The Girl With X-Ray Eyes

http://www.dream-detective.com latest on CSICOP

Actually, I said that. (Thanks.)

That well-known TV doctor in England still says Natasha Demkina's psychic diagnoses are amazingly accurate -- despite the fact that her reading of him scared him into getting a colonoscopy and other invasive and expensive medical tests, which showed the abnormalities Ms. Demkina "saw" were not there. I suspect they examined the wrong part of his anatomy.

Hi

I think things will become more clear when more cases like Natasha show up in the media. I'm certain that will happen sooner or later.

Bye bye

The Csicops statistics were flawed. Getting 4 right out of 7 is a 1 in 840 chance or 0.12%. That is better than significant odds by a large margin. This girl has been robbed.

I wonder if she would divine my apathy at this discussion. Here are my own psychic predictions. One: All of you who believe in her abilities will feel a wave of anxiety tomorrow when you look up in the sky and see that the Illuminati have been spraying chemtrails again in your neighborhood. Two: All of the rest of us will continue leading more productive and satisfying lives.

Puck T Benson's claim is clearly false. Not even paranormalist Nobel Laureate Brian Josephson, who has been leading the attack on the CSMMH-CSICOP test of Natasha Demkina, disagrees that Natasha had a 1-in-50 chance of matching 4 of the 7 medical conditions correctly. And those are the odds of Natasha blindly guessing. She was not blindly guessing. She had a multitude of normal clues to help her increase her number of correct matches. She said she couldn't do a truly blinded test so the investigators were forced to go with an incompletely blinded test. In any case, Puck Benson's claim is demonstrably false.

People need to learn more statistics because they are so easily taken in by supposed experts. Here is why ...

To get the first one precisely right (ie, matching the right person to precisely condition) is 1 in 7. You now have 6 people left. Picking another precisely is now 1 in 6, and so on ... 1 in 5 and 1 in 4.

A warning here: this is where they made their mistake This is NOT a lotto style draw. In which case she would only have had to pick the 4 people that had the 4 conditions, not which person had which condition.

Try it yourself. Take 7 scrabble tiles, letters 'A' through 'G'. Turn them face down and get someone to muddle them up. Pick a tile and predict what letter it is. Do that six more times. You may NOT guess the same letter twice. Record how many you get right out of 7. Now reset the game and try again and again, recording your results for each game. You could do that a hundred times and you'd never be able to get a count of 4 right (well, there is a slim chance).

The chance of getting 4 exactly right out of 7 in a blind non-return test is (1/7)*(1/6)*(1/5)*(1/4) = 1/840.

In addition to this rather rudimentary statistics, even if the result WERE 1 in 50 (which it is not) the result would still be significant, 2% likely, which is very low and not very likely by pure chance.

As I stated before, she was robbed.

Dear MR. Puck T Benson.....It is very clear many of us would really wonder about Natasha's ability.If natasha is capable of distinguishing even the tiniest pathology on a molecular level in the very deepest corners of our human body and if the rule of statistic is properly understood, statistic cannot be applied to medical diagnosis.I guess you know better than anyone else... In my honest humble opinion,there is no such things like a secondary vision so Nathasha did not get robbed.

Doesn't Puck T. Benson know that it's a crime to falsely report a robbery?

I asked Ray Hyman, Ph.D., professor of psychology at the University of Oregon in Eugene, to help me correct Puck's confused thinking. Prof. Hyman did the original calculation of the odds for the Natasha Demkina test. Those calculations were later confirmed by Prof. Richard Wiseman and others -- including one of our critics, Nobel laureate physicist Brian Jospehson. Here is Prof. Hyman's reply, including reference works for those who would like to verify the calculations for themselves:

"Statistics and combinatorial probabilities can mislead even the brightest people into terrible boo-boos. In the present case, the self-assured critic has made two serious blunders. He has misconstrued the problem. The problem we are dealing with is known as the matching problem. The mathematics for calculating the correct odds is not self evident. Indeed, it is very complicated. I painstakingly worked out the correct probabilities using the formulae in Frederick Mosteller's Fifty Challenging Problems in Probability With Solutions. I believe this is still available from Dover Books. The critic might find it useful to carefully follow the argument in this book. My other source was Hoel, P.G., Port, S.C., and Stone, C.J. (1971). Introduction to Probability Theory. This latter source provides some useful approximations for those who do not have the patience to calculate the exact probabilities. Richard Wiseman was able to check my probability calculations using tables provided by the Journal of the Society for Psychical Research. Our probabilities agreed.

The second mistake this critic makes is to use the probability for getting exactly four correct matches. The number that is relevant for our test is the probability of getting four or more correct matches. Contrary to this persons assertion, the probability of getting exactly four matches in our test is .01533 and not 1/840 (.0012) as he claims. The relevant probability is the probability of getting four or more correct matches which is .01899 (rounded to .02 or 1 in 50).

"I do not have time to give a lesson in probability theory and the matching problem, but let me give a simple heuristic example of how this person's approach provides a misleading answer. Assume we have three subject with conditions A,B, and C. And assume that Natasha's corresponding matchings can be designated a, b, and c. A correct match would be one where she assigns a to A, or b to B, or c to C. For this simple example, we can enumerate all the possible matching attempts. Once we get to four or more subjects, the enumerations become unwieldy."

[continued]

[continued]


"Here are all the possible matching attempts that Natasha could make in the present example:

Subjects:

Actually, there is an error in my previous calculations but there is an error in their calculations too. Here is the exact table of probabilities (used by them, reconstructed by me):

Correct Probability
0 36.79%
1 36.81%
2 18.33%
3 6.25%
4 1.38%
5 0.42%
6 0.00%
7 0.02%
-------
Total 100.00%

Note: There is no chance of getting 6 right because in that case you would actually get all seven right.

The probability of Natasha getting 4 or more exactly right is 1.38 + 0.42 + 0.02 = 1.82% or 1 in 55. But that is if she had 7 medical conditions and 7 people to assign them to. There were only 6 conditions and 7 people (one person had no condition at all). When you redo the calculations with that information you get the following table:

Correct Probability
0 42.09%
1 36.74%
2 15.77%
3 4.37%
4 0.89%
5 0.12%
6 0.02%
-------
Total 100.00%

The probability of Natasha getting 4 or more conditions exactly right is 0.89 + 0.12 + 0.02 = 1.03% or 1 in 97.

In addition to this, there is scientific and statistical method to be considered. First is the 'null hypothesis', which in this case would state that, all things considered, Natasha is no different from any other person. To check this we run a test and we use an alpha level (a cut off point) to excluded the null hypothesis. The usual alpha level (commonly used in normal statistical analysis) is 5%. CSICOPS believed they set their alpha level at 0.44% (which was actually 0.14%) (at least 5 right out of a possible 6) which is extremely harsh in my opinion in either case.

Natasha achieved 1,82% on their table, actually 1.03% on my table (at least 4 right out of 6). That is extremely unlikely by pure chance if she is an ordinary person. The expected number of correct answers for a normal person would be no more than 2. Try it yourself and you will see. Either way it is better than 5%.

The calculations posted previously were erroneous, but the ones here have been double checked and I am very confident in them.

Appendix:

This is the program used to generate the tables in this letter (please consider it public domain):

----------------------------------------------------------------------
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
int i;
int j;
int tmp;
int count;
int choice;
int actual[7];
int guess[7];
int result[8];

// Initialise
srand(time(NULL));
for (i = 0; i < 7; i++) {
actual = i + 1;
guess = i + 1;
result = 0;
}
result = 0;
// Play ten million guessing games
for (j = 0; j < 10000000; j++) {
// Randomise conditions (actual)
for (i = 0; i < 7; i++) {
choice = rand()%7;
tmp = actual[choice];
actual[choice] = actual;
actual = tmp;
}
// Randomise conditions (guess)
for (i = 0; i < 7; i++) {
choice = rand()%7;
tmp = guess[choice];
guess[choice] = guess;
guess = tmp;
}
// How many were right guesses?
count = 0;
for (i = 0; i < 6; i++) { // 6 conditions only
if (actual == guess)
count++;
}
// Record results
result[count]++;
}
for (i = 0; i < 8; i++) {
printf("=\t%8d\t%8.8g%%\n", i, result,
((float) result)/100000.0);
}
return 0;
}
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