Math question for LD experiment

Click here for the experiment details.

Out of curiosity, I have set up an experiment in which I am attempting to determine if a lucid dream can be used to "observe" physical things remotely...in this case a die that has a different color on each side (Red,Black,White,Blue,Green,and Orange). I haven't been in a math class for almost a decade now, which I am sure is part of the issue. Trying to figure out the overall probability of all of the trials combined (when some are "hits" and some are "misses" and when the individual probabilities of each trial can vary) makes me blank out. I cannot seem to make sense of how to do this...but I do know there is a way to do it correctly.



D = the number I observed in the lucid dream
R = the real number
* = a "hit"
X= a "miss"

X Trial 1: D = green, R = black (odds: 1/6)
* Trial 2: D = green, R = green (odds: 1/6)
* Trial 3: D = white or black, R = black (odds: 2/6)

I am aware its a very small sample of trials at this point, I did just start it after all. But I need to know how to calculate out the probabilities specifically in situations when the cube cannot be determined to be any specific color of two or more options. Any help would be greatly appreciated. I know its something simple I just do not remember what it is.

This is actually quite a complex question regarding statistical hypothesis testing, which I know little about.

One thing you can do however is work out the probability of getting, for example, 2 or more trials out of 3 correct, and try to make conclusions from that.

First you need the possibility of getting 2 out of 3 correct. There are 3 ways of doing this; you could get trials 1 and 2 right (and trial 3 wrong), trials 1 and 3 right (trial 2 wrong), or trials 2 and 3 right (trial 1 wrong).

The respective probabilities are

1/6 * 1/6 * 4/6 = 4/216 [note you work out the individual probability by multiplying the probabilities of all the events together; the last probability is the probability of getting tral 3 wrong which is 1 minus the probability of getting it right]

1/6 * 5/6 * 2/6 = 10/216 [here the 5/6 represents the probability of getting trial 2 wrong]

5/6 * 1/6 * 2/6 = 10/216

The total probability of getting 2 out of 3 right is hence (4 + 10 + 10)/216 = 24/216

Then we need to work out the probability of getting all 3 trials right, which there is only 1 way of doing:

1/6 * 1/6 * 2/6 = 2/216

So adding both of these together, the total probability of getting 2 or more out of 3 right is (24 + 2)/216 = 26/216 which is about 12%, which isn't really impressive enough to make any conclusions.

In the long term, with respect to the mathematics, it would be a lot easier to only guess once for each trial, because then I could work out a relatively simple general formula for you.

Good luck with the experiment, it's a neat idea. I guess you're shaking the die and hiding it under a cup or something?

You want to use the z-table to determine where your results lie in a normal distribution. For something to be viable, you want it in the 95th percentile of probability or sometimes 99th. Read up on z-tables, bell curves, distribution, and standard deviation.

That's actually a binomial distribution, not normal. You can approximate it with normal, but you need more trials.

The method Xei outlined should be sufficient to integrate the unique probability for trial 3 with the other trials. The probability it yields corresponds to the ubiquitous p-value, or the probability of obtaining a result at least as extreme as that observed, assuming that the null hypothesis is true (in this case, assuming that you cannot reliably predict the dice color). It is common practice to reject the null hypothesis if this probability is lower than 5%.

Note that this is doing things the long way, which is necessary if you want to retain the observation from trial 3. Widely available software programs (including web-based ones) make it trivially easy to compute the final probability if the expected probability of success under the null hypothesis is constant. However, satisfying this assumption would require excluding trial 3. (I wouldn't recommend revising the observation so as to be either white or black.)

One more thing about the 5% rejection thing. Do not simply stop collecting data as soon as your p-value up to that point falls below .05 (assuming that it does). For example, if your very next attempt turns up a hit, your cumulative p value will probably be below .05 at that point. However, it is unlikely that a sample of only 4 trials is representative of the potential population from which it comes (check out this article; it's a classic). You should decide in advance how many trials you will run. A good rule of thumb is to have at least 30 -- although the more trials you have, the greater are your chances of observing an effect if there really is one.

Quote:

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Originally Posted by Xei

I guess you're shaking the die and hiding it under a cup or something?

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Yeah, more or less. My first experiment was done with playing cards, but I found that this had too many smaller variables to consider given the instability of these experiences. This time I simplified it quite a bit, but still enough to where each trial still has a decent amount of statistical value. I could have done something like a coin that had a black side and a white side, but that would require far more trials, and LDing enough to make that method pay off would be a problem.

I will go over the math suggestions and try to re-teach myself what I need to do. Thanks all for the pointers, I am sure I will have more questions in the near future.

Ok, i have some questions about binomial distributions now that I have been teaching myself how to go about calculating the odds of my results of this experiment. Read over the following explanation and let me know if this is an accurate way of calculating the probability of a given number of "Hits" and "Misses" out of 20 trials:

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The number of trials you do will determine which row of pascals triangle you can reference. In this case there are 20 trials.

20th row of Pascal’s Triangle
1, 20, 90, 1140, 4845, 15504, 38760, 77520, 125970, 16790, 184756, 16790, 125970, 77520, 38760, 15504, 4845, 1140, 90, 20, 1

Sum of Row 20: 746036

The number of “Hits” will indicate which number you will use from the 20th row of Pascal’s triangle. For example:

0 Hits: 1
1 Hit: 20
2 Hits: 90
10 Hits: 184756
18 Hits: 90
19 Hits: 20
20 Hits: 1

I then apply this number (X) to the following equation because the odds of a getting a “Hit” per trial is 1/6, and the odds of getting a “Miss” is 5/6.

X(1/6)^H(5/6)^M

For example: If in 20 trials I get 8 hits and 12 misses, the equation would look like:

125970 * (1/6)^8 * (5/6)^12

This answer then becomes the numerator, and the denominator is the sum of row 20 (746036). This will give you the percentage of how likely it is to get exactly that number of hits and misses out of that number of trials.

Since the goal of this sort of experiment is getting 100% accuracy (ideally), I think it would be a good idea to calculate the probability of getting *at least* X number of hits out of 20 trials, which can be done by figuring out the probability of X hits plus the probabilities of the remaining (unfulfilled hits). For example, if I get 8 hits out of 20 trials, I would calculate out the odds of getting exactly 8 hits, exactly 9 hits, exactly 10 hits, and etc up to 20 hits and then add them all together for an over all probability of getting at least 8 hits.

__________________________

Does this seem right? If so, is there a simpler way to calculate the probabilities of dice related binomial distributions?

Also if you could help me make sense of the following equation, that would be helpful:

The probability of getting exactly x success in n trials, with the probability of success on a single trial being p is:

P(X=x) = nCx * p^x * q^(n-x)


probability of success (p): p
probability of failure: q
the number of trials: n
the number of successes out of those trials: x

They do not define what "C" stands for, though I know in this case it means n choose x, but I am not quite sure I grasp what exactly that means. If we assume that there have been 20 trials, 8 hits and 12 misses, the equation would look like:

20C8 * (1/6)^8(5/6)^(20-8)

which looks almost exactly like the equation I gave in my last post, so would I be safe to assume that "20C8" is equal to the "coordinates" of the number in regard to Pascal's Triangle (20th row, 8th element)?

If so, how exactly does one go about calculating nCx without referring to the Triangle itself? Also, would I still divide this number by the sum of the given row to find the distribution of an event?

Quote:

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Originally Posted by ethen

Does this seem right? If so, is there a simpler way to calculate the probabilities of dice related binomial distributions?

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It seems right.

There are applets on the web that do this for binomial distribution, but I don't know if they do the 'at least' calculation. You might need some real statistical software to calculate it. I'm sure you can find something good here:
http://www.freestatistics.info/stat.php

Quote:

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Originally Posted by ethen

If so, how exactly does one go about calculating nCx without referring to the Triangle itself? Also, would I still divide this number by the sum of the given row to find the distribution of an event?

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You would use factorials to calculate it yourself.
For example 4! = 4*3*2*1=24

The formula for the binomial coefficient (n choose x) is:
n!/(x!*(n-x)!)
Most calculators have this function, so you don't need to do it by hand.
You can even use Google to calculate it.

It's good to know what it means in terms of probability.
Say you are flipping a coin and are interested in 3 hits out of 4 trials.
You can get that in following ways(H:hit, 0:miss):
HHH0 or HH0H or H0HH or 0HHH
In other words, there are 4 ways to get 3 hits out of 4 trials. That's what '4 choose 3' means.
Check by Google:
http://www.google.com/search?ls=en&q=4+choose+3

So, what the probability formula actually tells you is:
Probability of getting 3/4 hits = number of possible arrangements of 3 hits in 4 trials * probability of getting 3 hits * probability of getting 1 miss

If you have a graphing calculator simply set up two matrices. One outlining the the details of the die, and the second outlining the 'points' to be provided for each hit or miss. Just simple highschool math.

Ok, just so that I am clear:

1. I would start off by determining nCx. This is done by using the formula n!/(x!*(n-x)!)

2. Once I have this amount, I then apply it to the formula nCx * p^x * q^(n-x)

Now, is this where I stop, or is there a third step (i.e. dividing the amount by the sum of that particular row in Pascal's triangle)?

When I was reading up on this stuff, I came across an example of flipping a coin 10 times, and they had the following figures all over 1024 (the sum of the row):


1 10 45 120 210 252 210 120 45 10 1


and so the odds of getting 3 heads and 7 tails would be 120/1024, or about 11.7%. Do I still so something like this?

Unless I'm mistaken you're overcomplicating things.

Here's an example of a working matrix.

-------------Wins----Loses----Ties
Flames------[1--------3---------4
Canucks------2--------4---------2
Oilers---------1--------0---------7]

That matrix shows win, ties, and loses for 3 teams in one season.

Now you want to find out their standings. So lets say a win = 3, a tie = 1, and a lose = 0.

Win-[3
Tie---1
Lose-0]

Now you simply multiply the matrices and that's that. Of course this is a simple example... which you could do without a calculator... but you'd need much larger matrices with fractions and whatnot to figure out your experiment. Same principle though.

For yours you'd set up like this

----------Tr. 1-------Tr. 2-------Tr. 3
Hit----[
Miss-----------------------------------]

And solving the issue of probability is easy.
Red,Black,White,Blue,Green,and Orange
So, the probability of correctly getting any one side is 1/6. But if it's multiple trials than the odds decrease. Say you try six times, then the probability is 1/36.

Once you've sorted out the probabilities and created an appropriate matrix set then a calculator will do the rest.

I am pretty sure that I will need to work with binomial distributions, and to do that I believe I would need to work with the previous equations. Of course, I haven't worked with matrices in quite a while so it may be the case that it could calculate out the values in the same way (but easier). In reality, solving the equations is not that complicated, I was more getting hung up on what the equations meant.

Fair enough...

Do whatever fits your needs ;)
It's just I remember learning matrices, probabilities, and pathways in highschool and thinking

Quote:

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"These might be the few things in math that I'll find myself using in the real world"

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which is why I remember them really well. They should do the job if you do them right, but like I said... do what you need to do.

Quote:

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Originally Posted by ethen

Ok, just so that I am clear:

1. I would start off by determining nCx. This is done by using the formula n!/(x!*(n-x)!)

2. Once I have this amount, I then apply it to the formula nCx * p^x * q^(n-x)

Now, is this where I stop, or is there a third step (i.e. dividing the amount by the sum of that particular row in Pascal's triangle)?

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This (2. in your post) is what gives you the probability for that specific number of hits, provided that you made them randomly.

The whole procedure goes:
1. Assume you guess the orientation of the die by chance alone. This is your null hypothesis. This gives you probability of guessing correctly to be 1/6

2. Decide you will do 30 trials

3. Do 30 trials. Say that you guess correctly 12 times.

4. Calculate the probability of guessing 12 out of 30 times under the null hypothesis. This means you are still assuming you got those guesses by pure luck.
You use the binomial distribution and the above formula for the calculation.

What you do in practice, you actually calculate the probability of getting 12 or more guesses correct by pure chance. This means you calculate probability for getting it right 12 times + probability of getting it 13 + ... + prob. of getting all 30 correct. This is called cumulative probability.
Don't do it by hand, you can use the following applet(cumulative probability will be displayed in the last box after you enter all the parameters):
http://stattrek.com/Tables/Binomial.aspx

5. Interpret the result. For our example (12/30), the cumulative probability is 0.002 or 0.2%. You have a very strong argument for rejecting the null hypothesis. You haven't really proven that you can see the die in your OBE, but you've proven that it's highly unlikely your guesses were due to pure luck.

YES, that app is lovely :banana:

Ok, one last final question. Suppose that I am doing 25 trials, the odds of a "success" is 1/6, and I get 10 successes by the end of the experiment.

According to binomial distribution, the odds of getting at least 10 success by chance alone is ~.5%. Alternately, there is a ~99.5% chance that something other than chance was at work. Does this then mean that these results fall into the 99% precentile, or is that something different I will need to consider?