You'd calculate the variance of a null distribution and then compare to this one. Get some kind of confidence/error calculation.

The question is, why are you doing this? Is it to truly determine if the thing is fair? You'll only ever get within some error, which will be quite large I would guess (pure guess). If you have data on file that shows a large uncertainty on whether it's fair, I'd say your ass is on the line. People don't understand math and variance but they do understand "there was large uncertainty whether it was fair".

Consequently, for a quick and dirty method, I'd do something like the below:

Take the mean of the balls (20), going on their face numbers. Calculate the standard deviation around that mean and their face values from your draw data.

Then number the balls from 1 to 40 based on their weight (lowest to highest). Calculate the standard deviation of these balls around the mean (take each draw as a data point, e.g. 38 20 38 14).

You can do both of these in about 15 minutes in Excel.

With luck these numbers will be very, very close. Then you'll look good and you can point to an extremely close similarity.

Let's see a plot. Weights on the x-axis and number of occurrences on the y- axis.

i used to know how to post images. not sure what changed.

image link

I don't want anyone to waste any time on this, having read your replies. I'm going to say that we have 300 trials where we would need tens of thousands for any meaningful analysis. I think that should suffice.

Thanks, all.

lol. I just learned our exposure is a billion. Thanks again, I'm out of this project.

Yeah for that kind of exposure you cover your ass by hiring a well regarded statistics professor. Heck, hire two. Have them tell you what they need to statistically call this fair.

We work with GLI on this sort of stuff all the time. Not sure why we're cutting corners for this. Crazy.

What one needs to do is have access to the balls and perform experiments for days and then we are in business. I am confident you will observe then something but very small. The problem of course is that nobody other than you can actually benefit from it by betting on the right balls because the public doesnt know what ball is what weight. The correct approach here is to have the balls replaced every few months.

If you are trying to beat the system though then there is the ethical/legal issue of having access to data and also the attempt to benefit from it is probably futile in the large scheme of things because of the small nature of the effect ( i mean given the volatility its entirely futile over the time periods one has available, there are way better ways to make money) .

I did happen to express some concerns about the mechanics and my involvement -- I think I've gotten to the bottom of things.

The ball machine is just an added layer of security. We're going to stuff secure envelopes 1 through 40 with various prize amounts, including the grand prize. The ball machine will determine the order and the number of the envelopes to be selected. Certainly we could have the contestant simply pick their own envelopes, but this wrinkle prevents one of our employees from potentially rigging the game.

There are other security measures in place but that is the basic concept. It also seems to be a one-time event, so I think it's unnecessary to worry about the public mining data.

I have the following Calc 3 problem:

Calculate the flux of the vector field:



through the surface described by:

, where

, .


I'm not sure I solved this one correctly. I used the divergence theorem, calculated the divergence, which is 3, then I converted the flux integral to a triple integral using spherical coordinates with rho from 0 to 2* sqrt(2), phi from 0 to pi/2 and theta from 0 to 2pi, and after solving I ended with 32*sqrt(2)*pi.

I see the surface described by ohmega as a spherical sector, where the base of the cone is z = 2 (after solving the system of equations, we get that these two surfaces intersect at z = 2, right?). I think what I should do is divide this shape into 3 integrals, the cone, the disk at z = 2 and what remains of the sphere, and then apply the divergence integral to each of these and add them up.

Any help is much appreciated, thanks.

Edit: by the way, Fr stands for frontier, I'm not sure what the exact terminology is in English, I just translated the text of the problem from my language. I guess it's the border of that surface? Fr = border, Fr = contour?

Isnt that like an ice cream cone (the lower one in in z<=0 range ) ie between the cone and the spherical cap? What you can then do if we understand by your notation that its the surface integral flux on the closed surface above is to take the divergence of the vector field and calculate using Gauss's divergence theorem the volume integral in eg spherical coordinates after first finding what Theta (polar angle) corresponds to the intersection of the cone with the sphere in z=-2. So r is from 0 to 8^(1/2) Theta from 3*Pi/4 to Pi if i did it right and phi from 0 to 2pi. This ought to be a very simple one time integral in spherical coordinates.


https://en.wikipedia.org/wiki/Spheri...rdinate_system

https://en.wikipedia.org/wiki/Divergence_theorem

Like the opposite of

I think i get 32*(√2-1)*Pi that way. Div is 3 indeed.

That is nothing like an ice cream cone

The virtual one is tastier.

It is in scientific society where profit is not above everything lol. It is an American scientific society also because it is massive and ready to make everyone fat if not careful.

It still isn't shaped like an ice cream cone.

If you make the angle smaller and the center of the circle higher with smaller radius all is possible. The object is topologically equivalent! So is a sphere actually to it but you know what i mean.

Still no. Ice cream cones, properly done, have a sort of frilly thing where cone meets ice cream. I will not stand for improperly modeled ice cream cones!

I said like an ice cream cone because you know;



The most important thing is to agree on the integral though. I am too lazy to do the actual flux to verify over the div calculation we did but anyone taking the class needs to do it that way too using the normal vector etc over 2 surfaces. I am tempted to claim that because there are 2 surfaces then it may make it more likely to get the mixed root plus integer result.


Also everyone calls it that lol



This by the way ends up being the same problem because div is 3 (of calculating the volume i mean). Their answer is same as mine if you scale it for radius^3. They have the same angle cone by the way.

There's a mistake in the problem's text, z should be greater or equal to zero, not less than zero... My bad.

Other than that, yes, that's exactly how it looks, like an ice cream cone (or spherical sector https://en.wikipedia.org/wiki/Spherical_sector).

You're right, masque, it's 32*pi*(sqrt(2) - 1). I took phi to be from 0 to pi/2 instead of pi/4. When I did rho-simple, I forgot to set x = 0 in the cone equation, I only did it for the sphere's equation.

Here is the solved problem, I skipped some obvious things: https://ibb.co/kXFOxF

What did you use to plot the surfaces, masque?

I simply web searched images of that type.

But if you have mathematica try

Plot3D[{Sqrt[8 - x^2 - y^2], Sqrt[x^2 + y^2]}, {x, -4, 4}, {y, -4, 4},
AspectRatio -> 1, Mesh -> None, PlotStyle -> Opacity[0.7]]

I will try to see how wolframalpha can do the same because it didnt work well there.

Eg you can put there

Plot3D[{Sqrt[8 - x^2 - y^2], Sqrt[x^2 + y^2]}, {x, -4, 4}, {y, -4, 4}]

http://www.wolframalpha.com/input/?i...y,+-4,+4%7D%5D

Tennis question that I'm stuck on

Player A vs Player B - the probability that player A wins any point is fixed at 0.46

The players are at a score of deuce, so the first player to be two points clear wins the game. What is the probability that player A wins the game?

Possible outcomes:

Player A wins the first two points and wins the game = 0.46^2 = 0.2116
Player B wins the first two points and wins the game = 0.54^2 = 0.2916
Each player wins one of the first two points = 2(0.46*0.54) = 0.4968 and we're back to the start

I'm not sure how to approach this one. With a binomial distribution would be my approach if we had a defined number of trials but without that I'm struggling.

All I can come up with is this. Make any sense?

We get a result 1-0.4968 = 0.5032

Of the 50.32% of the time that we get a result, player A wins 2116/5032 = 42.05% of the time and player B wins 57.95% of the time

So P(A win) = 42.05% and P(B win) = 57.95%?

Feels like I skipped a step. Any help much appreciated

Yes, there are several ways to get the answer.

(1) Set it up as an infinite series (find prob of A winning after 2 points, after 4 points, after 6 points, etc.)

(2) Write it as a recursion formula (this is essentially what you did)

Thanks for the reply

Can we define it as an infinite geometric sequence where in each cycle Player A is 0.2116 to win and each cycle differs by a factor of the players being tied after another two points = (0.46*0.54)+(0.54*0.46)=0.4968

a/(1-r) = 0.2116/(1-0.4968) = 0.4205

This is a small part of a larger question. I assumed that this was where I was going wrong but perhaps not. I'll post the full problem

P(A wins point) = 0.46 and is held constant

Find probability of Player A winning the set 6-3

Ways for A to win a game:

4 points played, A wins all 4
5 points played, A wins 4, B wins one
6 points played, A wins 4, B wins two
6 points played, A wins 3, B wins 3 and A wins from deuce

P(4-0) = dbinom(4,4,0.46) = 0.04477

P(4-1) = dbinom(4,5,0.46) = 0.12089

5C4 = 5 available combos but this must be done in exactly 5 points, so subtract the combo that wins the game in the first 4 points
Valid combos = 5C4 - 4C4 = 5-1 = 4
4 of 5 possible combos are valid so P(4-1 in exactly 5 points) = 0.12089*0.8 = 0.09671

P(4-2) = dbinom(4,6,0.46) = 0.19584

6C4 = 15 available combos but this must be done in exactly 6 points, so subtract the combos that win the game in 4 or 5 points
Valid combos = 6C4 - 5C4 - 4C4 = 15 - 4 - 1 = 10
10 of 15 possible combos are valid so P(4-2 in exactly 6 points) = 0.19584*0.66 = 0.12925

P(3-3) = dbinom(3,6,0.46) = 0.30654 and we go to Deuce

When at Deuce

P(2-0) = 0.2116
P(1-1) = 0.4968
P(0-2) = 0.2916

There is a result 0.2916+0.2116 = 0.5032

Of the instances where there is a result - 2116/5032 = 0.42051 wins for player A so P(A wins from deuce) = 0.4205

P(A wins game) = P(4-0) + P(4-1 in exactly 5 points) + P(4-2 in exactly 6 points) + (P(3-3)*P(winatdeuce))

= 0.04477 + 0.09671 + 0.12925 + (0.30654*0.4205) = 0.39963

To win exactly 6 out of 9 with success rate of 0.39963 = dbinom(6,9,0.39963) = 0.07404

Total number of combos where player A wins 6 of 9 games = 9C6 = 84

We need to eliminate the combos where player A reaches 6 games in less than 9 games played

Combos of 6-3 in exactly 9 games = 9C6 - 8C6 - 7C6 - 6C6 = 84-28-7-1 = 48

48 of 84 combos are valid = 0.57143

P(6-3 in exactly 9 games) = 0.07404*0.57143 = 0.04231 = 4.231%

The website that has this problem only has a box for the answer but it says that this is wrong. I have triple checked the arithmetic so its definitely not a fat fingers error