Coming Soon!

[FrediFizzx]

(Not that I have any new criticisms beyond the ones I already wrote about).

Your past criticisms were pure nonsense. No nonsense criticisms. They will be deleted very swiftly.

If you post nonsense on this thread it will be very swiftly and mercilessly deleted!
This applies to everyone not just Gill!

[FrediFizzx]

I maybe figured out a way to test the outcome pair probabilities. 1 million trials.



Not way out of whack but it would be nice if closer. Maybe more trials are needed. Ok, next we will run 90 degrees.
.

[Joy Christian]

Ok, now a question. Since all P(++)'s, etc. are equal to a 1/4 and the average of = 1/4, etc., does that prove that , etc. for our analytical situation? Keeping mind that P(++) and , etc. are actually averages.

So, I guess this question is still open. Joy seems to be trying to say no but not doing a very good job of it.

If f and g are two functions and < f > = 1/4 = < g >, then that does not prove that f = g. My examples above were just to demonstrate this fact. So the answer to your question is: No.
.

[FrediFizzx]

I maybe figured out a way to test the outcome pair probabilities. 1 million trials.



Not way out of whack but it would be nice if closer. Maybe more trials are needed. Ok, next we will run 90 degrees.

Nope! Of course I'm not doing it correctly. The (a-b) histogram is way different for the 3D vector scenario.



So, I'm not dividing by the correct number. Now, how to figure that out. BTW, the clue was that 90 degrees wasn't even close.
.

[FrediFizzx]

If f and g are two functions and < f > = 1/4 = < g >, then that does not prove that f = g. My examples above were just to demonstrate this fact. So the answer to your question is: No.

Well..., it also doesn't prove that they aren't equal. They could be equal to each other especially since their averages equal the same thing. So, I don't really get a NO from that either.
.

[FrediFizzx]

If you post nonsense on this thread it will be very swiftly and mercilessly deleted!
This applies to everyone not just Gill! So, be careful of what you spew.
.

[Joy Christian]

If f and g are two functions and < f > = 1/4 = < g >, then that does not prove that f = g. My examples above were just to demonstrate this fact. So the answer to your question is: No.

Well..., it also doesn't prove that they aren't equal. They could be equal to each other especially since their averages equal the same thing. So, I don't really get a NO from that either.

Sure. They could be equal. But < f > = 1/4 = < g > is not a very strong clue. And it is certainly not a mathematical proof of f = g. Far more likely is that it is just a coincidence. A logical or mathematical proof has to be beyond any doubt. So the answer to your question is almost certainly No as far as proof is concerned.
.

[FrediFizzx]

If f and g are two functions and < f > = 1/4 = < g >, then that does not prove that f = g. My examples above were just to demonstrate this fact. So the answer to your question is: No.

Well..., it also doesn't prove that they aren't equal. They could be equal to each other especially since their averages equal the same thing. So, I don't really get a NO from that either.

Sure. They could be equal. But < f > = 1/4 = < g > is not a very strong clue. And it is certainly not a mathematical proof of f = g. Far more likely is that it is just a coincidence. A logical or mathematical proof has to be beyond any doubt. So the answer to your question is almost certainly No as far as proof is concerned.

I don't think I will be buying that explanation. If they both equal a 1/4 then <f> = <g> so f = g. Seems like proof to me.
.

[Joy Christian]

If f and g are two functions and < f > = 1/4 = < g >, then that does not prove that f = g. My examples above were just to demonstrate this fact. So the answer to your question is: No.

Well..., it also doesn't prove that they aren't equal. They could be equal to each other especially since their averages equal the same thing. So, I don't really get a NO from that either.

Sure. They could be equal. But < f > = 1/4 = < g > is not a very strong clue. And it is certainly not a mathematical proof of f = g. Far more likely is that it is just a coincidence. A logical or mathematical proof has to be beyond any doubt. So the answer to your question is almost certainly No as far as proof is concerned.

I don't think I will be buying that explanation. If they both equal a 1/4 then <f> = <g> so f = g. Seems like proof to me.

More important question is: Will anyone buy the claim that f = g because < f > = 1/4 = < g >, where <...> is average over something?
.

[FrediFizzx]

Well..., it also doesn't prove that they aren't equal. They could be equal to each other especially since their averages equal the same thing. So, I don't really get a NO from that either.

Sure. They could be equal. But < f > = 1/4 = < g > is not a very strong clue. And it is certainly not a mathematical proof of f = g. Far more likely is that it is just a coincidence. A logical or mathematical proof has to be beyond any doubt. So the answer to your question is almost certainly No as far as proof is concerned.

I don't think I will be buying that explanation. If they both equal a 1/4 then <f> = <g> so f = g. Seems like proof to me.

More important question is: Will anyone buy the claim that f = g because < f > = 1/4 = < g >, where <...> is average over something?

Rewrite it < f > = < g > = 1/4 --> < f > = < g > --> f = g. That looks perfectly buyable to me.
.

[Joy Christian]

More important question is: Will anyone buy the claim that f = g because < f > = 1/4 = < g >, where <...> is average over something?

Rewrite it < f > = < g > = 1/4 --> < f > = < g > --> f = g. That looks perfectly buyable to me.

Not to me. In my view, the second implication in " < f > = < g > = 1/4 --> < f > = < g > --> f = g " is not valid. You will not find a single buyer.
.

[FrediFizzx]

More important question is: Will anyone buy the claim that f = g because < f > = 1/4 = < g >, where <...> is average over something?

Rewrite it < f > = < g > = 1/4 --> < f > = < g > --> f = g. That looks perfectly buyable to me.

Not to me. In my view, the second implication in " < f > = < g > = 1/4 --> < f > = < g > --> f = g " is not valid. You will not find a single buyer.

What??? Are you now claiming that < f > is not equal to < g >? ???
.

[Joy Christian]

Not to me. In my view, the second implication in " < f > = < g > = 1/4 --> < f > = < g > --> f = g " is not valid. You will not find a single buyer.

What??? Are you now claiming that < f > is not equal to < g >? ???

That is not second implication. The second implication in your argument above is this: < f > = < g > --> f = g. That is not a valid inference in any book of mathematics or logic.

You have the whole forum at your disposal. Find one person who agrees with you that < f > = < g > --> f = g is a valid inference, where <...> means average over something.
.

[FrediFizzx]

That is not second implication. The second implication in your argument above is this: < f > = < g > --> f = g. That is not a valid inference in any book of mathematics or logic.

You have the whole forum at your disposal. Find one person who agrees with you that < f > = < g > --> f = g is a valid inference, where <...> means average over something.

Quoting was screwed up on that last post. Say in 10,000 trials we have 2500 ++'s. The average is 0.25. We also have 2500 - -'s and the average is 0.25. So, < ++ > = 0.25 = < - - >. Are you trying to tell me that 2500 is not equal to 2500?
.

[Joy Christian]

That is not second implication. The second implication in your argument above is this: < f > = < g > --> f = g. That is not a valid inference in any book of mathematics or logic.

You have the whole forum at your disposal. Find one person who agrees with you that < f > = < g > --> f = g is a valid inference, where <...> means average over something.

Quoting was screwed up on that last post. Say in 10,000 trials we have 2500 ++'s. The average is 0.25. We also have 2500 - -'s and the average is 0.25. So, < ++ > = 0.25 = < - - >. Are you trying to tell me that 2500 is not equal to 2500?

No. That is not what I am saying. For any functions f and g, if < f > = < g >, then that does not imply f = g. It does not matter what f and g are. A specific example does not change this.
.

[FrediFizzx]

No. That is not what I am saying. For any functions f and g, if < f > = < g >, then that does not imply f = g. It does not matter what f and g are. A specific example does not change this.

Well, you are going to have to prove that. I'm not buying it.
.

[local]

I'll chime in to agree with Joy.

Suppose we have:

f = {2,4}
g = {0,6}

Then <f> = 3 and <g> = 3, but in no case is any f value equal to any g value.

[Joy Christian]

No. That is not what I am saying. For any functions f and g, if < f > = < g >, then that does not imply f = g. It does not matter what f and g are. A specific example does not change this.

Well, you are going to have to prove that. I'm not buying it.

It is a trivial fact and easy to prove using counterexamples.

One counterexample is sufficient to prove it.

Your claim is that < f > = < g > --> f = g.

So let f = 1 + cos(x) and g = 1 + sin(x), with 0 < x < 360 degrees.

Then < f > = 1 = < g >, where average is over all angles x.

This, according to your claim, implies f = g.

But it is self-evident that f =/= g.

So your claim is wrong (by reductio ad absurdum).
.

[FrediFizzx]

It is a trivial fact and easy to prove using counterexamples.

One counterexample is sufficient to prove it.

Your claim is that < f > = < g > --> f = g.

So let f = 1 + cos(x) and g = 1 + sin(x), with 0 < x < 360 degrees.

Then < f > = 1 = < g >, where average is over all angles x.

This, according to your claim, implies f = g.

But it is self-evident that f =/= g.

So your claim is wrong (by reductio ad absurdum).

10 million trials.



They are almost the same but not quite. If we call them the same then the totals for f and g are the same so f = g. Try again.
.

[Joy Christian]

It is a trivial fact and easy to prove using counterexamples.

One counterexample is sufficient to prove it.

Your claim is that < f > = < g > --> f = g.

So let f = 1 + cos(x) and g = 1 + sin(x), with 0 < x < 360 degrees.

Then < f > = 1 = < g >, where average is over all angles x.

This, according to your claim, implies f = g.

But it is self-evident that f =/= g.

So your claim is wrong (by reductio ad absurdum).

10 million trials.



They are almost the same but not quite. If we call them the same then the totals for f and g are the same so f = g. Try again.

No, I am not going to try again. I have already given you more than enough evidence that your claim is wrong. But you don't have to believe me. There are others who might believe you.
.