Conference rusure::math

nuisance answer 1: if x is odd, choose the other box, relying on the universal
law of mathematics that n is an integer.

nuisance answer 2: by induction, you can reach an unbounded payoff: choose the
left box, then, without opening it, choose the right box.  Repeat.

nuisance answer 3: like the Monty Hall problem, this really depends on the
behavior, motivation and/or decision rules of the host.


Ignoring the nuisance answers above, I think the usual decision theory applies:


Put your hand on one box.  The world is now in one of two states: 

	SAME = your hand is on the box that has the 2N 
	DIFF = your hand is on the box that has the N

Absent cheating, you have no reason to assume the probabilities of these two
states are different, so you can assign them both P=1/2.

You open the box and the host offers you the chance to change.  Ignoring 3
above, neither action changes the probabilities, still both P=1/2.

Now you have a choice of two strategies:

	KEEP = keep the money in the box you have opened
	CHANGE = take the money in the other box


If we construct the usual decision payoff table:

Strategy		State of World		Expected Value of Strategy

			SAME	DIFF
	Probability	1/2	1/2

KEEP			2N	N		3N/2
CHANGE			N	2N		3N/2

we see that the payoff of the two strategies is the same, matching the intuitive
answer.

In .0, we confuse ourselves by thinking in terms of X, which is 2N in one state
of the world and N in the other state.  We get a strange answer when we add up
these different X values.

    Re:     <<< Note 1947.1 by CSSE::NEILSEN "Wally Neilsen-Steinhardt" >>>
    
    Wally, 

    An interesting answer, and essentially correct, but it does little to
    explain the paradox to anyone who might be confused.

    >       -< a variation of the much discussed Monty Hall problem >-

    No it isn't, or at least, I don't see it that way. Read on ...

    Nuisance answer 1 (change if odd): This is a real answer, not a nuisance
    answer. It's a perfectly valid observation. So sometimes you *do* gain
    by swapping.

    Nuisance answer 2 (infinite payoff by swapping): Maybe you could, but
    that's not relevant. That's not the offer that's been made. You can
    only swap once.

    Nuisance answer 3 (depends on host's behaviour): Well, it depends on
    the probability distribution of N, but not on any other aspect of
    behaviour or psychology.

    Now to the main arugument:

Absent cheating, you have no reason to assume the probabilities of these two
states are different, so you can assign them both P=1/2.

    True.

You open the box and the host offers you the chance to change.  Ignoring 3
above, neither action changes the probabilities, still both P=1/2.

    False. See nuisance answer 1. Also, there's no need to worry about 3
    above. You know you always get a chance to change, it's specified in
    advance. Assume you know the prior distribution for N. You can still
    set up the apparent paradox by an appropriate choice.

In .0, we confuse ourselves by thinking in terms of X, which is 2N in one state
of the world and N in the other state.  We get a strange answer when we add up
these different X values.

    True. The N analysis is correct. But X is a perfectly valid random
    variable. You could set up the experiment and observe it empirically. It
    may be harder to describe things clearly in terms of X, but it can be
    done. The only excuse for strange answers is muddled thinking!
    
    Can you give a correct analysis based on the known prior distribution
    of N, and the observed value of X?
    
    Andy.

Andy,

>    Can you give a correct analysis based on the known prior distribution
>    of N, and the observed value of X?

	Try 1591.3.

	Example 1 corresponds to "nuisance answer 1", and I agree with you
that it is a real issue, not "nuisance". Moreover, it can be extended. If the 
game show hosts never set N odd (to avoid people switching when they pick X 
odd), then we always switch boxes when X = 2 mod 4. And so on. It's reminiscent
of the paradox of the prisoner's execution date.

Andrew.

>Also, there's no need to worry about 3
>    above. You know you always get a chance to change, it's specified in
>    advance.

As written in .0, I don't know in advance whether I will get zero, one or many
chances to change.

If I know in advance, that would change my thinking.

> Assume you know the prior distribution for N. You can still

As written in .0, I don't know the prior distribution of N.

If I know it, then that changes my thinking, and the analysis in 1591.3 looks
correct to me.

    Re .-2: Thanks, Andrew. I see that 1591 covers the same problem.
    
    There are still a couple of loose ends.
    
    It is possible to construct prior distributions which lead to a strategy
    of "always change". For instance, let p(i) be the probablility the the
    smaller of the two sums of money is i.
    
    	For i odd, define p(i) = (1/2^i)/4.
    
    	For i even, write i as 2^x * y, with y odd and x maximal, and
    	define p(i) = (3/4)^x * p(y).
    
    This gives a valid probability distribution, with the property that
    2*p(i/2) < p(2i) for all i. With this distribution you really do
    increase your expected gain by changing. However, as pointed out in
    1591, your expected payoff with this distribution is infinite.
    
    In fact, it is impossible to construct a distribution with finite
    expectation that gives the "always change" property.
    
    Excercise 1: Prove this.
    
    It is of course possible to construct finite distributions with the
    "sometimes change" property.
    
    As the game show host I would like to construct a distribution wich
    minimises my expected payout. "Always choose 1 and 2" looks like a good
    bet, with a certain loss of 2 against an intellegent contestant.
    
    Exercise 2: Can I do better than this? 
    
Re .-1:
    
>As written in .0, I don't know in advance whether I will get zero, one or many
>chances to change.
    
    I still can't see the ambiguity. Ah well, the English language is a
    slippery beast.
    
>As written in .0, I don't know the prior distribution of N.
    
    True. I was just pointing out that the problem is one of mathematics,
    (how does the prior distribution influence the strategy?) rather than
    of psychology as you seemed to be implying.
    
    Andy.

    PMJI, I was born in 1947.7 -- what a coincidence!

    Yup, an amazing coincidence. You see, the author of the base note of
    this string was also the author of the original birthday note. Gotta be
    the work of an alien intelligence ...
    
    Does anyone remember those "Dr. Matrix" articles that Martin Gardner
    used to do in Scientific American, debunking numerology? I can remember
    a fine one in which he predicted Humphrey would beat Nixon to the
    Whitehouse on the grounds of the large number of symmetries of his
    inititals HHH, which also transliterate to 666, the number of the
    beast.
    
    Hmmm. I'd better post some proofs here since there don't seem to be any
    takers.
    
    Andy.