Ask a physicist anything. (8)

[Zippy the Wonderslug]

If you know the center coordinates (x0,y0) and the inner and outer radius, you can generate any point within that annulus with the following formula:

(x,y) = (x0,y0) + r*(cos T , sin T)

Where r is randomized between the inner and outer radius.
Where T is randomized between 0 and 2*pi.

Edit:
Should you want it represented another way:

x = x0 + r*cos T
y = y0 + r*sin T

I can probably get close, if not exact, with the center coordinates using MS Paint and zooming in like 600%.

The center circle is all white though and it would be nice if I could get a random number that wouldn't click on that area.

Anyway, I kind of get your formula but how can I generate such numbers?

I usually go to RANDOM.ORG - True Random Number Service when looking to try something different.

Also, I'm using Google Spreadsheet and I'm not all too good at this.

 

[Elendur]

I can probably get close, if not exact, with the center coordinates using MS Paint and zooming in like 600%.

The center circle is all white though and it would be nice if I could get a random number that wouldn't click on that area.

Anyway, I kind of get your formula but how can I generate such numbers?

I usually go to RANDOM.ORG - True Random Number Service when looking to try something different.

Also, I'm using Google Spreadsheet and I'm not all too good at this.

I don't have a picture editing tool, unfortunately, so I can't investigate the center, but the other should work.

With google spreadsheet it seems you can use the function RAND() as found here:
https://support.google.com/docs/table/25273?rd=2
https://support.google.com/docs/answer/3093438

Since RAND() will return a number within the range of [0,1) (with a uniform distribution) you should be able to write as below.
_______________________________
Assuming the mid coordinates are (x0,y0), inner radius is r and outer radius is R:

length = (R-r)*RAND()+r
angle = RAND()*2*PI

x = x0 + length*cos( angle )
y = y0 + length*sin( angle )
_______________________________

You could also use CEILING(value, factor), with CEILING(value, 1) to round up to nearest integer, in order to get the x and y coordinates as integers.

Example:
CEILIING(x,1)

 

[Zippy the Wonderslug]

On my first two shots in an online pool game, I can sink all of these balls

or these

The rack is always random, except of course the eight ball.

My initial break is the same for both of these.

On my second shot, I have two possible choices, even though they look extremely similar.

Whatever colored ball goes in first on this second shot is what I'll be playing, either solids or stripes.

I highlighted in yellow which ball will go in first on this second shot in both pictures.

What are the odds of having all 7 same colored balls to go in that also match the highlighted ball, where I only have the eight ball left to sink?

I know that I need 7 out of the 9 balls to match in either scenario.

I also know that the highlighted ball needs to be part of the 7 that drops.

The math on this is just too difficult for me to understand at the moment.

I'm pretty sure the odds, if you add together both possibilities, are probably less than 10% at best.

It seems so unlikely to me.

Anyway, could I get one taker on this probability problem please?

Cheers and many thanks ahead if I don't respond back right away.

 

[Elendur]

On my first two shots in an online pool game, I can sink all of these balls

or these

The rack is always random, except of course the eight ball.

My initial break is the same for both of these.

On my second shot, I have two possible choices, even though they look extremely similar.

Whatever colored ball goes in first on this second shot is what I'll be playing, either solids or stripes.

I highlighted in yellow which ball will go in first on this second shot in both pictures.

What are the odds of having all 7 same colored balls to go in that also match the highlighted ball, where I only have the eight ball left to sink?

I know that I need 7 out of the 9 balls to match in either scenario.

I also know that the highlighted ball needs to be part of the 7 that drops.

The math on this is just too difficult for me to understand at the moment.

I'm pretty sure the odds, if you add together both possibilities, are probably less than 20% at best.

It seems so unlikely to me.

Anyway, could I get one taker on this probability problem please?

Cheers and many thanks ahead if I don't respond back right away.

Essentially, you'll be choosing 9 balls from a set of 15 balls, where there are three groups to consider.

1. Colored, size 7.
2. Striped, size 7.
3. Black, size 1.

However, your group is also determined by your initial ball. Not to mention, we want to avoid the black ball (or so I assume). However, I will initially avoid the argument regarding the black ball as you did not state it as a part of the problem (except for the implicit part of the first ball).

So, the probability to avoid the black ball as the first ball is 14/15. Then we have that we will choose 8 balls from a set of 15 balls, of which there are 6 sought after.

This ends up with the following equation:

Which evaluates to approximately 0.5%... Which means, that with both cases you have about 1% chance of the correct configuration (though it's important to note that the chance of success is not independent, which will affect the cumulative probability).

I noted that the problem ends up with being essentially identical to the probability (again, including the 14/15 initial chance) of choosing two from the unwanted group of nine. Which allows for the similar, but with the additional information of the black ball (where we reduced the sought after group to 8 balls), result:

Which evaluates to approximately 0.4%.

To summarize the answer:
If you don't care about the black ball, the probability for one of the shots to succeed is around 0.5%.
If you do care about the black ball, the probability for one of the shots to succeed is about 0.4%.

Disclaimer:
Hopefully I haven't done any mistakes. If someone finds an error, don't hesitate to point it out.
I've written this on the fly, so please do ask questions if there is anything I can clarify.

 

[Zippy the Wonderslug]

Sorry, but your math is off.

I highlighted in red all of the balls that would sink in both cases.

I went to this site:
Probability - theory of tossing coins

Apparently having 7 of 9 coin flips to be the same, the odds on that are only 7%.

Since the first ball that drops on the second shot should be the same as the majority, I'm assuming the odds are cut in half to now being just 3.5%.

So for either scenario, the odds would be this low.

I'm not sure how to collectively add them together though.

Six of the same balls drop in both cases and the only other big difference is which one falls on the second shot first because that determines if I play either solids or stripes.

Either way, I'm going to assume the chances of winning a game in just 3 shots is under 5%.

It would be nice to know the exact odds since I love math but I can't seem to solve for this one.

Thank you for the reply.

I hope you post another one.

 

[Elendur]

You're essentially incorrect in your approach, but I did find a mistake in my math. Will post(/edit this) as soon as I've done the dishes.

Edit:

Sorry, but your math is off.

I highlighted in red all of the balls that would sink in both cases.

I went to this site:
Probability - theory of tossing coins

Apparently having 7 of 9 coin flips to be the same, the odds on that are only 7%.

Since the first ball that drops on the second shot should be the same as the majority, I'm assuming the odds are cut in half to now being just 3.5%.

So for either scenario, the odds would be this low.

I'm not sure how to collectively add them together though.

Six of the same balls drop in both cases and the only other big difference is which one falls on the second shot first because that determines if I play either solids or stripes.

Either way, I'm going to assume the chances of winning a game in just 3 shots is under 5%.

It would be nice to know the exact odds since I love math but I can't seem to solve for this one.

Thank you for the reply.

I hope you post another one.

The method you are using is dependent on the binomial distribution, which samples n units which have a "win" rate of p (in the coin case p=1/2).

This is incorrect because the sampling can be infinitely large. You, in this case, only have 15 balls to work with. This is as described by the hypergeometric distribution.

Using their notation, as found here, we have:
N = 14 is the population size (due to the first ball already being chosen)
K = 6 is the number of success states in the population (as we've decreased the 'correct' balls by one with the first pick)
n = 8 is the number of draws (one already drawn)
k = 6 is the number of successes (as we want to down the entire set of balls)
(a b) = a!/(b!(a-b)!) is a binomial coefficient

Inserting this into their formula,

, and adding the multiplication representing the first choice, yields:

which, when evaluating, becomes approximately 0.009, i.e. almost 1%. If we are to repeat the argument where we add the criteria of not choosing the black ball, we get

which evaluates to approximately 0.007, i.e. about 0.7%.

So, correcting the error, you get about twice the chance as before. Still not high though.

 

[VProud]

What's your logical reasoning to make yourself accept the sum of all positive natural numbers being equal to -1/12?

 

[Subduction Zone]

What's your logical reasoning to make yourself accept the sum of all positive natural numbers being equal to -1/12?

ASTOUNDING: 1 + 2 + 3 + 4 + 5 + ... = -1/12 - YouTube

 

[VProud]

ASTOUNDING: 1 + 2 + 3 + 4 + 5 + ... = -1/12 - YouTube

I know the proof very well.
Also, I've watched pretty much every Numberphile and MinuitPhysics video going, they are very good.

Was just wondering if you have to do anything to make yourself accept that on a logical basis.

For instance, I just have to ascribe it as a 'trick', similar to the infinite room hotel problem, that involves shifting the numbers, to make myself accept that it is true.

 

[Justatruthseeker]

I know the proof very well.
Also, I've watched pretty much every Numberphile and MinuitPhysics video going, they are very good.

Was just wondering if you have to do anything to make yourself accept that on a logical basis.

For instance, I just have to ascribe it as a 'trick', similar to the infinite room hotel problem, that involves shifting the numbers, to make myself accept that it is true.

Trick, since only in mathematics is a negative number real. In reality one can never subtract 2 -apples from 2 existing apples, and get 4 apples. If we could we'd all be rich!

 

[Elendur]

Trick, since only in mathematics is a negative number real. In reality one can never subtract 2 -apples from 2 existing apples, and get 4 apples. If we could we'd all be rich!

You know what, I think you'll like my report which I'll post here in a week or two.

 

[Justatruthseeker]

You know what, I think you'll like my report which I'll post here in a week or two.

Will you perform magic tricks and take away negative non-existent things from existing things and end up with more than you started with?

 

[BaconWizard]

Will you perform magic tricks and take away negative non-existent things from existing things and end up with more than you started with?

Never been in debt then? My bank account currently contains negative money.

 

[BaconWizard]

Trick, since only in mathematics is a negative number real. In reality one can never subtract 2 -apples from 2 existing apples, and get 4 apples. If we could we'd all be rich!

Dude you've just abolished electricity and chemistry (and therefore physics as we know it in the observable universe) in one swell-foop.

That's amazing. Nothing physical can have a negative charge!

 

[Justatruthseeker]

Dude you've just abolished electricity and chemistry (and therefore physics as we know it in the observable universe) in one swell-foop.

That's amazing. Nothing physical can have a negative charge!

And yet you insist two positive charges and two negative charges = zero charge.



Yet when it comes to imagination land you insist it actually equals 4.

It's you that preaches pseudo-science. You know not of what you speak, as usual.

I don't think you know what electricity is, as a matter of fact, I think you think you know, but then every single solar theory you had fell, but then you ignore it in space too, don't you?

How's it going treating that plasma like solids, liquids and gasses????

Funny thing is you can't do that in the lab, but insist we do so in space, and wham, 95% Fairie Dust ..."insert here"...

Just outrageous the claims of mainstream religious supporters in Big Bang land.

 

[Elendur]

Will you perform magic tricks and take away negative non-existent things from existing things and end up with more than you started with?

Worse, perform magic tricks where I use imaginary things to demonstrate real results.

(Puns intended, if you don't understand now you'll probably understand when I post it )

 

[BaconWizard]

And yet you insist two positive charges and two negative charges = zero charge.

Yes. That's the point.

Yet when it comes to imagination land you insist it actually equals 4.

You assert. Now provide an actual instance of this being the case.

It's you that preaches pseudo-science. You know not of what you speak, as usual.

Well there's nothing ambiguous about it: Something is either science or it is not, and the requirements to be classed as science are very clear. I bet you don't even know what they are.

I don't think you know what electricity is, as a matter of fact, I think you think you know,

Movement of electrons. Next.

but then every single solar theory you had fell, but then you ignore it in space too, don't you?

Again, provide examples or it's just "blah blah blah"

How's it going treating that plasma like solids, liquids and gasses????

Who is doing so and again.. provide an example.

Funny thing is you can't do that in the lab, but insist we do so in space

Do WHAT in the lab/space?

Just outrageous the claims of mainstream religious supporters in Big Bang land.

Such as?

You claimed there is no such thing as a negative number in reality. Negative charges. Financial debt. Such things are what negative numbers were invented to represent. Fail.

 

[Maxwell511]

I am amazed this thread is still going. Well done Wiccan_Child.

 

[Justatruthseeker]

Worse, perform magic tricks where I use imaginary things to demonstrate real results.

(Puns intended, if you don't understand now you'll probably understand when I post it )

Ahhh, but there's a difference from hypothesizing things that can't be directly observed, and violating the laws of physics in order to support those hypothetical constructs. Personally, I've never seen a quark, but being a charged particle it certainly doesn't violate any physics to postulate it's existence.

Technically, the only thing I have ever seen is a photon, since again, technically, objects are observed by the light reflected, or emitted. The silly putty of the universe?

 

[Elendur]

Ahhh, but there's a difference from hypothesizing things that can't be directly observed, and violating the laws of physics in order to support those hypothetical constructs. Personally, I've never seen a quark, but being a charged particle it certainly doesn't violate any physics to postulate it's existence.

Technically, the only thing I have ever seen is a photon, since again, technically, objects are observed by the light reflected, or emitted. The silly putty of the universe?

You're so off the mark that you really need to read the report when I upload and post it here.

I won't spoil things too much for you but:
I'm no physicist.