Anyone heard about this dude? (pavelprorok on youtube)

[Thabiguy]

SO getting at least ONE exactly named out of 2, 9 times out of 10 beat odds 1.852 to 1... Does it allow me to have 3 trials?

No, sorry. If you have three trials, then getting "at least 9 times at least one right" in at least one trial is about 1 in 618. If you want to have three trials, you would need to go for 10 out of 10, in at least one trial.

There's also another option: if you're going to do 30 total sets anyway, why divide them into three trials? You could count them as one long run (with breaks inserted as needed), and claim success when you get 20 times at least one right. Then you only need an average 6-7 times in each 10-set "trial".

 

[pavel_do]

There's also another option: if you're going to do 30 total sets anyway, why divide them into three trials? You could count them as one long run (with breaks inserted as needed), and claim success when you get 20 times at least one right. Then you only need an average 6-7 times in each 10-set "trial".

thank you.
Good option, but that is in a way.. same as getting 20 out of 30 with sets of 3 where I pick one. I guess I have to try both ( as with sets of 3 I haven’t tried the one when I pick the one i am sure of) to find the best one.
But even if I would make run of 30 with 5 instead of 3 separate, still I would like to have 3 trials, as sometimes my best result from the first trial and sometimes from second like I am worming up or something. Beside that having only 1 “chance” putts allot of pressure on me and I will have it enough with stress any way.. so don’t really need one extra.

P.S.

Does 20 out of 30 allows me to have 3 trials? Or still only 1? ( 3 by 30) and how many pairs would I need to name exactly out of 30 to beat needed odds.. again including possibility of 3 trials?

 

[Thabiguy]

Gahhh... you just keep multiplying by three!

What I suggested was to reorganize 30 sets from 3*10 to 1*30. If you insist on having three trials, then my earlier post is of no importance.

Besides, who am I to argue? You'll have to negotiate your three trial requirement with JREF, not with us. If you want 3 trials, then here you go: the required results that allow for having 3 trials. (Whether you convince JREF to accept this, is up to you.)

10 sets in each trial: 5 times both right or 10 times at least one right
15 sets in each trial: 6 times both right or 13 times at least one right
20 sets in each trial: 7 times both right or 16 times at least one right
30 sets in each trial: 8 times both right or 21 times at least one right

 

[pavel_do]

No, sorry. If you have three trials, then getting "at least 9 times at least one right" in at least one trial is about 1 in 618. If you want to have three trials, you would need to go for 10 out of 10, in at least one trial.

There's also another option: if you're going to do 30 total sets anyway, why divide them into three trials? You could count them as one long run (with breaks inserted as needed), and claim success when you get 20 times at least one right. Then you only need an average 6-7 times in each 10-set "trial".

Gahhh... you just keep multiplying by three!

What I suggested was to reorganize 30 sets from 3*10 to 1*30. If you insist on having three trials, then my earlier post is of no importance.

Besides, who am I to argue? You'll have to negotiate your three trial requirement with JREF, not with us. If you want 3 trials, then here you go: the required results that allow for having 3 trials. (Whether you convince JREF to accept this, is up to you.)

10 sets in each trial: 5 times both right or 10 times at least one right
15 sets in each trial: 6 times both right or 13 times at least one right
20 sets in each trial: 7 times both right or 16 times at least one right
30 sets in each trial: 8 times both right or 21 times at least one right

Thank you very much for the calculations. I understood what you meant about 30 per trial… But I have my point in need of more than 1 trial as I explained… I will propose it to JREF, have spoken to one of the professors who is skeptic and was involved in research of the paranormal claims he said that a few trial in a reasonable way not 100 or 300 in not a problem for test same as to reveal the result immediately after I named it.. as to keep the rules of Double Blind test in order both sides should not know the order of the photos or the photo that was handled to me as it would be in case where 1 out of 3 in random would be passed to me. But to reveal the result immediately make no impact of BD test.

I think I have to work out the best option with number of sets, mean while tomorrow will try to propose the test just that the number of runs might be changed for example from 10 to 15, but the conditions would still remain the same. So maybe we can discuss with them at least basic thing of my propose. And during a week I will work with different variations of sets.. to find the best suitable for me.

Just a questions.. in your opinion to claim both as pairs same as single guess e.g. 15 sets in each trial: 6 times both right or 13 times at least one right. Is it possible? I guess it does not affect odds.. but either way beat odds that way more than a chance.

 

[Thabiguy]

Just a questions.. in your opinion to claim both as pairs same as single guess e.g. 15 sets in each trial: 6 times both right or 13 times at least one right. Is it possible? I guess it does not affect odds.. but either way beat odds that way more than a chance.

I'm sorry, usually I can figure out what you mean, but here I don't understand you at all. Can you ask your question again with fewer errors, or can you rephrase it in some other way?

 

[pavel_do]

I'm sorry, usually I can figure out what you mean, but here I don't understand you at all. Can you ask your question again with fewer errors, or can you rephrase it in some other way?

Sorry ….to make it simple.. do you think if I claim that I will “win” either way…E.g. with 15 runs per trial, I claim that I will get or 6 times or more of both right or 13 times at least one right. The question is stupid.. but do you think it make any difference in odds? If I claim both ways at same time?

 

[Thabiguy]

Sorry ….to make it simple.. do you think if I claim that I will “win” either way…E.g. with 15 runs per trial, I claim that I will get or 6 times or more of both right or 13 times at least one right. The question is stupid.. but do you think it make any difference in odds? If I claim both ways at same time?

Yes, it makes a difference in odds. If feat A has a probability P(A) and feat B has a probability P(B), then feat "A or B" has a probability P(A or B), which is generally greater than both P(A) and P(B). In this particular case, P(A or B) is a nightmare to calculate exactly, but we can at least establish an upper bound: P(A or B) <= P(A) + P(B). The required results in my previous post are calculated so that "A", "B" and "A or B" are all less likely than 1 in 1000, even if 3 trials are allowed.

However, as I said before, I do not recommend claiming "A or B" for your goal. I would recommend choosing either "A" or "B" instead of "A or B", for the sake of simplicity and clarity, because simplicity and clarity is what JREF will be pushing for in the negotiations.

You already want to ask some controversial things from JREF: You want to see the results immediately, not after the whole test (you say this is because of the way your ability works); you want to have 3 trials (you say this is because your performance may be irregular). You may be having a tough time just negotiating those. It's not a good idea to include additional complications that are not absolutely necessary.

 

[pavel_do]

Yes, it makes a difference in odds. If feat A has a probability P(A) and feat B has a probability P(B), then feat "A or B" has a probability P(A or B), which is generally greater than both P(A) and P(B). In this particular case, P(A or B) is a nightmare to calculate exactly, but we can at least establish an upper bound: P(A or B) <= P(A) + P(B). The required results in my previous post are calculated so that "A", "B" and "A or B" are all less likely than 1 in 1000, even if 3 trials are allowed.

However, as I said before, I do not recommend claiming "A or B" for your goal. I would recommend choosing either "A" or "B" instead of "A or B", for the sake of simplicity and clarity, because simplicity and clarity is what JREF will be pushing for in the negotiations.

You already want to ask some controversial things from JREF: You want to see the results immediately, not after the whole test (you say this is because of the way your ability works); you want to have 3 trials (you say this is because your performance may be irregular). You may be having a tough time just negotiating those. It's not a good idea to include additional complications that are not absolutely necessary.

Thank you, I will email propose of the test to JREF today anyway to see what they have to say. So we at least can start negotiation.

Will ask them to review it and start negotiations, just that the final number of photos per trial 10, 20 or 30 (and what shell I claim either pairs r single guesses if both will be not possible) I will figure out by the end of the week, as during thins week I will practice it every day and by evaluating and summing up the results I will be able to figure out what is the best for me.
Re to the immidiat results, that the way I see yesterday I tried one of 10 runs with my friend when I did not know the results till the end, and the result was way too bad.. even by chance u get more..

 

[pavel_do]

Just thought now.. as we all know when there is will there is a way.. so maybe we could make 6 trials.. where in 2 I would concentrate to get 5 pairs exactly ( with 10 sets per trial), in other 2 just s single ones 10 out of 10, and 2 where the results will be revealed after all envelopes named but in this case I doubt I can claim exactly either pairs or single guesses)
But if there would be way to have as Pairs same single guesses even if it has to be 2 and 2. than there would be no need for 30 sets per trial.. as I would do it with 10 knowing that if I don’t get 5 pairs at first than I will concentrate on getting singles at other one.. Cause problem with pairs.. that sometimes I get them right but in different order and I don’t know why and not really helpful.. same even with getting at least 1 correct of 2.. even that happens to me.. so when I have to get 10 out of 10.. and have no right even for 1 mistake or worng order.. that make it a bit tough..

 

[IXP]

Hi guys,

I did the tests with my friend over the week end, 2 times by 3 runs (6 in all) beside that also practiced a few times myself.
Here is the results of my tests with a friend.
+ is exact hit, - wrong, x is in wrong order.

1. + +, - +, x -, + +, + -, - -, + +, - -, - x, + +.
2. + -, + +, + -, - -, + -, x x, + -, + -, + -, - +.
3. + +, + -, + -, + +, - +, + -, x x, - -, x x, - +.

1. + -, + +, - +, x -, x x, + -, - -, - +, - x, + +.
2. - +, + -, + +, + +, - +, - +, x -, + +, + -, + -.
3. + -, + +, - +, x -, - -, + -, - +, - x, + +, + -.

A few times I was able to beat 1000 odds (I think). I am comfortable with this test more or less and I will practice it now and will drop all the rest and the practice with one test will help me to improve it same as the results. I hope JREF will like it unless they have there own to propose.
So I wrote a new proposal for the JREF and planning to send it tomorrow Monday morning. If you have any suggestions and comments. I will be glad to read them
Regards,
Pavel

Very interesting results Pavel.

When you first did this yourself you got these results for 3 runs (counting only the ones indentified correctly in the correct order):

run 1: 15/20
run 2: 15/20
run 3: 14/20

Now, with the help of a friend and reshuffling the pictures to different envelopes each time your get:

run 1: 10/20
run 2: 9/20
run 3: 9/20
run 4: 8/20
run 5: 12/20
run 6: 9/20

So tightening up the protocol dramatically reduced your power. You did not do was well on any of these 6 runs as you did on all of your original 3. Do you see what is happening? You tighten the protocol and your power diminished. The next time you tighten it more, it will fall some more. By the time you have an acceptable protocol for JREF and it is done by independent testers it will be to chance levels.

IXP

 

[pavel_do]

Very interesting results Pavel.

When you first did this yourself you got these results for 3 runs (counting only the ones indentified correctly in the correct order):

run 1: 15/20
run 2: 15/20
run 3: 14/20

Now, with the help of a friend and reshuffling the pictures to different envelopes each time your get:

run 1: 10/20
run 2: 9/20
run 3: 9/20
run 4: 8/20
run 5: 12/20
run 6: 9/20

So tightening up the protocol dramatically reduced your power. You did not do was well on any of these 6 runs as you did on all of your original 3. Do you see what is happening? You tighten the protocol and your power diminished. The next time you tighten it more, it will fall some more. By the time you have an acceptable protocol for JREF and it is done by independent testers it will be to chance levels.

IXP

Thank you.. but you judging it on 1 day results.. I will practice during the week and by the weekend will have results that I can evaluate and see more or lees what is “going on” will post them obviously, and by the end of the week will figure out what is the best number of photos per trial. If it is all seem just a chance.. well.. than good luck to repeat it with same or better results. But as I said to judge things on one test it is not really reasonable. I do have powers that I claim and I will demonstrate it.

 

[IXP]

9 times getting at least 1 correct is not enough if you get 3 tries. I calculate the probabailty at 1 in 500. It is enough if you get 1 try.

IXP

 

[pavel_do]

9 times getting at least 1 correct is not enough if you get 3 tries. I calculate the probabailty at 1 in 500. It is enough if you get 1 try.

IXP

I understood it, but still 1 in 500 odds that is just by chance?
Any way this week ( the results of my tests) will show what is my “limit”.

 

[Startz]

More on odds

In a private message, Pavel has asked me to do some further calculation of odds. I'm posting the calculations here for everyone to see (and to correct if needed).

In each case, the first column gives the minimum number of successes and the remaining columns tell how many times out of 1,000 one would expect to get that many successes or more.

Procedure: Pick two from five correctly and in order. Ten trials. Second column gives the number of wins in one trial. The third column gives the number of wins if one gets three tries.

0| 1000| 1000
1| 401.26306| 785.36124
2| 86.138356| 236.79475
3 | 11.503557| 34.115199
4 | 1.0284979| 3.0823215
5 | 0.063689831| 0.19105733
6 | 0.0027545826| 0.0082637251
7| 8.1983984e-005| 0.00024595193
8| 1.6050781e-006| 4.8152342e-006
9| 1.8652344e-008| 5.5956906e-008
10| 9.765625e-011| 2.9309888e-010

So getting it right 4 times out of 10 just misses the oft-quoted 1 in a 1,000 "rule." Best-of-three-tries requires a 5 out of 10 rule.

Procedure: Pick two from five correctly and in order. 30 trials. Second column gives the number of wins in one trial.

0| 1000
1| 785.36124
2| 446.45792
3| 187.82119
4| 60.771561
5| 15.63551
6| 3.2824856
7| 0.57348899
8| 0.084647494
9| 0.010678057
10| too small to report

Seven wins out of 30 does the job.

Procedure: Pick two from five, getting one right without regard to order. In ten trials the odds are 28/1000 of getting all 10 right and 149/1000 of getting 9 or more right.

Procedure: Pick two from five, getting one right without regard to order. 30 trials allowed.

23 | too many to report
24 | 159.52298
25 | 76.594752
26 | 30.154943
27 | 9.3165673
28 | 2.1131781
29 | 0.31233086
30 | 0.02253934

29 wins out of 30 required.

-Dick Startz

 

[Coveredinbeeees]

I've checked and confirmed your figures Startz. In so doing I discovered that I had made a mistake in my earlier analysis. Good thing I wasn't getting paid eh?

The key mistake in my earlier analysis was my saying that the odds of guessing one card right (with order unimportant) out of 2 drawn from 5 was 55% when, in fact, it is 70%. So all of my earlier data regarding odds for the case where order is not important were wrong.

Sorry for any confusion that may have arisen as a result.

Now, I'll have a go at analysing Pavel's more recent results (although I'll understand if no-one trusts my calculations from here on in).

If you were to class a hit as at least one photo guessed correctly with order unimportant in a run (so anything but -- is a hit) then Pavel scored at least 8 per trial. The odds of guessing at least one photo with order unimportant at least 8 times in 10 runs is 38.3%. The odds of doing at least this well 3 trials in a row (with 10 runs per trial) is 5.62% or about 1:18

If, on the other hand, you were to class a hit as at least one photo correctly identified and in the correct order in a run (so only ++, +- or -+ are hits) then pavel scored at least 6 hits per trial. The odds of guessing at least one photo in the correct order at least 6 times in 10 runs is 9.49%. The odds of doing at least this well 3 trials in a row is 0.0854% or 1:1170

Maybe someone could check those figures.

I know that you performed 6 trials but, since you want to do 3 on the day I thought you might find the odds for 3 trials performed with the consistency that you had yesterday useful.

To the odds given above I'll add,

If you were to make your protocol to something more along the lines of "For 3 trials with 10 runs per trial I will score at least 6 hits on each trial. A hit is defined as a run in which I correctly identify at least one of the photos from the two envelopes I select and which of the two envelopes it is in. If I score fewer than 6 hits on any trial the test will be considered a failure" The odds of accomplishing this by chance are about 1:1170 (someone please correct me if i'm wrong)

 

[pavel_do]

I've checked and confirmed your figures Startz. In so doing I discovered that I had made a mistake in my earlier analysis. Good thing I wasn't getting paid eh?

The key mistake in my earlier analysis was my saying that the odds of guessing one card right (with order unimportant) out of 2 drawn from 5 was 55% when, in fact, it is 70%. So all of my earlier data regarding odds for the case where order is not important were wrong.

Sorry for any confusion that may have arisen as a result.

Now, I'll have a go at analysing Pavel's more recent results (although I'll understand if no-one trusts my calculations from here on in).

If you were to class a hit as at least one photo guessed correctly with order unimportant in a run (so anything but -- is a hit) then Pavel scored at least 8 per trial. The odds of guessing at least one photo with order unimportant at least 8 times in 10 runs is 38.3%. The odds of doing at least this well 3 trials in a row (with 10 runs per trial) is 5.62% or about 1:18

If, on the other hand, you were to class a hit as at least one photo correctly identified and in the correct order in a run (so only ++, +- or -+ are hits) then pavel scored at least 6 hits per trial. The odds of guessing at least one photo in the correct order at least 6 times in 10 runs is 9.49%. The odds of doing at least this well 3 trials in a row is 0.0854% or 1:1170

Maybe someone could check those figures.

I know that you performed 6 trials but, since you want to do 3 on the day I thought you might find the odds for 3 trials performed with the consistency that you had yesterday useful.

To the odds given above I'll add,

If you were to make your protocol to something more along the lines of "For 3 trials with 10 runs per trial I will score at least 6 hits on each trial. A hit is defined as a run in which I correctly identify at least one of the photos from the two envelopes I select and which of the two envelopes it is in. If I score fewer than 6 hits on any trial the test will be considered a failure" The odds of accomplishing this by chance are about 1:1170 (someone please correct me if i'm wrong)

Thank you. Seems like I decided what I am going to do.. and what I will claim.. will wait till the end of the week to sum up my results but still I guess I know already.. hopefully JREF would be satisfied with it too… Yesterday I sent them my propose and just asked to start negotiations if it is possible on the conditions.. despite the numbers for now as by the end of the week I will decide how many photos per run and how many trials I would go for. So discussing at least the conditions for now, would save allot of time and could make testing possible sooner.

Have a nice day guys, I will have my coffe and then my Photos

 

[Marcus]

I understood it, but still 1 in 500 odds that is just by chance?
Any way this week ( the results of my tests) will show what is my “limit”.

If you were to beat 500 to 1 odds in a controled JREF test, you might not pass but you would certainily arouse some interest.

The results of your self tests are interesting, but I think you understand now why they must be taken with a grain of salt.

 

[pavel_do]

If you were to beat 500 to 1 odds in a controled JREF test, you might not pass but you would certainily arouse some interest.

The results of your self tests are interesting, but I think you understand now why they must be taken with a grain of salt.

Yes I do understand

 

[Akhenaten]

This is what sets you apart from many people claiming extraordinary abilities, Pavel. The way you listen to people, take their advice and adjust your testing, all with openness and good humour.

As Marcus said, even if you don't beat the MDC, any result significantly better than chance will still arouse great interest, and may present future opportunities for you.

In the meantime, I remain impressed with your approach to the Challenge, no matter what the result.

Cheers,

Dave

 

[hopfen]

Small point, but if I were as sure as you seem to be, I don't think I would be using the word "guess" in my proposal.