Question I have a weird question for you. I am trying to come up with a way of testing the validity of an Ouija board—-whether that heart-shaped “planchette” I have my fingers on is actually being moved by a supernatural spirit or is being subconsciously moved by me.
Here is one idea I have: I sit at a table with the Ouija board in front of me, and to the side of it is a cardboard “screen” or wall. I have five 6-sided dice. I cast the dice behind the screen where I cannot see them. Then I call out to the spirits! I implore any spirit to move the planchette to indicate the numbers that are on top of the five dice. Lets make the wild assumption that the planchette actually moves, pointing to five numbers printed on the Ouija board. With pen & paper I write down those numbers, then I look on the other side of the screen to see the results of the dice cast.
What are the odds that the 5 numbers I have written on the paper match the 5 numbers on the tops of the dice? (The SEQUENCE of the numbers pointed to would be irrelevant, as would be any sequential pattern the dice happen to fall into.) Is it 1 out of 6 x 6 x 6 x 6 x 6, i.e., 7776?
I want to keep the test simple lest the spirits are mathematically challenged or are disinclined to be tested. Yet I want the test to be decisive. Of course it's not foolproof; my subconscious could get lucky or I might have some paranormal power that allows me to directly ascertain the numbers on the dice (which would itself be spectacular).
Can you think of a better test?
Answer jim -
your answer is correct if you take the sequence order of the dice into account - as if you were to roll 1 die 5 times, rather than 5 dice all at once.
if you just look at the set of outcomes for the 5 dice, without any regard to order, the correct probability is more complicated.
if, for instance, the 5 numbers showing are all different, the correct chance is 1 out of 6^5/5! = 7776/5!.
the 5! divisor accounts for the fact that a particular unordered outcome of 5 different numbers corresponds to any of 5! ordered outcomes. so the chance of such an unordered outcome is 5! times the chance of any of the corresponding ordered outcomes - all of which are 1/7776.
if the pattern that is showing has repeats in it - for example {1,1,2,3,4} - the probability is a different multiple of 1/7776. one has to count how many ordered outcomes there are with the numbers {1,1,2,3,4} in it - in some order.
for this example, the counting is as follows (think about tossing 1 die 5 times here - to get an ordered outcome): if we number the throws from 1 to 5, the two "1"s can appear in any two of the 5 throws. there are C(5,2) = 5 choose 2 = 5!/(2!3!) = 10 pairs of positions for the "1"s. the 2,3, and 4 will then occur on the 3 remaining tosses - and can be arranged in any of 3! = 6 ways.
so the number of ordered outcomes giving the unordered outcome {1,1,2,3,4} is C(5,2)x3! = 60. and the chance of the unordered outcome {1,1,2,3,4} is 60/7776 or 1 chance in 7776/60, if you prefer to put it that way.
you can see that the chance of a particular unordered outcome depends on the multiplicities of the numbers in it. the computation for {1,1,2,3,4} works for any other unordered outcome with just one number repeated ONCE - such as {1,,2,3,3,6}.
if the outcome were {1,1,1,2,3} - the pattern of multiplicities is different from the preceding and the computation would be different - and a still different computation has to be made for an unordered outcome such as {1,1,2,2,3}.
i hope this gives you an idea of what is involved in computing the chance that your "ouija guess" matches the actual outcome exactly.
what would you say if your "ouija guess" gets 4 of the 5 numbers correct? for example - if the actual outcome is {1,1,2,3,4} and your guess is {1,1,2,3,5} - or {1,2,3,3,4}? would that be a confirmation that the ouija board knows something?
