Anyone heard about this dude? (pavelprorok on youtube)

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That is exactly what I expected, you are using the same pictures in the same envelopes each time. This is no good. You could easily be determining the photo by some subtle differences in the envelopes. You might even be doing this unconsciously.

You have to either use new envelopes each time (this could get expensive!) or at least, take out all the photos and replace them randomly with lot's of shuffling so you don't know which photo goes into which envelope, or better, have a friend shuffle them and re-place them in the envelopes out of your sight.

IXP

thank you.
I understand that... I will do a few trials with my friend’s help over the weekend, will ask him change envelopes after each trial (will use 20 different ones as I have them from the tests with 20 sets, so every time random 5 envelopes will be used to place 5 photos for each run)
 
thank you.
I understand that... I will do a few trials with my friend’s help over the weekend, will ask him change envelopes after each trial (will use 20 different ones as I have them from the tests with 20 sets, so every time random 5 envelopes will be used to place 5 photos for each run)
Great! This will make the test a lot more credible.

IXP
 
In the real test, it seems likely that you will not be allowed to handle the envelopes between tests. I would imagine that the JREF would insist that they are taken away and shuffled out of your sight, and replaced in fresh envelopes.

Would that be a problem?
 
In the real test, it seems likely that you will not be allowed to handle the envelopes between tests. I would imagine that the JREF would insist that they are taken away and shuffled out of your sight, and replaced in fresh envelopes.

Would that be a problem?

Well, I think all the sets will be prepared upfront and after each run will be replaced.. New will be passed to me and the currently used will be removed off the table. That will save allot of time.
 
odds on recent tests

Hi Pavel,

Thanks for answering my earlier question. It is good to know that you do not drop runs which are not going well.

IXP's point made above about using different envelopes each time is a good one. Opening and closing the envelope then reusing it, particularly when there are only 5, could explain your accuracy.

I'm afraid I disagree with IXP regarding the odds of getting your results. They are very good and better than your 1:1000 goal but not quite so astronomical.

Thabiguy gave a similar analysis to this for your earlier protocol.

There are 20 possible outcomes when selecting 2 cards from a set of 5.

If order is not important then the odds of selecting two particular cards (++ or XX) is 2:20.

If order is not important then the odds of selecting two cards but only one of them being "right" (+- or -+ or X- or -X) is 9:20

So the odds of getting at least one card right is 11:20 or 55% if order is not important.

In your most recent test neglecting order you were able to identify at least one card in each pair (++, XX, -+, +-, -X or X-) at least 9 times out of 10 over the course of 3 trials. The odds of getting at least 9 out of 10 in a trial with a 55% chance of a hit on each run is 2.33%. The odds of achieving this 3 trials in a row is 0.0000126%. Even achieving it two times in a row would beat your goal of 1:1000 odds.

If we consider order to be important then the odds of an exact hit (++) is 1:20 on each run. The odds of an inexact hit (+- or -+) is 6:20. So the chance of getting at least one card correct and in the right order (++ or +- or -+) is 7:20 or 35%

If we judge your runs by this standard then, in your most recent test you were able to identify at least one card in each pair (++ or +- or -+) at least 8 times out of 10 over the course of 3 trials. The odds of getting at least 8 out of 10 in a trial with a 35% chance of a hit on each run is 0.48%. To achieve this 3 trials in a row would have odds of 1 in ten million. Even achieving it for two runs would beat 1:43402 odds.

It looks like you have found a protocol which can be performed in a short time and is highly unlikely to be achieved by chance alone.

Do continue to take the advice of other forumites when it comes to eliminating all mundane clues for yourself within the protocol. I would love to see this episode reach a satisfying conclusion.

I made use of an on-line binomial calculator for some of the analysis above.
 
thank you

Hi Pavel,

Thanks for answering my earlier question. It is good to know that you do not drop runs which are not going well.

IXP's point made above about using different envelopes each time is a good one. Opening and closing the envelope then reusing it, particularly when there are only 5, could explain your accuracy.

I'm afraid I disagree with IXP regarding the odds of getting your results. They are very good and better than your 1:1000 goal but not quite so astronomical.

Thabiguy gave a similar analysis to this for your earlier protocol.

There are 20 possible outcomes when selecting 2 cards from a set of 5.

If order is not important then the odds of selecting two particular cards (++ or XX) is 2:20.

If order is not important then the odds of selecting two cards but only one of them being "right" (+- or -+ or X- or -X) is 9:20

So the odds of getting at least one card right is 11:20 or 55% if order is not important.

In your most recent test neglecting order you were able to identify at least one card in each pair (++, XX, -+, +-, -X or X-) at least 9 times out of 10 over the course of 3 trials. The odds of getting at least 9 out of 10 in a trial with a 55% chance of a hit on each run is 2.33%. The odds of achieving this 3 trials in a row is 0.0000126%. Even achieving it two times in a row would beat your goal of 1:1000 odds.

If we consider order to be important then the odds of an exact hit (++) is 1:20 on each run. The odds of an inexact hit (+- or -+) is 6:20. So the chance of getting at least one card correct and in the right order (++ or +- or -+) is 7:20 or 35%

If we judge your runs by this standard then, in your most recent test you were able to identify at least one card in each pair (++ or +- or -+) at least 8 times out of 10 over the course of 3 trials. The odds of getting at least 8 out of 10 in a trial with a 35% chance of a hit on each run is 0.48%. To achieve this 3 trials in a row would have odds of 1 in ten million. Even achieving it for two runs would beat 1:43402 odds.

It looks like you have found a protocol which can be performed in a short time and is highly unlikely to be achieved by chance alone.

Do continue to take the advice of other forumites when it comes to eliminating all mundane clues for yourself within the protocol. I would love to see this episode reach a satisfying conclusion.

I made use of an on-line binomial calculator for some of the analysis above.


Thank you very much for the great answer, calculations and explanations. It does seem to me that this test would be the best out of all considered and tried by me lately hopefully the JREF would like it too. I will work over the weekend with it, make some trials with the controls and help of my friend. Will prepare the protocol and will post it afterwards together with the results of the trials that I will have chance to do over the weekend. And if it all be fine I will send new protocol propose to the JREF on Monday morning, as hopefully we could start negotiating it and come to the agreement in order to have a test done in some weeks and while I am in US.
 
I noticed a mistake in my earlier post.

If we judge your runs by this standard then, in your most recent test you were able to identify at least one card in each pair (++ or +- or -+) at least 8 times out of 10 over the course of 3 trials. The odds of getting at least 8 out of 10 in a trial with a 35% chance of a hit on each run is 0.48%. To achieve this 3 trials in a row would have odds of 1 in ten million. Even achieving it for two runs would beat 1:43402 odds.

The last sentence should read "To achieve this 3 trials in a row would have odds of 1 in ten million. Even achieving it for two trials would beat 1:43402 odds."

It was probably clear from the context but it never hurts to be certain.
 
Thanks for all of your answers Pavel. Describing what you see must be a bit like describing a colour, so I can understand that it's not easy :)

I think you chose well in deciding to test your picture reading ability, rather than reading people. The latter would obviously be a more useful talent, but coming up with a self-evident test would be a nightmare.

This latest protocol is showing a lot of promise though, so best of luck with your practice runs.


Cheers,

Dave



PS The description of your readings of the University people reminded me of Deanna Troi, the Betazoid from Star Trek. If the MDC doesn't work out I reckon you'd make a fine Starship Counselor :) Empathy is a power I have no trouble believing in.
 
weekend practise.

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
 
new test propose.

Propose of the protocol to test my claiming powers.
Based on the testing with 10 sets of photos (by 5 photo in each set.)​

Without looking in any possible- physical ways in side of the sealed envelope, just by holding it between my hands, I will be able to identify the photo that will be inside and will name it.
In order to be able to demonstrate such abilities I need to be able to perform a minimum of three trials.
That would give me enough time to demonstrate it. Also having 3 trials significantly reduce stress level for me in comparence to 1 and only trial for the testing of claimed by me abilities.
During 3 trials with 10 sets in each trial, in one of them I will get at least 9 correct out of 20 photos to which I will offer my guess. If I get less than 9 correct in any of 3 trials, I accept that I was not able to perform my abilities successfully under/according the conditions and the protocol that was agreed before the test by both sides such as Myself and JREF and that the test results are negative. With 9 or more correct photos in any of three trials, the test result is positive.

Materials required:

Materials for each set of 5 photos: 5 highly distinguishable photographs, with a unique name for the image. 5 identical small envelopes. 5 identical Manila envelopes.
50 photographs per trial 3.5 x 5 inches printed from the files that I will assemble in advance. (150 for all 3 trial)

50 letter sized envelopes ( 150 in all)

50 larger manila envelopes (150 in all)

Paper sheets with the list of the photos that will be used in sets.

Drawing book with plain pages. (In order to be able if necessary to draw the visions that I will see during the test)

Environment stipulations:

No talking or any kind of distraction with light or sound or any other ways of interruption during the test. No photography during the test will be allowed as the flash or the sound produced by the camera can be distracting.

No trickery, or cheating from either Pavel or JREF or other parties present in room during the test. The environment should be established to give me the best chance of success and JREF the assurance that my abilities are real.
No substance with a strong odor will be allowed in the room same, include people who might use perfume that has strong odor and that can cause distraction during the concentration and a headache due to the strong smell.

It is important that in the room where the test will be held, there should be NO photos or pictures of any kind on the walls or anywhere near me and visible to me.

What Will Constitute a Successful Test:

If I am able to identify 9 photos out of 20 sealed envelopes (drawn from 10 sets of 5, by two from each set) just by holding it between my hands, the test will be successful.

In order to beat the 1/1000 odds, that is necessary according the JREF expectations to pass the preliminary test I should be able to perform the results of either one of two feats:

1. Guess both exactly right at least 5 times out of 10.

2. Guess at least one right at least 9 times out of 10.


List of the photographs. (I might still change some of the photos)

1. The cross
2. Ship “Titanic”
3. Egyptian pyramids
4. Clock
5. Landscape with sunset (in color).


Preparation:

By a third party in a separate room all the pictures that are used in the testing will be placed into letter sized envelopes and then place the letter sized envelopes into larger manila envelopes.

For each run there, shell be prepared separate set of envelopes, in order to save time in replacing photos in the new envelopes and reshuffling them.
If there will be agreed 3 trials by 10 sets each, in this case 3 identical sets of 10 sets (of five agreed photos in each) have to be prepared in advance.

For each run:

1. Each of the five photos is placed in a different small envelope and sealed.
2. These 5 small envelopes are shuffled sufficiently in order to lose track of which is which.
3. Each of these small envelopes is placed in a different manila envelope and sealed.
4. The 5 manila envelopes are shuffled sufficiently in order to lose track of which is which.

Ten sets according to conditions described above have to be prepared for each trial.
Three sets of 10 sets of five, shell be prepared for whole test.

The envelopes will be checked for marks, folds, creases etc by JREF representatives before being brought into the test room.


Test settings:

1. The photos will be placed in envelopes in separate room by a third party with the supervision of a JREF representatives who will make sure that there will be no interference in the process of preparation of the photo pairs for the test.

2. After I am situated, ready, and comfortable, I will cover my eyes with a mask and my ears with the plugs and will have 10 minutes to get ready and relax.
After that, the testing shell begin, and one random envelope from the first/any set shell be handed to me.

3. I would need up to 2 minutes to hold each envelope. (Though usually it take about 40-45 minutes for all 10 runs including brake of 5 minutes between every 5 sets)

The test.

The envelopes arranged by the sets, will be prepared to be used in test.

The person will give me the envelopes of the current set of 5, with no words or any other sound signs that would in any way suggest or define the order of the envelopes of the set that will be handed to me.

I will hold each of the five sealed manila envelopes in turn, and choose the two that I get the strongest image on. (Each envelope I might hold up to 2 minutes).

I will take the mask off and offer my guess for the two photos that will be chosen by me.
I will offer my guess by writing the name of the image on each of these two envelopes or draw the image on top of each envelope and name it a laud.

After the writing/ drawings will be finished and the answers was announced, I will hand the envelopes back to the third party. The third party will open the envelopes in front of everyone including me* and will reveal the photo that was inside. Then the result will be recorded as either correct or incorrect by the JREF or any other party who will be in the room.

The same procedure will be done with all 10 sets of photos in current trial.

And during all trials that will be agreed, until the last envelope will be identified by me and my “guess” will be announced.

I would need 5 minutes brake between each 5 photo sets.
I would need 10/15 minutes brake between every trial.

After all the tests-runs will be complete, the JREF representative will announce the results for the Preliminary Test as either Positive or Negative.

Remarks:

* I need to know the result of my “guess” immediately after I name the photos. I prefer this because, when I hold the envelopes I am trying to foresee what will I actually see when the photo will be revealed. By asking myself – “What picture will I see when the envelope will be open”. For this reason, it is important for me and necessary to see the actual photo immediately after I offer my “guesses” during the test.


Notes:

1. No talking (or signaling!) during the test from any person who will be present in room.

2. No text messaging or phone calls allowed, as the sms and call alerts can distract my concentration and affect the result of my answer. (The smsing during the test is not permitted either, as the sound of the constantly clicking buttons can be very irritating and distracting). For that reason, all mobile phones in the room, must be switched to a silent mode to eliminate interruptions of the testing. If anyone needs to make calls or sms, it can be done during the brake between runs

3. Any person that is present at the testing may raise his or her hand during the demonstration to inform the others of suspicious behavior of anyone in the room. However, no shouting or interruptions will be allowed during the process of identifying the photo. If there are any doubts or suspicions, the person shall raise his/her hand silently and inform the others after I have finished with identification of the current photo and announced my answer.
 
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Propose of the protocol to test my claiming powers.
Based on the testing with 10 sets of photos (by 5 photo in each set.)​


In order to beat the 1/1000 odds, that is necessary according the JREF expectations to pass the preliminary test I should be able to perform the results of either one of two feats:

1. Guess both exactly right at least 5 times out of 10.

2. Guess at least one right at least 9 times out of 10.

Pavel:

If I have done the math right, the second feat is problematic. The probability of randomly guessing at least one of five photos correctly in two tries is 0.7. I think the chance of doing this at least 9 times out of 10 is about fifteen percent, which isn't going to fly.

But perhaps someone could check my math.
 
Pavel:

If I have done the math right, the second feat is problematic. The probability of randomly guessing at least one of five photos correctly in two tries is 0.7. I think the chance of doing this at least 9 times out of 10 is about fifteen percent, which isn't going to fly.

But perhaps someone could check my math.


Ok, maybe some one can check it.. as I was going (and going to) sent this propose to JREF, as still seem like good test.
What does your calculations say about my results, is it possible to calculate odds for each trial?
Again, if 9 out of 10 don’t beat the odds needed, how many would beat then? Cause we could make say run of 15 instead of 10 sets.
 
Ok, maybe some one can check it.. as I was going (and going to) sent this propose to JREF, as still seem like good test.
What does your calculations say about my results, is it possible to calculate odds for each trial?
Again, if 9 out of 10 don’t beat the odds needed, how many would beat then? Cause we could make say run of 15 instead of 10 sets.

If you stick to the first feat, correctly identifying both photos in order, then the proposal is fine. Just identifying one of the two is too easy, even with 15 in a run.
 
If you stick to the first feat, correctly identifying both photos in order, then the proposal is fine. Just identifying one of the two is too easy, even with 15 in a run.

too easy? we can make 20 instead of 10 or 15. How many should I get correct to beat 1.000 odds (guessing at least 1 of 2) if 20 runs per trial done? well I think than easier if I go with test where I pick 1 out of 3 ( not 5) and have to get 19 out of 30 as that beat 1.000 if it is correct.
I am sure I can do that if I do it this new way where I pick 1 out of 3 and name it (instead of when 1 given to me in random as I proposed before)

I will practice to get 5 exact pairs.. to beat 1.000 odds but sometimes I get them in a wrong order and I don’t know why and how to eliminate it. But to be honest I would prefer not to claim only 5 pairs exactly cause I might get 4 pairs and say 5 by 1 each correct and even 1 time both to be wrong.. and that will mean I am failed. Cause I would need to get 5 not less… when with getting what ever the number of 1 out of 2 I am sure I will pass it in 1 of 3..even if it should be 20 per run instead of 10.

any suggestions? :)
 
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too easy? we can make 20 instead of 10 or 15. How many should I get correct to beat 1.000 odds (guessing at least 1 of 2) if 20 runs per trial done? well I think than easier if I go with test where I pick 1 out of 3 ( not 5) and have to get 19 out of 30 as that beat 1.000 if it is correct.
I am sure I can do that if I do it this new way where I pick 1 out of 3 and name it (instead of when 1 given to me in random as I proposed before)

I will practice to get 5 exact pairs.. to beat 1.000 odds but sometimes I get them in a wrong order and I don’t know why and how to eliminate it. But to be honest I would prefer not to claim only 5 pairs exactly cause I might get 4 pairs and say 5 by 1 each correct and even 1 time both to be wrong.. and that will mean I am failed. Cause I would need to get 5 not less… when with getting what ever the number of 1 out of 2 I am sure I will pass it in 1 of 3..even if it should be 20 per run instead of 10.

any suggestions? :)


You'd have to get all 20 right.

If you can get both right, 4 out of 10 might do.

And only one try, not three.
 
If I have done the math right, the second feat is problematic. The probability of randomly guessing at least one of five photos correctly in two tries is 0.7. I think the chance of doing this at least 9 times out of 10 is about fifteen percent, which isn't going to fly.

But perhaps someone could check my math.

The probability of guessing at least one correct is 0.7 only when order is ignored (i.e. if "AB" vs. "BE" counts as one correct).

If order is important (i.e. the guess is only correct if it matches the envelope it is written on, not the other evnelope), then the probability is 0.35, and the likelihood of doing it 9 times out of 10 is about 1 in 1852, which is fine.

But it should be clarified that order is indeed important.

One other thing:

If I am able to identify 9 photos out of 20 sealed envelopes (drawn from 10 sets of 5, by two from each set) just by holding it between my hands, the test will be successful.

No, this is wrong. To get 9 out of 20 is much more likely than to get 9 times at least one correct. The exact probability is difficult to calculate, but it's somewhere between 1 in 24 and 1 in 100.

This is because getting 9 out of 20 includes many cases when you don't get at least one correct 9 times. For example, ++/++/++/+-/-+/+-/--/--/--/-- is 9 right out of 20, but it has only 6 times at least one correct (not enough), and only 3 times both correct (not enough).

So you should drop this incorrect formulation from the proposal.

Furthemore, for the sake of clarity, consider choosing only one of the two feats (either both right 5 times, or at least one right 9 times) as your goal. It would make the evaluation and odds calculation more straightforward, and would also make the goal clearer for you.
 
The probability of guessing at least one correct is 0.7 only when order is ignored (i.e. if "AB" vs. "BE" counts as one correct).

If order is important (i.e. the guess is only correct if it matches the envelope it is written on, not the other evnelope), then the probability is 0.35, and the likelihood of doing it 9 times out of 10 is about 1 in 1852, which is fine.

But it should be clarified that order is indeed important.

One other thing:



No, this is wrong. To get 9 out of 20 is much more likely than to get 9 times at least one correct. The exact probability is difficult to calculate, but it's somewhere between 1 in 24 and 1 in 100.

This is because getting 9 out of 20 includes many cases when you don't get at least one correct 9 times. For example, ++/++/++/+-/-+/+-/--/--/--/-- is 9 right out of 20, but it has only 6 times at least one correct (not enough), and only 3 times both correct (not enough).

So you should drop this incorrect formulation from the proposal.

Furthemore, for the sake of clarity, consider choosing only one of the two feats (either both right 5 times, or at least one right 9 times) as your goal. It would make the evaluation and odds calculation more straightforward, and would also make the goal clearer for you.

Pavel:

Thabiguy makes several good points here, both about statistics and about clarity.
 
The probability of guessing at least one correct is 0.7 only when order is ignored (i.e. if "AB" vs. "BE" counts as one correct).

If order is important (i.e. the guess is only correct if it matches the envelope it is written on, not the other evnelope), then the probability is 0.35, and the likelihood of doing it 9 times out of 10 is about 1 in 1852, which is fine.

But it should be clarified that order is indeed important.

One other thing:



No, this is wrong. To get 9 out of 20 is much more likely than to get 9 times at least one correct. The exact probability is difficult to calculate, but it's somewhere between 1 in 24 and 1 in 100.
Thank you for correction will fix it but I did mean 9 times at least one correct out of 10 runs with 20 photos in all.

I know it is great to claime 5 pairs as make all simple, but what if I get 4 pairs and then the rest will be at least 1 correct… I will fail it. so it is easier to claim 20 out of 30 (where 30 sets by 3 photo each and I pick 1 that I am sure the most) cause as I said if I will be picking one my self as I do with 5, I am sure I am able to get 20 out of 30 (though need to try 3 trials to see the results, but still like this I am not “pressed” by pairs)
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? Why cant I use then both like if during the run I get eather 5 pairs.. or 9 times out of 10 at least one EXACTLY named in both cases it beat the odds? But I am on the safe side and not pressed by the fact of pairs.. cause I can get either way or both in one.
 
Still think that 2 out of 5 is a good test, just need to figure out best number per trail, like Starz calculated, 6 exactly named pairs out of 15 (instead of 10) would beat the odds.. sound reasonable.. better that 5 out of 10.
Just in case, how many times at least one exactly named out of 2 needed to beat 1000 with 15 sets, and same with set of 20. how many exact pairs I need and how many one of two?
Cause looking at my results with my friend and I will practice more I am sure I can get 6 out of 15.. or what is needed for 20. Though the practice can make it stable 5 out of 10.. as I did on my own and 9 out of 1 by one..
The point is I would prefer to be with both as exact pairs same with 1 out of 2 exactly named.
 
Any one can suggest any variations?
Cause looking at my results and if it is impossible to go for Both… as pairs same for “single” guesses, than it is still better to go just for 20 out of 30 where only 1 out of 3 pulled out.. or still with 2 out of 5.. but either I pull only 1 out of 5 and not 2, or I claim just single “hits” and if I get pairs than it will be just more evidence.
Would appreciate your suggestions.
 
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