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From: bdm@anucsd.OZ (Brendan McKay)
Newsgroups: sci.math
Subject: Re: a mathematical problem in psi (read it anyway, ok?)
Message-ID: <234@anucsd.OZ>
Date: Sun, 16-Nov-86 21:04:22 EST
Article-I.D.: anucsd.234
Posted: Sun Nov 16 21:04:22 1986
Date-Received: Mon, 17-Nov-86 08:56:46 EST
References: <8250@watrose.UUCP>
Distribution: sci
Organization: Computer Science, Australian National Uni, Canberra
Lines: 37
Summary: solution

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.