Need help calculating the odds on a test

[SimpleIrony]

Hello one and all,

I have a friend who claims that she has paranormal powers. To my amazement, she is interested having me construct a test to see if she actually has the powers in question.

Here is my problem: I am not sure how many times she would have to be successful for it to be significant.

What she can do, allegedly, is like dowsing. The way I was going to construct the test was to have 100 objects of her choosing, and she is going to divide them into ten groups. When she has left the room, I'm going to go in and hold one object out of each of the ten groups. When I'm done, she will return to the room, and tell me which objects that I've held. She thinks that she will be able to identify all ten.

I know that I can safely expect for her to get 1 or 2 of the objects right, but how many does she have to get right before it's statistically significant?

If anyone can explain how I'd calculate the odds of her getting as many as six or seven right, I'd really appreciate it.

[Holler Hoojer]

Well, it's pretty simple. If a third party, unknown to either of you, picks the objects, and you stay out of the room while she selects, the odds for the first pick are 0.1, the odds for the first and second are 0.01, and so on. However, if she's any good as a carnie, she'll get at at least nine of the objects when she picks the objects and you stay.

[Rodney]

Hello one and all,

I have a friend who claims that she has paranormal powers. To my amazement, she is interested having me construct a test to see if she actually has the powers in question.

Here is my problem: I am not sure how many times she would have to be successful for it to be significant.

What she can do, allegedly, is like dowsing. The way I was going to construct the test was to have 100 objects of her choosing, and she is going to divide them into ten groups. When she has left the room, I'm going to go in and hold one object out of each of the ten groups. When I'm done, she will return to the room, and tell me which objects that I've held. She thinks that she will be able to identify all ten.

I know that I can safely expect for her to get 1 or 2 of the objects right, but how many does she have to get right before it's statistically significant?

If anyone can explain how I'd calculate the odds of her getting as many as six or seven right, I'd really appreciate it.
I wouldn't call that ability dowsing, but if she could get even three correct, that would be unlikely and four or more would be statistically significant. Here is how the calculation is done:

She has a 90% chance of being wrong for each of the 10 groups, so the odds of getting all 10 wrong is 90% to the 10th power or .9^10 = 34.9%.

Getting 1 right and 9 wrong can happen in ten different ways (10!/9!), so the odds of that are 10 times (.1^1) times (.9^9) = 38.7%.

Getting 2 right and 8 wrong can happen in 45 different ways [10!/(8!2!)], so the odds of that are 45 times (.1^2) times (.9^8) = 19.4%.

Getting 3 right and 7 wrong can happen in 120 different ways [10!/(7!3!)], so the odds of that are 120 times (.1^3) times (.9^7) = 5.7%.

Accordingly, the odds of getting 0, 1, 2, or 3 right is the sum of the above, or 98.7%. That, in turn, means the odds of getting 4 or more right is 1.3%. The odds of getting 3 or more right is that plus the odds of getting 3 right, or 1.3% + 5.7% = 7.0%.

So, it would be pretty impressive if she were to get 3 right, statistically significant if she were to get 4 right, and very impressive if she were to get 5 or more right. Provided, that is, you select your object from each of the ten groups randomly and don't give her any cues as to which objects you selected.

[ejk]

Provided, that is, you select your object from each of the ten groups randomly and don't give her any cues as to which objects you selected.

I don't pretend any sort of expertise on the subject, but a couple precautions suggest themselves.

1. Determine which object to pick from each group by a random method (coin toss, roll of dice, etc.) or have someone she doesn't know make the selection.

2. Don't have anyone who knows which objects were chosen/held in the room while she identifies them. Or at least find a way to reduce their reactions to her selections, such as facing away and not knowing what she's picked until later.

Any clues about which objects are most likely to be picked or unconscious feedback would tilt the odds in favor of a right guess, enough that 4 out of 10 would have far less significance as an outcome.

[VisionFromFeeling]

SimpleIrony, this has been helpful to me in the past.
Originally referred to me by EHocking, and authored by a Joseph Czapski,
http://www.automeasure.com/chance.html

[SimpleIrony]

I wouldn't call that ability dowsing, but if she could get even three correct, that would be unlikely and four or more would be statistically significant. Here is how the calculation is done:

I said it was like dowsing, but I could be wrong. I didn't really know what to call it. Perhaps I should have said she had the power to... guess.

She has a 90% chance of being wrong for each of the 10 groups, so the odds of getting all 10 wrong is 90% to the 10th power or .9^10 = 34.9%.

Getting 1 right and 9 wrong can happen in ten different ways (10!/9!), so the odds of that are 10 times (.1^1) times (.9^9) = 38.7%.

Getting 2 right and 8 wrong can happen in 45 different ways [10!/(8!2!)], so the odds of that are 45 times (.1^2) times (.9^8) = 19.4%.

Getting 3 right and 7 wrong can happen in 120 different ways [10!/(7!3!)], so the odds of that are 120 times (.1^3) times (.9^7) = 5.7%.

Accordingly, the odds of getting 0, 1, 2, or 3 right is the sum of the above, or 98.7%. That, in turn, means the odds of getting 4 or more right is 1.3%. The odds of getting 3 or more right is that plus the odds of getting 3 right, or 1.3% + 5.7% = 7.0%.


Thank-you for explaining this so clearly. I understood everything except why you used the exclamation marks.


So, it would be pretty impressive if she were to get 3 right, statistically significant if she were to get 4 right, and very impressive if she were to get 5 or more right. Provided, that is, you select your object from each of the ten groups randomly and don't give her any cues as to which objects you selected.

Oh, I forgot to mention that part. I'd pick which objects I was going to handle by rolling a ten sided die, so it would be entirely random. Also, when I leave the room, I'll be leaving by a door that doesn't connect to the room that she will be entering from. She won't see me until after she has written down her choices.

[wollery]

Oh, I forgot to mention that part. I'd pick which objects I was going to handle by rolling a ten sided die, so it would be entirely random. Also, when I leave the room, I'll be leaving by a door that doesn't connect to the room that she will be entering from. She won't see me until after she has written down her choices.Sounds like a good test protocol, but I have on quick suggestion - if she's going to select the 100 objects and separate them out into each group of ten, then you pick up one object, she may be able to tell which object of the ten has moved. I'd suggest putting each group of ten on a tray that you can pick up and shake after the selected item has been returned. That way all of the objects in each group will have changed position.

[rjh01]

Otherwise note the position it was before you held it and put it back in exactly the same position as before. It is an old party trick. Put all 10 objects in a certain pattern and the slightest movement of any objects means it has been touched. I would not let her even see any of the objects after you have touched one of them. Or at least move them without touching them. You can do this if there is a tablecloth under the objects.

[SimpleIrony]

Sounds like a good test protocol, but I have on quick suggestion - if she's going to select the 100 objects and separate them out into each group of ten, then you pick up one object, she may be able to tell which object of the ten has moved. I'd suggest putting each group of ten on a tray that you can pick up and shake after the selected item has been returned. That way all of the objects in each group will have changed position.


Otherwise note the position it was before you held it and put it back in exactly the same position as before. It is an old party trick. Put all 10 objects in a certain pattern and the slightest movement of any objects means it has been touched. I would not let her even see any of the objects after you have touched one of them. Or at least move them without touching them. You can do this if there is a tablecloth under the objects.

Good points.

I wouldn't trust myself to put it back accurately enough. I was thinking of having them on trays that I could just tip into some pillow cases or some other sort of bag.

I agree that they should all be moved.

[SimpleIrony]

SimpleIrony, this has been helpful to me in the past.
Originally referred to me by EHocking, and authored by a Joseph Czapski,
http://www.automeasure.com/chance.html

Great link, thanks!

[HansMustermann]

Well, ideally it should also be a double-blind test, or else you shouldn't be in the room when she decides which objects have been touched. Otherwise, some people can just get body language hints from you.

[69dodge]

When she has left the room, I'm going to go in and hold one object out of each of the ten groups.

Holding an object in your hand could warm it up a bit.

I recall a story in one of Feynman's popular books where he was able to tell by smell which book someone had removed from a large bookcase, handled, and then replaced while he was out of the room.

[AntiTelharsic]

Thank-you for explaining this so clearly. I understood everything except why you used the exclamation marks.

"n!" is notation for "n factorial", being the product n * (n-1) * (n-2) * ... * 1. Thus 10! is 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3628800. When necessary, 0! is defined to be 1.

Rodney uses it above for "combinations" -- the number of distinct ways to choose r objects from n objects is n!/(r!(n-r)!).

[Fredrik]

1. Determine which object to pick from each group by a random method (coin toss, roll of dice, etc.) or have someone she doesn't know make the selection.

This one is very important, and it's not a good idea to let anyone choose the object based on what they feel like choosing. The objects should definitely be chosen at random. Assign a number to each object and then use e.g. the random number generator at www.random.org.

[Rodney]

"n!" is notation for "n factorial", being the product n * (n-1) * (n-2) * ... * 1. Thus 10! is 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3628800. When necessary, 0! is defined to be 1.

Rodney uses it above for "combinations" -- the number of distinct ways to choose r objects from n objects is n!/(r!(n-r)!).
Right, so 10!/9! is 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 divided by: 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1. Because the terms in the numerator are the same as those in the denominator, except for the 10 in the numerator, this simplifies to 10, which is the number of ways that your friend could get one right in ten attempts. In other words, she could be right on her first guess and then wrong on the next nine guesses; wrong on the first guess, right on the second guess, and then wrong on the next eight guesses; etc.

Calculating the number of ways that your friend could get two or more right involves the same principle. For example, for two right, she could get the first two right, then miss the next eight; get the first one right, miss the second one, get the third one right, then miss the next seven; etc. So, the formula for the number of ways of getting two right in 10 attempts is 10!/(8!2!), which simplifies to 10 times 9 divided by 2 equals 45.

[SimpleIrony]

Right, so 10!/9! is 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 divided by: 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1. Because the terms in the numerator are the same as those in the denominator, except for the 10 in the numerator, this simplifies to 10, which is the number of ways that your friend could get one right in ten attempts. In other words, she could be right on her first guess and then wrong on the next nine guesses; wrong on the first guess, right on the second guess, and then wrong on the next eight guesses; etc.

Calculating the number of ways that your friend could get two or more right involves the same principle. For example, for two right, she could get the first two right, then miss the next eight; get the first one right, miss the second one, get the third one right, then miss the next seven; etc. So, the formula for the number of ways of getting two right in 10 attempts is 10!/(8!2!), which simplifies to 10 times 9 divided by 2 equals 45.

Ah, I see!

That actually makes it all pretty easy to calculate.

I'm not expecting her to get more than 1 or 2 correct, but it is nice to know the odds.

[SimpleIrony]

Holding an object in your hand could warm it up a bit.

I recall a story in one of Feynman's popular books where he was able to tell by smell which book someone had removed from a large bookcase, handled, and then replaced while he was out of the room.

By smell?

Okay, wearing gloves probably won't fly. I'll just have to trust in washing my hands with non-scented soap and having her wait for a couple of minutes before she comes in to make her choices.

I'm pretty sure she isn't going to be that sophisticated about it. That reminds me though, she does have a dog. I should probably make it explicit that the dog can't come in to the room we'll be testing in. Just in case he leads her to any objects. I don't want this to be a test of her dog's sense of smell after all.

[blutoski]

My recommendation from trying to design protocols for dowsing (or similar abilities that must be compared to guessing) is to keep it as simple as possible and mitigate against luck by having just two outcomes per trial, but multiple trials. Rather than many outcomes but only one trial.

The math is simpler, and it reduces the risk false positive due to a one-off lucky guess. I have seen this happen. You do *not* want to be there when a WAG hits home.

The modification I suggest is to have only two objects and to do ten runs. Eight or more hits out of ten trials has a p value <=.05, which means it could only happen one time in twenty, which is a CI of 95% and pretty common for preliminary trials in medicine and social sciences.

However, the precedent established at Skeptics Canada for dowsing tests is to take more of the hard sciences model (physics, chemistry, biology), and make the CI of 99.992% which would require 18/20 positive hits.

[SimpleIrony]

My recommendation from trying to design protocols for dowsing (or similar abilities that must be compared to guessing) is to keep it as simple as possible and mitigate against luck by having just two outcomes per trial, but multiple trials. Rather than many outcomes but only one trial.

The math is simpler, and it reduces the risk false positive due to a one-off lucky guess. I have seen this happen. You do *not* want to be there when a WAG hits home.

The modification I suggest is to have only two objects and to do ten runs. Eight or more hits out of ten trials has a p value <=.05, which means it could only happen one time in twenty, which is a CI of 95% and pretty common for preliminary trials in medicine and social sciences.

However, the precedent established at Skeptics Canada for dowsing tests is to take more of the hard sciences model (physics, chemistry, biology), and make the CI of 99.992% which would require 18/20 positive hits.

In my test, she would need to pick the right 1 out of 10, ten times.

The test you seem to be proposing she would need to pick the right 1 out of 2, ten times.

Please explain how your test protects against lucky guesses better than my test?

[bpesta22]

I agree with blut; seems like 100 objects just makes a mess and complicates things.

Fewer items just requires higher accuracy to be impressive, but the probability calculation's the same as the one demonstrated above.

[Jeff Corey]

That's the way I would test it. Ten objects, you randomly choose to touch some of them (not necessarily 5) and see how many she gets correct.

[SimpleIrony]

Sigh...

I need a large number of objects because once I've touched the object... I've touched the object.

I can't repeat the test with the same 2, 5, or 10 objects, because I would have already imbued one with my "life essence".

I appreciate your concern that she won't be able to successfully put 10 objects on ten trays, and your further concern that I will somehow be confused by having to pick one object out of ten rather than picking one object out of 2... but I think we'll be able to muddle through somehow.

[blutoski]

In my test, she would need to pick the right 1 out of 10, ten times.

The test you seem to be proposing she would need to pick the right 1 out of 2, ten times.

Please explain how your test protects against lucky guesses better than my test?

It depends on what confidence interval you're seeking, and whether you want to hedge against the predictable renegging.

Specifically, with a 1/10 per trial, two hits in a run of ten trials exceeds the 95% confidence interval I mentioned above, whereas one hit does not. It's vulnerable to *one* lucky hit over random chance. Whereas, a binary x 10 requires 8 hits (3 lucky hits over random chance).

Regarding renegging, take the Demkina demonstration as an example of a protocol that was unfortunately very open to spin because the math was too complex for laypeople. With a binary protocol, most people can understand with the analogy of coin-flips, and the tables and in-depth discussions are available in any stats 101 text.

Also: I don't see any need to re-use objects with a binary protocol. Use A&B in the first trial, B&C in the second trial, and so on. You only need 11 objects in total this way.


Illustration:
You can also turn a 10-object test into a binary by looking at it a different way: any one of the 10 objects is either touched or it isn't. Agree to either touch or not touch objects. You don't have to touch exactly five for this to work - you could touch zero or all or 7 or whatever. If she correctly guesses 8+ of the 10 objects in one run, she's exceeded p<=.05 just as if you did 10 runs of two objects. Same math, and you only need 11 objects and one run's worth of time.

Mathematically identical alternative: run 10 one-object tests where you either touch an object or you don't. One at a time. Math identical. Just takes more time.

Suggestion: flip a coin for randomization.


Summary: keep it simple.

[Jeff Corey]

'I need a large number of objects because once I've touched the object... I've touched the object.
'
No you don't. You only need 10 objects, some of which you have touched, others you have never touched. She makes her guesses and you report the outcome here.
There are a large number here who could do the Sign Test on the data.
http://www.graphpad.com/quickcalcs/binomial1.cfm

[Uncayimmy]

Two decks of cards ought to cover it, yeh?

[Jeff Corey]

Why? It can be tested more easily.

[bpesta22]

SI-- i wasn't trying to be condescending but a few posters here have good backgrounds in experimental design and are perhaps just sharing wisdoms learned by doing it to help make your test fair, efficient, and internally valid.

Do report the results here!

[Uncayimmy]

Why? It can be tested more easily.

Maybe in your fantasy world it could work more easily, but out here in the real world it's hard enough to get these "special" people to agree to a test at all. If she says she'll do it with 10 trials of 10 objects, then do it with 10 trials of 10 objects. It's not "easier" if you have to argue and explain several times why you're changing things around after the agreement was made. There's nothing inherently wrong with what he's doing.

[Rodney]

Specifically, with a 1/10 per trial, two hits in a run of ten trials exceeds the 95% confidence interval.
No. As I detailed in post #3 on this thread, under SimpleIrony's proposed protocol, the subject would need to get four out of ten right to exceed the 95% confidence level. Getting only two right would be expected to happen by chance 26.4% of the time. Even three right would be expected to happen by chance 7.0% of the time. So, the proposed protocol appears fine to me.

[SimpleIrony]

Maybe in your fantasy world it could work more easily, but out here in the real world it's hard enough to get these "special" people to agree to a test at all. If she says she'll do it with 10 trials of 10 objects, then do it with 10 trials of 10 objects. It's not "easier" if you have to argue and explain several times why you're changing things around after the agreement was made. There's nothing inherently wrong with what he's doing.

Thank-you!

I'd be willing to change the protocol if there is a flaw in it, but all I keep hearing is that people think it will be easier. Not seeing it.

She has four sets of rune stones (quite close on the deck of cards by the way). And we have both agreed that we will only consider it a hit if she gets six in a row.

If anyone has an serious objection, other than it's not the way they would do it, I'm all ears, but otherwise it will be a done deal by tomorrow evening.

[SimpleIrony]

No. As I detailed in post #3 on this thread, under SimpleIrony's proposed protocol, the subject would need to get four out of ten right to exceed the 95% confidence level. Getting only two right would be expected to happen by chance 26.4% of the time. Even three right would be expected to happen by chance 7.0% of the time. So, the proposed protocol appears fine to me.

We've opted for 6 tries out of 10 being as success. We were originally debating between six and seven, thanks to your excellent breakdown of the expected values, I'm feeling confident that the test is sufficiently robust.

Thanks again.

[SimpleIrony]

.

Specifically, with a 1/10 per trial, two hits in a run of ten trials exceeds the 95% confidence interval I mentioned above, whereas one hit does not. It's vulnerable to *one* lucky hit over random chance. Whereas, a binary x 10 requires 8 hits (3 lucky hits over random chance).

As Rodney pointed out, no 2 hits wouldn't be over 5%. Second point, even in the first post I'm asking about 6 or 7 correct hits. Third point, in the first post I discuss that I can safely expect for her to get 1 or 2 right which should lead you to expect that we weren't going to use 1 or 2.


Regarding renegging, take the Demkina demonstration as an example of a protocol that was unfortunately very open to spin because the math was too complex for laypeople. With a binary protocol, most people can understand with the analogy of coin-flips, and the tables and in-depth discussions are available in any stats 101 text.

Most people are also capable of understanding "more than half".

In your test, I might be in the unenviable position of telling her that she got more than half right and that it's still not significant. No thanks.


Also: I don't see any need to re-use objects with a binary protocol. Use A&B in the first trial, B&C in the second trial, and so on. You only need 11 objects in total this way.

You may notice that B appears twice in your experiment. That's called reusing.

Last minute edit: I'm sorry about being so snarky, but I just can't see the point in debating something unless there is a real problem with the experiment as it stands.

[Rodney]

We've opted for 6 tries out of 10 being as success. We were originally debating between six and seven, thanks to your excellent breakdown of the expected values, I'm feeling confident that the test is sufficiently robust.

Thanks again.
You're welcome. My only additional comment is that, if she should come up short of 6 right, but gets at least 3, it would be worth doing some more testing on another day. In all likelihood, if she's simply guessing, she'll get no more than 2 right.

[SimpleIrony]

You're welcome. My only additional comment is that, if she should come up short of 6 right, but gets at least 3, it would be worth doing some more testing on another day. In all likelihood, if she's simply guessing, she'll get no more than 2 right.

Certainly sounds like a good idea. Although, that sounds like something I shouldn't bring up with her until after the experiment is over.

[blutoski]

Most people are also capable of understanding "more than half".

In your test, I might be in the unenviable position of telling her that she got more than half right and that it's still not significant. No thanks.

I don't see how making it more complex makes it easier to explain either. But you will know your friend better, of course. I'm just going on my personal experience from doing this for 20 years. I find the binary far easier to convey - especially to media - and also try to avoid comparing against predictions of accuracy.



You may notice that B appears twice in your experiment. That's called reusing.

You said you couldn't reuse the ones you touched. B was the untouched one. It's likely I misunderstood.



Last minute edit: I'm sorry about being so snarky, but I just can't see the point in debating something unless there is a real problem with the experiment as it stands.

Probably my fault for sounding pedantic.

I'm just passing along my suggestions based on having done quite a few trials, watched them crash, mitigated the post-failure rationalizations, &c.

[SimpleIrony]

I don't see how making it more complex makes it easier to explain either. But you will know your friend better, of course. I'm just going on my personal experience from doing this for 20 years. I find the binary far easier to convey - especially to media - and also try to avoid comparing against predictions of accuracy.

Perhaps you should do more tests with people who play dice based games.

Experienced Dungeons and Dragons players will have many memories of people rolling three 20's in a row.


[QUOTE]You said you couldn't reuse the ones you touched. B was the untouched one. It's likely I misunderstood.


This would require revealing the results of every test before the next test. I want her to have a free choice, unbiased by previous results. Also this would require her to reset the test multiple times. Certainly not easier than having the whole test set up once.

Having ten that I either touch or not touch would work, but why do I want to risk having her get the first 4 right, and then have something "interfere".

More objects means that she is less likely to get on a lucky run.

I'm just passing along my suggestions based on having done quite a few trials, watched them crash, mitigated the post-failure rationalizations, &c.

Do you really think that there aren't going to be any post failure rationalizations, no matter how the test is done?

[blutoski]

I don't see how making it more complex makes it easier to explain either. But you will know your friend better, of course. I'm just going on my personal experience from doing this for 20 years. I find the binary far easier to convey - especially to media - and also try to avoid comparing against predictions of accuracy.

Perhaps you should do more tests with people who play dice based games.

Experienced Dungeons and Dragons players will have many memories of people rolling three 20's in a row.

I take 'em as the come.




This would require revealing the results of every test before the next test. I want her to have a free choice, unbiased by previous results. Also this would require her to reset the test multiple times. Certainly not easier than having the whole test set up once.

Having ten that I either touch or not touch would work, but why do I want to risk having her get the first 4 right, and then have something "interfere".

More objects means that she is less likely to get on a lucky run.

Possible. It's up to you how much setup investment you want to put into it.



[Do you really think that there aren't going to be any post failure rationalizations, no matter how the test is done?


Not at all.

The value of simplicity is somewhat dependent on the audience. If the claimant is planning to engage media, then your challenge is to explain to the reporter as quickly and succinctly as possible which of you is correct about interpreting the results. Everybody has experience flipping coins, so the analogy is a bit more accessible.

Generally, I find that there's no point in expecting to convince the claimant - it's about making something accessible to a target audience who hopefully are less invested.

[SimpleIrony]

The test will commence in about 30 minutes. I'll keep you all posted. Thank-you to everyone for the tips.

Even those to whom I reacted hostile toward. :o

[Vorticity]

I know it's too late, but for future reference you might want to consider this:

Suppose she does lousy on the test, and then says: "Oh, I understand! These 10 objects were too close together on the tray. When you picked object #7 up, you got close enough to all ten objects to imbue all of them with your essence. So that's what buggered up the results!"

That's why the first try can't be a blind try. She should first be given a chance at a "warm-up", in which you'll pick one of ten objects in her presence and under conditions identical to those to be used in the actual test. Then, knowing which object you handled she'll test her "power" and make sure that she really can "feel" your "essence" on the correct object. In this way, the procedure can be proofed against any confounding factors that might "block" her "power", and it makes it much more difficult to post-hoc rationalize a failure in the real, blind test. Obviously, the initial warm-up try will not count toward anything.

[SimpleIrony]

And the results are in...

Here's how it went down. She arranged four sets of rune stones in groups of ten on ten plates. She had a few extras to make it a full 100 stones. (I am personally agog that she had 5 sets of rune stones, but that's beside the point.)

She left the room, I took out my handy-dandy 10 sided die and used it to pick a stone at random. I noted the symbol on the stones on a notebook, which I also used to cover my hand as I rolled the die and pick the stone. (I didn't think she would have secret cameras, but I thought it would be best to not to risk it.)

I waited two minutes after noting all of the stones to make sure that any heat had dissipated. I gave each of the plates a little shake so all the stone had moved a little. I knocked on the door to the room that she was waiting in, and I left the room. 15 seconds later, as agreed. She came in to make her selections and record her choices.

I collected the choices, and I compared them in private. Ticking off all of the ones that she got right on her list, and then giving it back to her to review.

I made a total of zero ticks. Naught for ten.

She told me that she knew that she had made at least a couple of mistakes. I asked her which ones she knew she had gotten wrong, and what she thought they were. She was right on one out of the two. I pointed out that if she had gotten that one right, she still would have been far below the six out of ten that would have been success. She accepted that part anyway.

She decided the problem was that my energy was so scattered it didn't really stick to one stone, but kind of jumped all around. Apparently, my energy is so nervous, that the entire room is saturated with it.

I suggested that we could arrange to do the same test with someone else more grounded, but she said that she felt bad about doing the test now. She is starting to feel that it is really inconsiderate and ungrateful of her to challenge her abilities this way, and that she thinks that it somehow cheapens them to use them on demand like that.

At least she considers it a fair test. By that I mean that she doesn't think I tried to trick or swindle her. She accepts that she wasn't able to do what she said she should be able to do. She still thinks she has her powers though.

Thanks again, again, for all the help.

[Vorticity]

Thanks very much for the report.


She decided the problem was that my energy was so scattered it didn't really stick to one stone, but kind of jumped all around. Apparently, my energy is so nervous, that the entire room is saturated with it.

Did I call it or what?

Evidently, I have the paranormal ability to predict details of post-hoc rationalizations about failed paranormal abilities.

All we need now is to get ten groups of ten failed paranormalists, and see how I score.

[Vorticity]

She decided the problem was that my energy was so scattered it didn't really stick to one stone, but kind of jumped all around. Apparently, my energy is so nervous, that the entire room is saturated with it.

Imagine how hard it would have been for her to claim this if she had first been given an unblinded practice run, after which she had proclaimed the room and tester's energies to be just fine.

[SimpleIrony]

I know it's too late, but for future reference you might want to consider this:

Suppose she does lousy on the test, and then says: "Oh, I understand! These 10 objects were too close together on the tray. When you picked object #7 up, you got close enough to all ten objects to imbue all of them with your essence. So that's what buggered up the results!"


Are you sure you're not psychic?

Good suggestion. Actually, it occurred to me as I was doing the tests that blutoski's suggestion of just having a couple of objects would have worked really well in a non-blind test.

The thing is, being a friend of mine, she has "demonstrated" her powers in front of me and my friends before. One time that comes to mind is when she sensed which of three origami cranes was made by whom when they were all sitting on a table together. In that case, the cranes were touching and one of them was mine, so today's test wasn't substantially different.

I didn't want to try and argue the point further than I already did, because she is a friend. She's incredibly sweet even if she is a little flaky.

[SimpleIrony]

Imagine how hard it would have been for her to claim this if she had first been given an unblinded practice run, after which she had proclaimed the room and tester's energies to be just fine.

It would have been harder, but not impossible. Perhaps she would have been tired out by the practise run? Or my energy had longer to accumulate in the room?

If I do another test with anyone, I'll do the practise run though.

[Vorticity]

Are you sure you're not psychic?

I knew you'd say that.

[SimpleIrony]

I knew you'd say that.

Touché.

[Uncayimmy]

Thanks for sharing the results of the test. You're right that you can't prevent every avenue for post-test excuses.

[Doug Higley]

You sure her name isn't Sylvia?

[Jeff Corey]

Another test in Dr. Rhine's file drawer, right next to Davy Jones' locker.

The unblinded pretest would have been great, I've seen Randi do it with auras and dowsing before and don't know why I forgot it.

One point about the binary test that was suggested, it would not have so completely disappointed the claimant. She would have most likely gotten 4, 5 or 6 right out of 10, not a stunning 0. Extinction at the beginning of the testing is quite discouraging in the real world.