I can do telepathy EASY! - NSA

[Arouet]

It should be roughly correct if the hit ratio is 25% - 50%, which is what we're both expecting.

If the hit ratio is 50%-99% then it will skew the ODDS higher, but if that's the case the paranormal bias is so high the amount of trials is fairly small and exponential growth kicks in on the ODDS.

There is no 50-99% in this protocol. It's either 0/4, 1/4, 2/4 or 4/4

Considering
A/ there is no contract, so the data can be sought before the stats

B/ there is no set ODDS to demonstrate, 100:1? 1000:1? 10000:1?
C/ we don't know the hit ratio or what the minimum should be
D/ the exact stats are unwieldy
E/ I don't even know if the protocol gets the channels in the right order yet
F/ Using an 'ordering' stats, I don't know whether to make 4/4 or 2/4 the pass mark
G/ If you use 2/4 as the pass mark, you will need 40 batches not 10
H/ It's a simple case of adding a few batches to overcompensate for the odds

It's like getting out your protractor to divy up the apple pie!
Just add up the hits![/quote]

You can't just add up the hits. Because this isn't a ganzfeld type protocol you've set up where there is one target and three controls and the odds at chance are 1/4. Here, you're given all the words, and asked to sort them. This is apples and oranges. You set out the odds:

The stats of guessing the order of 4 words.

There are 4! or 4X3X2X1 = 24 ways to order the words

1234 - 4 right
1243 - 2 right
1324 - 2 right
1342 - 1 right
1423 - 1 right
1432 - 2 right
2134 - 2 right
2143 - 0 right
2314 - 1 right
2341 - 0 right
2413 - 0 right
2431 - 1 right
3124 - 1 right
3142 - 0 right
3214 - 2 right
3241 - 1 right
3412 - 0 right
3421 - 0 right
4123 - 0 right
4132 - 1 right
4213 - 1 right
4231 - 2 right
4312 - 0 right
4321 - 0 right

TOTALS
0 right = 9
1 right = 8
2 right = 6
3 right = 0
4 right = 1

This above is fine, you lose yourself below:

Average is (8 + 12 + 4) / 24 = 1

2+ right = 7/24 or 3.4:1 ODDS (less than 1 chance in 3)
4 right = 24:1 ODDS (1 chance in 24)

Let's make the PASS CLAIM
AVERAGE SCORE OVER MANY TESTS >= 2/4.

So I will try to get ON AVERAGE, 2 right every test repeatable.
Now I have just have to work out how many trials to break 1000:1.
It will be roughly 3_/1000 or 10 trials.

Sound about right? 10 trials and I average 2 right every test to pass 1000:1.
I'll get the exact number of trials later.[/quote]


This doesn't seem to be the correct way to go about it at all. Here are how the odds break down:

At chance, the expectation is as follows:

0 right = 9 - 9/24 = 0.375
1 right = 8 - 8/24 = 0.333
2 right = 6 - 6/24 = 0.25
3 right = 0 - N/A
4 right = 1 - 1/24 = 0.0417

total = 0.9997 (doesn't = 1 due to rounding)

So you guessed one trial 1/4 (33.3% chance) and another at 2/4 (25% chance).

If you combine the stats and do 3/8 that would be 37.5% which obviously can't be right. What you want to test is are you getting these stats at a rate that is significantly better than expectation. I don't know exactly how to calculate that, but I do know that you just can't add them together.

For obvious reasons, you don't want to just consider getting a 2/4 to be significant since this is going to happen very often. What you want to figure out is whether you are getting 2/4 significantly more often than chance. If you accept the above, I'm willing to go and post on a probability forum to see if anyone has any suggestions on how to figure out what significant stats would be here and what sample size would be needed.

It is complicated, which is why you might want to consider going back to a protocol where you've got to guess the real word from the fakes. Then we can add up the hits and it will probably be much easier to figure out what a significant result would be.

You want to figure out if you've got a special ability right? You've been trying to establish this for 10 years. Why not do it right?

[G. ADAM]

Dude, whether you consider each batch individually or add up the totals the expected average is 25%.

Considering each batch individually, the odds are about 2/7 of getting >=2 hits.
Considering the total hits / total words, the odds are 1/4 (2/8) of getting >=2 hits.

The error is very small (1/8) if my TOTAL RATIO is <=50%.

Do you get that far? Statistics use approximate distributions more times than not.

You want to move from a BATCH PROTOCOL to a TEST PROTOCOL and we don't have any numbers yet.

BATCH PROTOCOL - While score>average repeat as necessary
TEST PROTOCOL - You need half a dozen figures to work out the number of trials.

If you INSIST on predeclaring how many trials you are going to do, then there is no point in continuing as we haven't even *tested* the BATCH PROTOCOL.

Your meaning of RIGHT is merely - odds calculated to 100% precision.

I can probably break 24:1 ODDS 10 batches runnning! That's 6,000,000,000,000:1 ODDS.

We don't know at this point.

If there's no $1,000,000 on offer why do you insist on working out the exact ODDS?

Concentrate on getting REPEATABLE OBJECTIVE RESULTS - NOTHING MORE AT THIS POINT.

[G. ADAM]

For obvious reasons, you don't want to just consider getting a 2/4 to be significant since this is going to happen very often. What you want to figure out is whether you are getting 2/4 significantly more often than chance.

DON'T YOU THINK I KNOW THAT?

Why are you SHIFTING THAT ARGUMENT into a contractual protocol instead of an investigative protocol?

We had the ball rolling and you have a mid test crisis.

[Arouet]

Dude, whether you consider each batch individually or add up the totals the expected average is 25%.

The odds of what is 25%?

Considering each batch individually, the odds are about 2/7 of getting >=2 hits.

How do you figure: There are 6 combos of two hits out of 24 options. 6/24 = 1/4

Considering the total hits / total words, the odds are 1/4 (2/8) of getting >=2 hits.

I'm not clear on what you are getting at here. What is the 2/8 based on?

The error is very small (1/8) if my TOTAL RATIO is <=50%.

The error was not small.

Do you get that far? Statistics use approximate distributions more times than not.

Where do you get that idea? We don't need to rely on approximate distributions here, we can find exact distributions.

You want to move from a BATCH PROTOCOL to a TEST PROTOCOL and we don't have any numbers yet.

BATCH PROTOCOL - While score>average repeat as necessary

The average score means nothing. You are trying to figure out if you can predict these orders better than chance. Your averaging doesn't allow us to do that. You think 2/4 is significant. I think its only significant if you are doing it better than chance.

TEST PROTOCOL - You need half a dozen figures to work out the number of trials.

Yes, we need to figure out the propper number of trials.

If you INSIST on predeclaring how many trials you are going to do, then there is no point in continuing as we haven't even *tested* the BATCH PROTOCOL.

The way you are testing the batch protocol doesn't make any sense. You criteria doesn't make any sense. It only makes sense if compared to chance. Otherwise, what are we testing?

Your meaning of RIGHT is merely - odds calculated to 100% precision.

I can probably break 24:1 ODDS 10 batches runnning! That's 6,000,000,000,000:1 ODDS.

If you could get 10 trials in a row at 4/4 that would indeed be significant. If you got 2/4 50% of the time, I'm sure that would be pretty significant as well over a proper sample. If you got 1/4 correct the other 50% of the time that would also be significant over a proper sample.

But not based on your pass protocol at 2/4. That doesn't tell us anything. I'm willing to play along, but we should do this semi-proper. I'm not saying we need a rock solid protocol (the one we have set up is certainly not rock solid). But we should have something set up that will give us some indication of whether something interesting is going on. I would think you would want that too.

We don't know at this point.

If there's no $1,000,000 on offer why do you insist on working out the exact ODDS?

Concentrate on getting REPEATABLE OBJECTIVE RESULTS - NOTHING MORE AT THIS POINT.

[G. ADAM]

For Christ's sake.

The expected average is 1 out of 4 right. Both methods agree at that expected average.

Just to give you something objective, I'll set my paranormal bias at +50%.

That's 1.5 right per batch ON AVERAGE.

Can you proceed now?

Exact stats are a little tricky, e.g. since I can repeat guesses how do you measure 3/4?

YOU WON'T FIND ANY STATISTICIAN WHO WILL WORK IT OUT FOR YOU HOW TO TALLY BATCHES.

*I* can ask sci.math because I know how to word the question, but it's not worth my time!

[G. ADAM]

REALLY?!?!?!... are you blaming me!??!
oh wow..

dude !! i am supporting you
i want this to be true
it would be so awesome to be part of this IF it was true

No worries! I have a hard time telling the difference between here and the 10 years of delay tactics from the Skeptics Competitions.

[Arouet]

For Christ's sake.

The expected average is 1 out of 4 right. Both methods agree at that expected average.

Just to give you something objective, I'll set my p aranormal bias at +50%.

That's 1.5 right per batch ON AVERAGE.

Can you proceed now?

Exact stats are a little tricky, e.g. since I can repeat guesses how do you measure 3/4?

YOU WON'T FIND ANY STATISTICIAN WHO WILL WORK IT OUT FOR YOU HOW TO TALLY BATCHES.

*I* can ask sci.math because I know how to word the question, but it's not worth my time!

The expected average of what? The expected rate of getting 2/4 is 1/4. You will be expected to get 1/4 about 1/3 of the time, and 0/4 37.5% of the time. This is what you'll get if you use an RNG, or just guess. So you getting 2/4 or even 4/4 means nothing unless you're doing it significantly more than expected. Don't you see that? You took a stats class or two, right? I've never taken a stats class and I know this.

Of course its worth your time to consult the math forum - at least if you want to actually do it right. Why would you want to do it half-assed? You've been at this for a decade - don't you think its worth your while to actually develop something that makes sense? You should if you want to be taken seriously. You should if you want to prove something to yourself.

I doubt the math is that hard for someone who actually knows what they are doing.

[Arouet]

REALLY?!?!?!... are you blaming me!??!
oh wow..

dude !! i am supporting you
i want this to be true
it would be so awesome to be part of this IF it was true

No worries! I have a hard time telling the difference between here and the 10 years of delay tactics from the Skeptics Competitions.

If by "delay tactics" you mean "working out a proper protocol" then I can see why you never reached an agreement. You seem quite opposed to actually trying to figure out a proper test. I'm mystified as to why that is. It really shouldn't be that hard, and will be well worth your time.

[G. ADAM]

Arouet for the 5th time.

I do realise that breaking 3.4:1 ODDs is not significant.

I never said it was, I used the term PASS FOR A SINGLE BATCH IS 2/4.

[G. ADAM]

If by "delay tactics" you mean "working out a proper protocol" then I can see why you never reached an agreement. You seem quite opposed to actually trying to figure out a proper test. I'm mystified as to why that is. It really shouldn't be that hard, and will be well worth your time.

What is your objective Arouet? to beat 1,000,000:1 ODDS?

I'm just trying to demonstrate paranormal, James Randi won't even look at your stats he'll just look at how many hits over how many words I got right.

[G. ADAM]

Arouet here is my challenge to you.

This is my channel from Batch 3 Word 1

Word 1 channel is ((Potro del Mantelete a coin and told him to telephone Peregil from the))

Go and find that word and post it with 4 new random options.

See how I go with a simpler protocol.

You here to argue or do a paranormal test?

[G. ADAM]

Your 1 hour is up!

This is all a big charade to conceal the WORD 1 from batch 3.

Skeptics always bail when the power is evident.

Arouet The Skeptic - BUSTED!

[Arouet]

Arouet for the 5th time.

I do realise that breaking 3.4:1 ODDs is not significant.

I never said it was, I used the term PASS FOR A SINGLE BATCH IS 2/4.

Why would you call it a pass to get 2/4 in a single back when you will get that 25% of the time? Don't you see that it only matters how often you get that 2/4?

[Arouet]

Your 1 hour is up!

This is all a big charade to conceal the WORD 1 from batch 3.

Skeptics always bail when the power is evident.

Arouet The Skeptic - BUSTED!

What's wrong with you?

[Arouet]

Arouet here is my challenge to you.

This is my channel from Batch 3 Word 1

Word 1 channel is ((Potro del Mantelete a coin and told him to telephone Peregil from the))

Go and find that word and post it with 4 new random options.

See how I go with a simpler protocol.

You want to pick out of 5 now?

You here to argue or do a paranormal test?

In this case its one and the same.