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How Are Paranormal Test Odds Calculated?

Luke T.

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In the Paranormal portion of this forum, a test of a young Russian girl by the name of Natasha Demkina was brought up. And this led to me to wonder how Randi, and others, calculate the odds of success for various paranormal tests. Some I can figure out, but the one with Natasha has me stymied.

Basically, there were seven patients. Six had medical conditions which could be verified by an X-ray. The girl was told of the six medical conditions and had to tell which patient had which condition just by looking at them. She got four right.

So what are the odds of getting four out of six conditions correct for seven subjects, and how do you calculate that?

I don't wish to debate here whether this girl has paranormal abilites. That is what the Paranormal forum is for. I just want to see the math. Thanks.

Also, if anyone else can think of math problems along these lines that they would like answered, please feel free to add on your own questions.
 
Just wanted to add that what is throwing me on the Demkina Problem is that in that situation, you can't have just one wrong. There has to be either all correct, or at least two wrong. Or am I wrong?
 
C = Condition, A-G.
G = Guess.

C: ABCDEFG
G: ABCDEF.

You can't have 6 correct and 1 wrong, because when you guess the 6th correctly, the 7th also has to be correct.
 
CFLarsen said:
C = Condition, A-G.
G = Guess.

C: ABCDEFG
G: ABCDEF.

You can't have 6 correct and 1 wrong, because when you guess the 6th correctly, the 7th also has to be correct.
Click to expand...

Okay, so I was right about that part. So what are the odds of getting 4 of 7 correct?
 
I believe it is 35/5040, so 1 in 144.

ETA: Rereading the OP, I think you wanted to see the math, so:
To get the total number of possible outcomes, 7! or 7P7 = 5040.
To get the number of possible outcomes with 4 right: 4C7= 35

Just in case somebody needs it: xPy is permutations and xCy is combinations.
 
I don't know the chances of getting four right off the top of my head but here are the chances of getting them all right.

To get them all right you have to get the 1st one right, and then get the 2nd one right, etc. If there are 7 things then the probability of getting them all right (assuming you're just guessing) is

(1/7) * (1/6) * (1/5) * (1/4) * (1/3) * (1/2) * (1/1)

That is 1 in 5,040.
 
This is the 'matching problem'. For n possible values, the probability of matching k times is given by the following:
<html><p class="equation"><var>P</var>(<var>N<sub>n</sub></var> = <var>k</var>) = (1 / <var>k</var>!)
bigSigma.gif
<sub><var>j</var>= 0, ..., <var>n - k</var></sub> (-1)<var><sup>j</sup></var> / <var>j</var>! for <var>k</var> = 0, 1, ..., <var>n</var>.</p></html>

i.e. to answer your question, in the case of 7 potential matches, the probability of matching 4 times is:

P(N=4) = (1/4!) * (1/0! - 1/1! + 1/2! - 1/3!) = 0.0138

so about 1 in 70 of getting four right by chance. If you read more about the case you'll see how she clearly had more than chance to go on in at least two of the subject's cases.
 
Donks said:
I believe it is 35/5040, so 1 in 144.

ETA: Rereading the OP, I think you wanted to see the math, so:
To get the total number of possible outcomes, 7! or 7P7 = 5040.
To get the number of possible outcomes with 4 right: 4C7= 35

Just in case somebody needs it: xPy is permutations and xCy is combinations.
Click to expand...
This can't be right because 7C0 + 7C1 + ... + 7C7 = 128, which means the chances of getting 0, 1, 2, 3, 4, 5, 6 or 7 right is 128 / 5,040. But we know that the chances of getting one of those has to be 1 = 5,040 / 5,040.
 
In case it's not totally clear (I forget how much math people may or may not know), here's the calculation for each of the possible outcomes in the case of n=7 (the one we're discussing):

P(N=0) = (1/0!) * ((1/0!) - (1/1!) + (1/2!) - (1/3!) + (1/4!) - (1/5!) + (1/6!) - (1/7!)) = 0.3677
P(N=1) = (1/1!) * ((1/0!) - (1/1!) + (1/2!) - (1/3!) + (1/4!) - (1/5!) + (1/6!)) = 0.3679
P(N=2) = (1/2!) * ((1/0!) - (1/1!) + (1/2!) - (1/3!) + (1/4!) - (1/5!)) = 0.183
P(N=3) = (1/3!) * ((1/0!) - (1/1!) + (1/2!) - (1/3!) + (1/4!)) = 0.0625
P(N=4) = (1/4!) * ((1/0!) - (1/1!) + (1/2!) - (1/3!)) = 0.0139
P(N=5) = (1/5!) * ((1/0!) - (1/1!) + (1/2!)) = 0.0042
P(N=6) = (1/6!) * ((1/0!) - (1/1!)) = 0
P(N=7) = (1/7!) * (1/0!) = 0.000198

Hopefully you can see the pattern. It's kinda cool how k=n-1 always comes out to be zero (as others above have noted).
 
Number Six said:
This can't be right because 7C0 + 7C1 + ... + 7C7 = 128, which means the chances of getting 0, 1, 2, 3, 4, 5, 6 or 7 right is 128 / 5,040. But we know that the chances of getting one of those has to be 1 = 5,040 / 5,040.
Click to expand...
I'm pretty rusty on this stuff :)
 
I don't know the chances of getting four right off the top of my head but here are the chances of getting them all right.

To get them all right you have to get the 1st one right, and then get the 2nd one right, etc. If there are 7 things then the probability of getting them all right (assuming you're just guessing) is

(1/7) * (1/6) * (1/5) * (1/4) * (1/3) * (1/2) * (1/1)

That is 1 in 5,040.
 
And if it changes to getting at least four right? What is the probability then?

I thought I had it because it seems that it should be equal the probability for exactly 4 right plus the probability for exactly 5 and exactly 7.

When I do that from the list above I get 0.018298, but I'm having difficulty transforming that into a ration which, if others are correct, should be around 1 in 50.
 
Donks said:
I'm pretty rusty on this stuff :)
Click to expand...

Nowhere near as rusty as I am. I took Calculus I and II back in the 80s and haven't used it since. I'm just glad SpaceFluffer elaborated on his post. :)
 
Garrette said:
And if it changes to getting at least four right? What is the probability then?

I thought I had it because it seems that it should be equal the probability for exactly 4 right plus the probability for exactly 5 and exactly 7.

When I do that from the list above I get 0.018298, but I'm having difficulty transforming that into a ration which, if others are correct, should be around 1 in 50.
Click to expand...
By inspection we know that 0.02 is 1 in 50 so 0.018298 must be 1 in X where X is a little larger than 50.

And plugging the numbers in I get X is 55 (rounded to the nearest integer).
 
Duh. I really could have seen that myself,
but thanks Number Six.

So the "about 1 in 50" was correct.
 
For large n, the expression (for the matching scenario) given can be approximated by

e^(-1)*SUM(1/k!), k=4 to 7, which gives .0189
 
Unrelated to the X-ray girl:

In another thread, when someone said the Randi test couldn't work for "weak" psi, I gave a hypothetical example: Someone who can make a coin come up heads 50.0001% of the time (I imagine correctly predicting the coin flip with a similar percentage would also do) and said that that could be tested, though it'd probably take a while to go through the requisite number of trials.

Anyone know how to calculate the number of trials it would take to test that claim?
 
BronzeDog said:

In another thread, when someone said the Randi test couldn't work for "weak" psi, I gave a hypothetical example: Someone who can make a coin come up heads 50.0001% of the time (I imagine correctly predicting the coin flip with a similar percentage would also do) and said that that could be tested, though it'd probably take a while to go through the requisite number of trials.

Anyone know how to calculate the number of trials it would take to test that claim?
Click to expand...


hmm

what comes to mind first is solving

1.96*sqrt(.25/trials) = .0001, for trials.
 
jzs said:
hmm

what comes to mind first is solving

1.96*sqrt(.25/trials) = .0001, for trials.
Click to expand...

I agree, that's what I was going to post in a different way. Comes up 100,000,000 if you use 2 standard deviations instead of 1.96, which is what I was doing. Comes up about 96,000,000 with the 1.96.

Of course, this is the number of trials using a 95% confidence interval. If the requirement was a 1 in 1000 probability that the results weren't due to chance, the number of trials would be more.

EDITED TO ADD: looks like about 270 million trials for a 1 in 1000 chance - I get 3.29 standard deviations instead of 1.96 for that, then solve the equation above.

Also, you'd need to do the same number of trials first as a control without someone 'affecting' the coin as a control to determine the bias in the coin.
 
bmillsap said:
Also, you'd need to do the same number of trials first as a control without someone 'affecting' the coin as a control to determine the bias in the coin.
Click to expand...

Yeah. I remembered that after I posted. Think I read somewhere on Snopes that a penny is slightly biased towards tails because the head side is slightly heavier.

Was doing the math wrong for a while, since it's been some time since I've done the old algebraic gymnastics. I get 96,040,000. That would take quite a while.

Another possible difficulty with that test: Over the course of nearly 100,000,000 flips, testers might begin to slip up in the protocols.
 
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