The statistics of 'psychic challenge'

In summary, the conversation is about a problem that the speaker thought they had solved many years ago, but upon further examination, they realized there were still many unclear aspects. The problem involved a TV program where a psychic had to match five couples without any prior knowledge about them. The speaker's interpretation involved finding the probability of correctly guessing a certain number of couples in a specific order. They were able to find a formula for guessing at least k specific couples correctly, but the original problem was to find the probability of guessing k couples and not the remaining ones. The speaker also found that the expected value for a large number of trials was always 1, regardless of the number of couples. They then asked some questions about the problem and its solutions.
  • #1
lavoisier
177
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This is a problem that I thought I 'solved' many years ago.
In actual fact there are many things about it that are not clear to me, and I would like to hear your opinion, please.

Very briefly, there was this TV programme where a (supposedly psychic) guy had to match 5 (husband-wife) couples, obviously without knowing anything about them.
My interpretation was: if we have N distinct objects (say, the letters A, B, C, D, E) that must be placed in one specific order, how likely are we to place k = 0, 1, 2 ... N of them correctly by choosing the order randomly?

I was pretty sure the problem had long been studied, but I wanted to have some fun figuring it out for myself.
I found quite easily the probability to guess at least k specific couples correctly, regardless of what happens to the other couples (that's (N-k)!/N!, I believe, in this case with N=5).
However, the original problem was to find the probability P(k,N) to guess k (no matter which) couples and not the remaining N-k. Things got a bit tougher then, at least for my limited maths skills, and sums with factorials and alternating signs started appearing, but in the end I seemed to find a formula that accounted for the explicitly enumerated cases:

[itex]P(k,N) = \frac{1}{k!} \sum_{i=0}^{N-k}\frac{(-1)^{i}}{i!}[/itex]

One nice feature of this result is that it correctly tells you that P(N-1,N)=0 (it's impossible to guess 4 couples right and not the 5th one). And it also tells you that for odd values of N, it is more likely to guess 1 couple right than 0 right, meaning perhaps you're more of a 'psychic' if you get them all wrong than if you guess 1 right(!).

Another nice thing about it (which I can't prove, but so far has worked numerically) is that it sums up to 1:

[itex]\sum_{k=0}^{N}P(k,N) = 1 [/itex]

As required, I believe, because the events for k = 0, 1, 2... N are mutually exclusive and, taken together, they constitute the whole set if possible outcomes of this experiment, so the probability that at least one of them happens must be 1.

What I wanted to know next (and I did this only a few days ago, years after finding the above formula) was the expected value, i.e. if we did a large number of trials, what would be the average number of correctly guessed couples?
Somewhat to my surprise, I found that, at least for the values of N I tried (2, 3, 4, 5, 10, 20}, it's always 1:

[itex]\sum_{k=0}^{N}k \cdot P(k,N) = 1 [/itex]

as in: regardless of how many couples there are, on average you're likely to guess only 1 right if you choose at random.
I plotted the probabilities for the above numerical cases, and it seems that, after some oscillations for N=2, 3 and 4, already from N=5 the curve gets very close to a limiting case where the probabilities of k = 0 or 1 are always the highest and both very close to 0.36788 (if anything is special about this number, I don't know), and the rest of the curve .

Now, I'm no mathematician - it may be that all of the above is particularly obvious to an expert. But it's not to me.
So here are my questions/doubts:

1. when I found the formula I wanted to check if it was correct, but I could find no website describing it - does anybody know if this kind of problem has a particular name?
2. can the formula be reduced to a closed form?
3. is there a way to prove that the probabilities for k = 0 ... N sum up to 1, and that the expected value is independent of N and is also 1?
4. if we wanted to test whether a guy is a psychic (OK, nobody is, but bear with me), what statistical method would we use? Chi squared, based on the expected and observed number of successes in NT trials?

Thank you
L
 
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  • #2
Just noticed, 0.36788 is close to 1/e !
Now I'm even more intrigued...
 
  • #3
Interesting problem. I tried with mathematica it gives a closed form expression for sum of (-1)^n/n! But its a recuresively defined function.

However $$ exp (-1)=\sum_{i=0}^\infty \frac {(-1)^i}{i!} $$ and its a rapidly converging series
 
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  • #4
Thank you @jk22.
That explains indeed why for high values of N the curves get all very similar: the first two probabilities k=0 and k=1 are close to 1/e, and the terms for k>=2 approach some value dominated by 1/k! and the first few terms of the sum.

pkN_vs_k.JPG


P(k,10) is there but it's overlapping so well with P(k,20) that we don't see it!

Does mathematica say anything about the expected value summing up to 1?
 
  • #5
Mathematica can do only numerically for a given value of N but does not prove for all N.
 
  • #6
OK, thank you.
So it's not as obvious as it sounds.
Have a good weekend!
 

1. What is the purpose of a psychic challenge?

The purpose of a psychic challenge is to test the validity and accuracy of claims made by individuals who claim to have psychic abilities. It is a way to objectively evaluate their abilities and determine if they have any actual psychic powers.

2. How is the success or failure of a psychic challenge determined?

The success or failure of a psychic challenge is determined by statistical analysis of the results. This involves comparing the actual results to what would be expected by chance alone. If the results significantly deviate from chance, it may indicate the presence of psychic abilities.

3. What is the significance of statistical significance in a psychic challenge?

Statistical significance is used to determine if the results of a psychic challenge are due to chance or if they are a result of actual psychic abilities. A result is considered statistically significant if it is unlikely to have occurred by chance alone, typically with a p-value of less than 0.05.

4. How are control groups used in a psychic challenge?

Control groups are used in a psychic challenge to provide a baseline for comparison. They are made up of individuals who do not claim to have psychic abilities and are used to determine what results would be expected by chance alone. This allows for a more accurate evaluation of the results from individuals claiming to have psychic abilities.

5. What are some common criticisms of psychic challenges?

One common criticism of psychic challenges is that they do not take into account the complex nature of psychic abilities and the potential influence of external factors. Additionally, some argue that the strict protocols and controlled environments of psychic challenges do not accurately reflect real-life situations where psychic abilities may be used. Others believe that the presence of skeptics or a lack of belief in psychic abilities can affect the results of a challenge.

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