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- From: daryl@oracorp.com (Daryl McCullough)
- Subject: Re: Games with Nonmeasurable Sets
- Message-ID: <1992Nov4.191956.26318@oracorp.com>
- Organization: ORA Corporation
- Date: Wed, 4 Nov 1992 19:19:56 GMT
- Lines: 71
-
- In article <thompson.720739075@daphne.socsci.umn.edu>,
- thompson@atlas.socsci.umn.edu (T. Scott Thompson) writes:
-
- >I am puzzled by the implicit definition of "certainty" in the
- >discussion of these problems.
-
- If the probability of an outcome is 1, then the outcome is said to be
- certain (or almost certain to be more precise).
-
- >Based on the original game and the followup versions it appears that
- >the "paradox" arises from the fact that the sets in which either
- >player wins are nonmeasurable. This leads me to question the sense in
- >which each player is "certain" that she will win. The answer to this
- >question appears to be that each player knows that the outer measure
- >of the set in which she wins is one (and the inner measure of its
- >complement is zero).
-
- You are right, that before the game starts, the probability for each
- player that she will win is undefined. However, I am talking about the
- probabilities each player calculates *after* she is dealt a card. For
- definiteness, let me assume that player 1 is randomly dealt a card, and
- then player 2. Let's then look at how the probabilities work from the
- point of view of player 1.
-
- There are three stages in the game: Stage 1: neither player has been
- dealt a card, Stage 2: player 1 is dealt a card, and Stage 3: player 2
- is dealt a card.
-
- At Stage 1, player 1 tries to calculate her odds of winning as
- follows: The probability of player 1 winning = the measure of the set
- of all pairs <x,y> such that LT(x,y) = undefined, since this is not a
- measurable set. At this point, player 1 has no confidence that she
- will win.
-
- At Stage 2, player 1 has been dealt a card labelled with the real r1.
- She is allowed to reassess her odds of winning based on the
- information that has been dealt r1. Just as in the case with the game
- "High Card Wins" after being dealt the King of Spades, she is allowed
- to recompute her odds based on this additional information. She now
- computes her odds as follows: Let C(r1) = set of all reals r2 in [0,1]
- such that LT(r2,r1). Then
-
- Probability of player 1 winning after being dealt r1
- = Probability that player 2 will be dealt a card greater than r1 in
- the ordering LT
- = 1 - Probability that player 2 will be dealt a lower card than r1 =
- = 1 - measure of C(r1)
- = 1
-
- (Since C(r1) is a countable set, it has measure 0).
-
- So, while the probability of player 1 winning starts off undefined, it
- changes to probability 1 after seeing what card she was dealt.
-
- >If I were to be told "the probability that you have won is no less
- >than zero and at most one" then I don't think that I would be very
- >"certain" that I had won.
-
- After seeing your card, you calculate that your probability of winning
- is 1. Period. Inner and outer measures agree that the measure of a
- countable set is always zero, and complements of countable sets have
- measure 1. So assuming "probability 1" means "certain", then you
- should be certain you will win. Of course, the fact that your opponent
- will also calculate 1 as her probability of winning, there is
- something strange about identifying "certain" with "probability 1".
-
- Daryl McCullough
- ORA Corp.
- Ithaca, NY
-
-
-