rpjday@watrose.UUCP (rpjday) (11/10/86)
Now that the furor about my original posting requesting analog models of computation has died down, how about another problem -- the use of statistics in psychic experiments (this is a MATHEMATICAL article, so just keep reading). I'm sure that we are all familiar with the 25-card deck that is commonly used to test psychic ability. There are 5 shapes used, with 5 of each shape in the deck (wavy lines, cross ... whatever). Since the subject knows the shapes and their frequency distributions, we would expect that, just by guessing, any subject should get 5 out of 25 correct. Nothing deep here. But this takes into account that the subject does not get any feedback. We could imagine that there are two kinds of feedback he may receive: 1) He may be told whether he is correct or not. 2) He may actually be shown the card. Based on either of the above types of feedback, what is the expected number of "hits" now? I'm more interested in this value for case 2). The strategy in case 2) seems clear (at least to me). The subject keeps track of which cards have come up, and always guesses the shape for which there are the most cards left in the deck. This, however, does not supply the expected number of hits. Anyone have an answer and the accompanying justification? On a different note, psyshic experimenters also like to bandy about the phrase "psi missing", when the subject does particularly poorly (eg. 1 or 2 hits out of 25). Given the feedback in case 2), how can a subject get the LOWEST possible score and what is the expected value of this score? DISCLAIMER: Yeah, sure, it's mine, so what?
bdm@anucsd.OZ (Brendan McKay) (11/17/86)
In article <8250@watrose.UUCP>, rpjday@watrose.UUCP (rpjday) writes: > (paraphrased by bdm): > Take a deck containing 5 cards each of 5 different shapes: 25 cards in all. > "The subject" guesses which shape is on each card, one at a time. > If no feedback occurs, and the subject isn't psychic, the expected number > of correct guesses is 5 out of 25. > > One the other hand, if the subject is shown each card after guessing its > value, that information can be used to raise or lower the expected score. > By how much? If the subject always guesses a shape for which the number in the deck is least, the expected number of correct guesses out of 25 is about 2.296063 (surprisingly small!). If the subject always guesses a shape for which the number in the deck is greatest, the expected number of correct guesses is about 8.646753. I think the exact values are 2048941091/892371480 and 23148348523/2677114440, but I wouldn't swear to that. Outline of method: At an arbitrary point of time, let r[1],...,r[5] be the number of cards of shapes 1,...,5 left in the deck. Thus exactly 25-(r[1]+...+r[5]) cards have been dealt so far. The probability of having exactly these shape counts at this point is B(5,r[1]) * B(5,r[2]) * B(5,r[3]) * B(5,r[4]) * B(5,r[5]) P(r)= --------------------------------------------------------- , B(25,r[1]+r[2]+r[3]+r[4]+r[5]) where B(n,k) is the binomial coefficient n choose k. The probability of the subject making a correct guess at this point is G(r), where either G(r) = min(r[1],...,r[5])/(r[1]+...+r[5]) or G(r) = max(r[1],...,r[5])/(r[1]+...+r[5]), depending on the strategy. Now sum P(r)*G(r) over all possible values of r[1],...,r[5]. The full distributions are rather harder to work out, although the variance wouldn't be all that difficult.
lambert@mcvax.uucp (Lambert Meertens) (11/19/86)
In article <234@anucsd.OZ> bdm@anucsd.OZ (Brendan McKay) writes: > In article <8250@watrose.UUCP>, rpjday@watrose.UUCP (rpjday) writes: > (paraphrased by bdm): >> Take a deck containing 5 cards each of 5 different shapes: 25 cards in all. >> "The subject" guesses which shape is on each card, one at a time. [...] >> One the other hand, if the subject is shown each card after guessing its >> value, that information can be used to raise or lower the expected score. >> By how much? > [...] I think the exact values are 2048941091/892371480 and > 23148348523/2677114440, but I wouldn't swear to that. I obtained the same exact values using a different computational strategy (computing the result "bottom up", using a recurrence relation over the shape distributions). -- Lambert Meertens, CWI, Amsterdam; lambert@mcvax.UUCP