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From: egs@ulysses.UUCP
Newsgroups: net.math
Subject: Re: Re: Another counterintuitive statistics problem
Message-ID: <582@ulysses.UUCP>
Date: Mon, 29-Aug-83 06:07:33 EDT
Article-I.D.: ulysses.582
Posted: Mon Aug 29 06:07:33 1983
Date-Received: Tue, 30-Aug-83 01:27:33 EDT
Organization: Bell Labs, Murray Hill
Lines: 43

I believe I may have come up with another strategy for the "largest of 10
numbers problem" that differs a little bit from the optimal M-strategy described
previously. My argument may be specious, however, so I would appreciate
corrections if any.

Essentially, you apply the M-strategy to only the positive numbers of the 10
chosen ones. In the case of 10 numbers, I think this strategy yields a 0.4137
chance of getting the highest number using M=1 (as opposed to a 0.3987 chance
following the straight M-strategy with M=3).

I arrive at this result as follows:

	The ten numbers may be grouped into positive and negative sets with the
	following chances:

		case 0.	10 neg, 0 pos		C(10,0)/2^10	1/1024
		case 1.	9 neg, 1 pos		C(10,1)/2^10	10/1024
		case 2.	8 neg, 2 pos		C(10,2)/2^10	45/1024
					.
					.
					.
		case 10.0 neg, 10 pos		C(10,10)/2^10	1/1024
	
	Applying the modified M-strategy to each of these cases, you see that
	when case 0 occurs, the modified strategy will win by default 1/10 of
	the time since the last number will always be chosen. In case 1, you can
	get it right 1/10! of the time, since the positive number will occur
	in last place this many times (this case can be ignored for all
	practical purposes). For case j>1, you will guess right F(j,1) times
	(where F(N,M) was defined in my previous note as the chance of the
	straight M strategy working on N numbers).

	Multiplying these coefficients with the probability for each case and
	summing them gives a 0.4137 chance of guessing the right number using
	this modified M-strategy for N=10.

I haven't given much thought to what happens as N gets large, but my gut feeling
is that the optimal modified M-strategy for N numbers will perform no better
than the straight M-strategy, and will most likely get close to a 1/e chance as
well.

							Ed Sheppard
							BTL @ MH