Anyone heard about this dude? (pavelprorok on youtube)

[Startz]

Hi Dick,

Success in what sense? As a test in general? Well as I claimed a minimum of 65%... than getting 6/7 correct out of 10 runs as cant get 6 and a half ) would be a success for me I guess... or you mean what do I consider as a “hit”? Well in this case it is obvious I need to get the photos right.. at least one out of 2 that was chosen out of 10
... snip

Pavel,

If I've done the calculation right--and I'm not sure if I have--getting 6 or more right is better than 1 in 100, and getting 7 or more right is better than 1 in 1,000. (if you want three separate shots at it, as you've sometimes mentioned in the past, you'd need 8 or more right.)

I'm pretty sure that if you can reliably hit 9, as you did in practice, then the statistical design should satisfy JREF.

-Dick

 

[pavel_do]

Pavel,

If I've done the calculation right--and I'm not sure if I have--getting 6 or more right is better than 1 in 100, and getting 7 or more right is better than 1 in 1,000. (if you want three separate shots at it, as you've sometimes mentioned in the past, you'd need 8 or more right.)

I'm pretty sure that if you can reliably hit 9, as you did in practice, then the statistical design should satisfy JREF.

-Dick

Thank you for the calculations.
Yes I would like to have a minimum 3 trials… though even better 5 . But not one as with a few trials I would be more comfortable and confident and that would help me allot.

 

[Thabiguy]

There are 90 possible ways to pull 2 envelopes out of 10.

If order does not matter, then out of these 90 ways, 2 combinations are an exact hit (both pictures correct), 32 are a partial hit (one picture correct) and 56 are a miss (no pictures correct).

If order does matter, then out of these 90 ways, 1 combination is an exact hit (both pictures correct), 16 combinations are a partial hit (one picture correct) and 73 are a miss (no pictures correct).

Therefore, if order does not matter, probability of getting at least one picture correct is 34/90, or about 1 in 2.65. This feat would have to be repeated at least 9 times out of 10 to beat the 1:1000 odds.

If order does matter, probability of getting at least one picture correct is 17/90, or about 1 in 5.29. This feat would have to be repeated at least 7 times out of 10 to beat the 1:1000 odds.

 

[Marcus]

If I understand you correctly, Pavel, you have 10 envelopes, with a different picture in each envelope.

If you make a single identification, and are correct, the odds are 10 to 1 against that.
If you make 2 identifications, and both are correct, the odds are 100 to 1 against that. If you make 3 identifications, and all 3 are correct, the odds are 1000 to 1 against that. Assuming you don't open any envelopes until you are finished. I think that is the way it works, anyway. I'm not claiming any expertise.

Note that we are not talking about making 10 identifications and getting 3 correct, we are talking about making 3 and being correct on all 3.

 

[pavel_do]

Note that we are not talking about making 10 identifications and getting 3 correct, we are talking about making 3 and being correct on all 3.

I understand that. I don’t name 10 and then match the one I was sure the most.. I pick 2 that I am sure the most of out of 10 and than open ONlY the ones I named.

 

[Paul2]

There are 90 possible ways to pull 2 envelopes out of 10.

Why the 2 out of 20?. He guessed twenty times as to whether a picture was one out of ten, and got 13 correct. It doesn't matter that the runs were given to him in sets of ten.

We need a statistician, stat.

 

[Thabiguy]

Why the 2 out of 20?.

2 out of 10. He guessed 2 pictures out of 10 every time.

He guessed twenty times as to whether a picture was one out of ten, and got 13 correct.

No, he did not. He pulled out 2 out of each 10 and guessed at those 2. He got both correct 4 times, one correct 5 times and no correct 1 time.

It doesn't matter that the runs were given to him in sets of ten.

Guessing 2 pictures out of 10, ten times, is not the same as guessing 1 picture out of 10, twenty times. In the former case, the two pictures guessed together can never be the same, while in the latter case, any two pictures can be the same.

 

[Startz]

There are 90 possible ways to pull 2 envelopes out of 10.

If order does not matter, then out of these 90 ways, 2 combinations are an exact hit (both pictures correct), 32 are a partial hit (one picture correct) and 56 are a miss (no pictures correct).

If order does matter, then out of these 90 ways, 1 combination is an exact hit (both pictures correct), 16 combinations are a partial hit (one picture correct) and 73 are a miss (no pictures correct).

Therefore, if order does not matter, probability of getting at least one picture correct is 34/90, or about 1 in 2.65. This feat would have to be repeated at least 9 times out of 10 to beat the 1:1000 odds.

If order does matter, probability of getting at least one picture correct is 17/90, or about 1 in 5.29. This feat would have to be repeated at least 7 times out of 10 to beat the 1:1000 odds.

Why the 2 out of 20?. He guessed twenty times as to whether a picture was one out of ten, and got 13 correct. It doesn't matter that the runs were given to him in sets of ten.

We need a statistician, stat.

I'm pretty sure that Thabiguy's analysis is statistically correct.

 

[Paul2]

Let's be clear on what happened:

I made 10 runs, with 5 minute brake between each 5. Out of each 10 I pulled out 2 that I was most sure off. Here is the results. Out of 10 runs, 4 times I got both 2 correct, 5 times only 1 out of 2 correct and 1 time both of them was wrong. . . .

As I re-read this in order to continue the discussion, I find it now not clear exactly what happened. What constitutes a "run?" Is a run one time or ten times that Pavel is asked, "What picture is in this envelope?"

Pavel, please clarify:

How many times, in total, did you make a guess?

How many times, in total, did you have an opportunity to make a guess but didn't (the other eight times out of ten in which you did not make a guess)? This number is actually irrelevant, but may help to clarify the relevant numbers.

For each guess, Pavel had a one in ten chance of being correct by luck. The problem is to calculate the odds of being right by luck for however many times he made guesses.

Whatever the number of opportunities Pavel had to make a guess and didn't is irrelevant; the only thing that matters is how many guesses he made, and the odds of him being correct by chance for any one guess. It's as if those opportunities to make a guess, and for which didn't, he didn't never existed.

 

[pavel_do]

Let's be clear on what happened:


As I re-read this in order to continue the discussion, I find it now not clear exactly what happened. What constitutes a "run?" Is a run one time or ten times that Pavel is asked, "What picture is in this envelope?"

Pavel, please clarify:

How many times, in total, did you make a guess?

How many times, in total, did you have an opportunity to make a guess but didn't (the other eight times out of ten in which you did not make a guess)? This number is actually irrelevant, but may help to clarify the relevant numbers.

For each guess, Pavel had a one in ten chance of being correct by luck. The problem is to calculate the odds of being right by luck for however many times he made guesses.

Whatever the number of opportunities Pavel had to make a guess and didn't is irrelevant; the only thing that matters is how many guesses he made, and the odds of him being correct by chance for any one guess. It's as if those opportunities to make a guess, and for which didn't, he didn't never existed.

The test was like this.

I have set of 10 deferent photos, I go through it one by one, and trying to “see” in my way what is the photo inside. As soon as I had a bright vision and I am sure of I put the envelope a side.. despite how many envelopes out of 10 I put a side.. out of them still I chouse only 2 that I am sure the most.. (though some times I put a side up to 5 and then went through them ones again and chouse 2). So when 2 out of 10 finale chosen and named them, after that I open them ONLY those 2 to see if I was right or wrong.

Same procedure was done 10 times ( with the same set of 10, obviously every time reshuffled).

Every time is deferent how many envelopes I go through, some times first ones are bright and I put them a side straight a way and name them. Still doesn’t matter how many I go through in the end only two I named and opened to reveal the results.

 

[Crundy]

As someone else mentioned, you would also need to pick two envelopes and say "Envelope 1 is the ship, Envelope 2 is the bird", rather than "These two envelopes contain the bird and the ship" so we can get the mismatched order odds eliminated.

Is it possible for you to pick three and get all three right? If you could do this three times out of 5 then that's probably good enough odds.

 

[Paul2]

So Pavel made 20 guesses. Each guess had a one in ten chance of being correct by luck. He got 13 of the correct.

Can someone calculate these odds? Let's assume that he addressed Crundy's point and identifies each envelope specifically, eliminating the mismatched order odds.

 

[pavel_do]

As someone else mentioned, you would also need to pick two envelopes and say "Envelope 1 is the ship, Envelope 2 is the bird", rather than "These two envelopes contain the bird and the ship" so we can get the mismatched order odds eliminated.

Is it possible for you to pick three and get all three right? If you could do this three times out of 5 then that's probably good enough odds.

I can try with 3 out of 10. Will do that later today and will post the results, also I will make one trial same as yesterday with 2 out of 10.

 

[IXP]

So Pavel made 20 guesses. Each guess had a one in ten chance of being correct by luck. He got 13 of the correct.

Can someone calculate these odds? Let's assume that he addressed Crundy's point and identifies each envelope specifically, eliminating the mismatched order odds.

Odds of getting 13 or more correct out of 20 with probability for each guess 1/10 is 3.9e-9 or about 250,000,000:1.

So we can rule out the possibility that Pavel did this by chance. If he can do anywhere near this well in a JREF test he gets $1M.

So how did he do this? There are several possibiliies.

1. He is lying.
2. He got the information in some non-paranormal way, without knowing it.
3. He is psychic.
4. We are not understaning correctly exactly what he did.

IXP

 

[chillzero]

So Pavel made 20 guesses. Each guess had a one in ten chance of being correct by luck. He got 13 of the correct.

Can someone calculate these odds? Let's assume that he addressed Crundy's point and identifies each envelope specifically, eliminating the mismatched order odds.

Where did you get 13 from? I thought he decided on 2 only.

ETA: and out of 10, not 20

 

[IXP]

Where did you get 13 from? I thought he decided on 2 only.

ETA: and out of 10, not 20

Out of 10 runs, 4 times I got both 2 correct, 5 times only 1 out of 2

10 runs, each one with 2 guesses so 20 guesses.

4 x 2 + 5 x 1 = 13 correct.

IXP

 

[IXP]

A couple of questions for Pavel.

Who put the pictures in the envelopes?
If someone other than you, was this done out of your view?
Were different envelopes used for each run?
Were the envelopes the kind intended to obscure the contents?

IXP

 

[pavel_do]

A couple of questions for Pavel.

Who put the pictures in the envelopes?

If someone other than you, was this done out of your view?

Were different envelopes used for each run?

Were the envelopes the kind intended to obscure the contents?

IXP

I put them my self ( when there is chance my friend does it for me) in white first and than to manila. So each photo is in double envelopes just as on youtube videos.

The envelopes all the same, with no bands or signs or anything, beside I cover my eyes with mask each time. If you ask because there is a chance that I mark envelopes and cheat.. well I see no point in cheating any ways and waist my time, if I would than I would say I got 100 out of 100 and would not even apply for Challenge.

I must say that when photo in one envelope I see it brighter.. but placing them in double envelopes to avoid suspect that I see through the envelopes that can be transparent.

 

[IXP]

I am not suggesting that you are deliberately cheating, only that there might be some way to distiguish the envelopes without psychic powers. If there is any difference in the envelopes and you have put the pictures inside yourself, then you have proved nothing.

I would suggest that you do this test again with a friend helping you. You can use the same envelopes, but each run, your friend must remove the photos from the envelopes and put them back in randomly selected envelopes in another room where you are not present. See if this reduces your astounding results any. This would be much close to the way it will be done in a JREF test.

IXP

 

[pavel_do]

I am not suggesting that you are deliberately cheating, only that there might be some way to distiguish the envelopes without psychic powers. If there is any difference in the envelopes and you have put the pictures inside yourself, then you have proved nothing.

I would suggest that you do this test again with a friend helping you. You can use the same envelopes, but each run, your friend must remove the photos from the envelopes and put them back in randomly selected envelopes in another room where you are not present. See if this reduces your astounding results any. This would be much close to the way it will be done in a JREF test.

IXP

I understand all that you saying and I know that test with JREF will be sort of different. But I am doing my best each time to make it as close as possible… so all the envelopes are the same I am always careful open them to make sure no fold or creases would appear, for each run I reshuffle them a few times etc. So when I have 10 envelopes in front of me in a pile.. I have no idea, which is which photo. I will practice now and will post the results of 10 runs. And over the weekend will do the same with my friend.