HSC 2015 MX2 Permutations & Combinations Marathon (archive) (4 Viewers)

braintic · Sep 30, 2015

Re: 2015 permutation X2 marathon

Drsoccerball said:
I dont see the last case

What is his chance if he fires at nothing - a deliberate miss?

InteGrand · Sep 30, 2015

Re: 2015 permutation X2 marathon

Drsoccerball said:
I dont see the last case

A could shoot C and hit – and then he's dead.

Edit: wait never mind, I think I misunderstood your solution before.

Drsoccerball · Sep 30, 2015

Re: 2015 permutation X2 marathon

InteGrand said:
A could shoot C and hit – and then he's dead.

But then B kills A ?

Drsoccerball · Sep 30, 2015

Re: 2015 permutation X2 marathon

braintic said:
What is his chance if he fires at nothing - a deliberate miss?

You cant add extra rules

InteGrand · Sep 30, 2015

Re: 2015 permutation X2 marathon

braintic said:
What is his chance if he fires at nothing - a deliberate miss?

Lol is that allowed? They could all just keep doing deliberate misses and then everyone survives.

InteGrand · Sep 30, 2015

Re: 2015 permutation X2 marathon

Drsoccerball said:
But then B kills A ?

Yeah I meant 'he' as in person A. But I think I might have misunderstood your solution before.

turnerloos · Sep 30, 2015

Re: 2015 permutation X2 marathon

Meanwhile me... "An unbiased coin is tossed. What is probability of landing a head?"

Drsoccerball · Sep 30, 2015

Re: 2015 permutation X2 marathon

InteGrand said:
Yeah I meant 'he' as in person A. But I think I might have misunderstood your solution before.

Im considering all cases in which A survives not where he dies

braintic · Sep 30, 2015

Re: 2015 permutation X2 marathon

InteGrand said:
Lol is that allowed? They could all just keep doing deliberate misses and then everyone survives.

We are assuming they don't conspire just to survive.
He is allowed to shoot wherever he wants - he can fire it into his mother-in-law as a bonus if he wants.
But this gives him the greatest chance of surviving.

Drsoccerball · Sep 30, 2015

Re: 2015 permutation X2 marathon

braintic said:
We are assuming they don't conspire just to survive.
He is allowed to shoot wherever he wants - he can fire it into his mother-in-law as a bonus if he wants.
But this gives him the greatest chance of surviving.

You never said he could shoot blanks... hmmm well i guess if he purposefully misses a shot he has a 3/10 chance of winning which is higher than the others...

braintic · Sep 30, 2015

Re: 2015 permutation X2 marathon

Drsoccerball said:
You never said he could shoot blanks... hmmm well i guess if he purposefully misses a shot he has a 3/10 chance of winning which is higher than the others...

I said "... shooting once at the shooter's choice of target." There was nothing to say the target had to be one of the other two men. It could be a bird in the tree, a particular blade of grass, ....... or the person who last challenged him on his wording in a maths problem.

Paradoxica · Sep 30, 2015

Re: 2015 permutation X2 marathon

braintic said:
I said "... shooting once at the shooter's choice of target." There was nothing to say the target had to be one of the other two men. It could be a bird in the tree, a particular blade of grass, ....... or the person who last challenged him on his wording in a maths problem.

lol. that would make a great sig

braintic · Sep 30, 2015

Re: 2015 permutation X2 marathon

"On a given circle, a chord is selected at random. What is the probability that the chord is longer than the radius?"

Explain the ambiguity in this question. (This is my main point in asking this question)

Remove the ambiguity, then answer your particular version (or versions) of the question.

_______________________________________________________________________________

A related question:

You wish to randomly sample points on a unit disc by choosing points mathematically (NOT physically by throwing a dart or similar method). Perhaps the disc is Flatland, and you wish to find a representative figure for the average temperature over all of Flatland by sampling the temperature at a number of randomly chosen points.

One method is to centre the disc at the origin of the Cartesian plane, randomly choose a value from -1 to 1 for the x value, and another value from -1 to 1 for the y-value, then exclude points which are more than one unit from the origin. This method does indeed give a random smattering of points over the disc, but I want to debar this method for a reason I will explain later.

A second method is to randomly choose a number between 0 and 1 for the radius, then randomly choose an angle of rotation between 0 and 2π (in other words, polar coordinates).

(a) Explain why this method does not give an even distribution of points over the disc.
(b) How would you adjust the randomly chosen numbers (r and/or θ) to give such an even distribution?

braintic · Sep 30, 2015

Re: 2015 permutation X2 marathon

No takers? I'm happy for non 2015ers to join in.

InteGrand · Oct 1, 2015

Re: 2015 permutation X2 marathon

braintic said:
"On a given circle, a chord is selected at random. What is the probability that the chord is longer than the radius?"

Explain the ambiguity in this question. (This is my main point in asking this question)

Remove the ambiguity, then answer your particular version (or versions) of the question.

_______________________________________________________________________________

A related question:

You wish to randomly sample points on a unit disc by choosing points mathematically (NOT physically by throwing a dart or similar method). Perhaps the disc is Flatland, and you wish to find a representative figure for the average temperature over all of Flatland by sampling the temperature at a number of randomly chosen points.

One method is to centre the disc at the origin of the Cartesian plane, randomly choose a value from -1 to 1 for the x value, and another value from -1 to 1 for the y-value, then exclude points which are more than one unit from the origin. This method does indeed give a random smattering of points over the disc, but I want to debar this method for a reason I will explain later.

A second method is to randomly choose a number between 0 and 1 for the radius, then randomly choose an angle of rotation between 0 and 2π (in other words, polar coordinates).

(a) Explain why this method does not give an even distribution of points over the disc.
(b) How would you adjust the randomly chosen numbers (r and/or θ) to give such an even distribution?

$Haven't seen the second part you added yet, but for the first one, it's ambiguous because we get different answers depending on our choice of selecting a ``random'' point - it is not specified what method should be used.$

$One method can be we fix one endpoint of the chord, say on the point $(1,0)$ on the unit circle, and then we choose an angle from the uniform distribution on $\left(-180^\circ,180^\circ \right]$, $\mathcal{U}(-180,180)$, and the other endpoint of the chord corresponds to the ray with this angle. It is easy to show that the chord length equals or is less than the radius if and only if the angle picked is between $-60^\circ$ and $60^\circ$. So the chord will be longer than the radius iff we pick the angle in an interval of ``length'' $240^\circ$. Hence the required probability will be $\frac{240}{360}=\frac{2}{3}=0.666...$.$

$Another method would be to say that we will pick a height $h\in [0,1]$ \textit{above} the centre of the circle (from the uniform distribution on $[0,1]$, $\mathcal{U} (0,1)$), and take the chord that corresponds to the horizontal line at this height (i.e. only in the upper semicircle, due to symmetry of the circle). One may show that the chord is longer than the radius iff $0\leq h < \frac{\sqrt{3}}{2}$. Hence the required probability is $\frac{\sqrt{3}}{2}\approx 0.866$.$

braintic · Oct 1, 2015

Re: 2015 permutation X2 marathon

InteGrand said:
$Haven't seen the second part you added yet, but for the first one, it's ambiguous because we get different answers depending on our choice of selecting a ``random'' point - it is not specified what method should be used.$

$One method can be we fix one endpoint of the chord, say on the point $(1,0)$ on the unit circle, and then we choose an angle from the uniform distribution on $\left(-180^\circ,180^\circ \right]$, $\mathcal{U}(-180,180)$, and the other endpoint of the chord corresponds to the ray with this angle. It is easy to show that the chord length equals or is less than the radius if and only if the angle picked is between $-60^\circ$ and $60^\circ$. So the chord will be longer than the radius iff we pick the angle in an interval of ``length'' $240^\circ$. Hence the required probability will be $\frac{240}{360}=\frac{2}{3}=0.666...$.$

$Another method would be to say that we will pick a height $h\in [0,1]$ \textit{above} the centre of the circle (from the uniform distribution on $[0,1]$, $\mathcal{U} (0,1)$), and take the chord that corresponds to the horizontal line at this height (i.e. only in the upper semicircle, due to symmetry of the circle). One may show that the chord is longer than the radius iff $0\leq h < \frac{\sqrt{3}}{2}$. Hence the required probability is $\frac{\sqrt{3}}{2}\approx 0.866$.$

A third possibility - and this connects with my second problem:

Pick a point at random within the circle. Provided this point is not the centre, this point can be the midpoint of only one chord.
If the points are chosen perfectly randomly inside the circle, the probability can be shown to be 3/4.

Which leads to the second question. The reason I am asking that is that it leads on to another question which I don't know the answer for. I asked it a year or two ago and didn't get a satisfactory answer, mainly because no-one seemed to understand what I was asking. I am hoping this question will provide a good lead-in to the other question.

InteGrand · Oct 1, 2015

Re: 2015 permutation X2 marathon

braintic said:
A third possibility - and this connects with my second problem:

Pick a point at random within the circle. Provided this point is not the centre, this point can be the midpoint of only one chord.
If the points are chosen perfectly randomly inside the circle, the probability can be shown to be 3/4.

Which leads to the second question. The reason I am asking that is that it leads on to another question which I don't know the answer for. I asked it a year or two ago and didn't get a satisfactory answer, mainly because no-one seemed to understand what I was asking. I am hoping this question will provide a good lead-in to the other question.

$Well a start on the second one before I head off; by \textsl{even} distribution we mean that the probability of picking a point in any region of given area in the unit circle is proportional to that area. Clearly by picking the $r$ of the chosen point from the uniform distribution $\mathcal{U} (0,1)$, we do \emph{not} achieve this property. This is easy to see by considering annuli of equal given area $A$ within the circle. Annuli close to the circle centre have a larger width $\Delta R$ than those further from the centre, so would be more likely to be picked than those further away, despite the equal area.$

$So maybe we need a distribution for the random variable chosen radius $R$ that has higher probability density for higher values of $r$, as opposed to uniform. Maybe we need the following: for all $0\leq r_1, r_2, r_3, r_4 \leq 1$, if $r_2^2 -r_1^2 = r_4^2 - r_3 ^2 \text{ }(\geq 0)$ (condition for two annuli to have equal area), then $\Pr (r_1 \leq R \leq r_2) = \Pr (r_3 \leq R \leq r_4)$ ($R$ is the random variable that is the distance of chosen point from centre). I haven't considered much else though yet, nor the role of $\theta$.$

seanieg89 · Oct 1, 2015

Re: 2015 permutation X2 marathon

Judging by the way you have worded things you would like a way to randomly choose a point inside the unit disk given that you have a way of randomly choosing a points inside any given interval. (Ignoring the method of randomly choosing points inside the circumscribing box and rejecting those that lie outside the disk.)

The way to do this is to choose your radial coordinate to not have any distortion.

In standard polar coordinates, the infinitesimal area element is $r\, dr\, d\theta$ , which can by found by considering the area of the the tiny sector of an annulus where $r$ and $\theta$ lie in some tiny intervals respectively. This region is approximately rectangular, and multiplying its side lengths together give you the quantity claimed. To do this more rigorously you use Jacobians to change variables in a double integral, which is a higher dimensional analogue of integration by substitution.

The r dependence is the problem here, so we would like to use a differently defined radial variable (say s) such that
$ds \, d\theta=r\, dr\, d\theta\Rightarrow \frac{ds}{dr}=r\Rightarrow s=\frac{r^2}{2}+C.$ We might as well take C=0 to keep the origin at s=0.

Now the process of choosing a point at random in the unit disk is by randomly choosing a pair of real numbers $(s,\theta)\in [0,1/2]\times [0,2\pi]$ and taking the point $(\sqrt{2s}\cos(\theta),\sqrt{2s}\sin(\theta)).$

To summarise in a way that is applicable to choosing points randomly on more general regions/surfaces/manifolds, we would like to find a function F that maps a rectangular region to our shape in question such that F does not distort area. This distortion is quantified by the Jacobian determinant of F.

seanieg89 · Oct 1, 2015

Re: 2015 permutation X2 marathon

(And quantities like dx are understood rigorously as things called differential forms.)

braintic · Oct 1, 2015

Re: 2015 permutation X2 marathon

seanieg89 said:
Judging by the way you have worded things you would like a way to randomly choose a point inside the unit disk given that you have a way of randomly choosing a points inside any given interval. (Ignoring the method of randomly choosing points inside the circumscribing box and rejecting those that lie outside the disk.)

The way to do this is to choose your radial coordinate to not have any distortion.

In standard polar coordinates, the infinitesimal area element is $r\, dr\, d\theta$ , which can by found by considering the area of the the tiny sector of an annulus where $r$ and $\theta$ lie in some tiny intervals respectively. This region is approximately rectangular, and multiplying its side lengths together give you the quantity claimed. To do this more rigorously you use Jacobians to change variables in a double integral, which is a higher dimensional analogue of integration by substitution.

The r dependence is the problem here, so we would like to use a differently defined radial variable (say s) such that
$ds \, d\theta=r\, dr\, d\theta\Rightarrow \frac{ds}{dr}=r\Rightarrow s=\frac{r^2}{2}+C.$ We might as well take C=0 to keep the origin at s=0.

Now the process of choosing a point at random in the unit disk is by randomly choosing a pair of real numbers $(s,\theta)\in [0,1/2]\times [0,2\pi]$ and taking the point $(\sqrt{2s}\cos(\theta),\sqrt{2s}\sin(\theta)).$

To summarise in a way that is applicable to choosing points randomly on more general regions/surfaces/manifolds, we would like to find a function F that maps a rectangular region to our shape in question such that F does not distort area. This distortion is quantified by the Jacobian determinant of F.

Hmmm .... I've HEARD of a Jacobian - I guess I learned it 30 years ago.

For this particular question, do we really need calculus?

I figured that θ didn't need to be modified - the points would already be evenly distributed around the circle.
It should only be r that needs to be modified.

If I pick an initial value of r between 0 and 1, I want to turn it into a new value R so that:

(Area enclosed by circle of radius R) / (Area of disc) = r

$\frac{\pi R^2}{\pi}=r$

$R=\sqrt{r}$

You might recall my follow-up question from when I asked it a year or two ago. I don't think you understood what I was asking at the time.

I want to do a similar thing on the surface of the earth.

That is, I want to randomly pick a number between -90 and 90 [OR -π/2 and π/2 OR -1 and 1] and turn it into a latitude so that I can sample the earth's surface with evenly distributed points. It should be clear that trying to pick a random latitude directly will cause a greater density of sample points near the poles than near the equator.

I tried to do this in a similar way to the disk question. I looked up the surface area of a spherical cap (yes I could have integrated to find that). Then I tried to turn the randomly chosen latitude α into a new latitude β such that the surface area between the equator and the parallel of latitude β was α/90 that of the entire hemisphere.
But I got a nonsensical answer - α=90 should give β=90, and it didn't.

Any thoughts about my line of thinking? I'd prefer to avoid calculus if possible. (ie. I'm happy to accept the formula for the SA of a spherical cap)

HSC 2015 MX2 Permutations & Combinations Marathon (archive) (4 Viewers)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

TAFE Executive Officer

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 4)