a point from a uniform distribution over the semicircle and an angle uniformly distributed $(0,\pi)$ then look at the distance d that path is from the origin you will not get a uniform distribution for d in (-D,D), so the probability of a goal will be different from what we previously calculated.

the distance of the point from origin will also be the distance of the line from origin as the line is perpendicular to the line joining origin and that point so you don't need to look at $d$ again since you already fixed that while picking the point