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# A paradox? of two chests

Go to solution Solved by bonanova,

## Question

You enter a room with two chests.  You know that one chest has a lot of money (but you are unsure as to which).  You know the other chest has half as much.  Being the greedy person you are you want the most money but the chest are indistinguishable from each other outside of opening and counting the contents.  You picked the first chest.  Just before you open it, the owner of the chests offers you an opportunity to switch. Should you?

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• Solution

Spoiler

Switching is not something you should do. Nor should you keep your first choice. You have no knowledge of which chest will reward you more. So flip a coin and enjoy what you get.

There is a simplistic argument that suggests you should switch. You've chosen \$x. There are equal odds that a switch doubles or halves it. So, the reasoning goes, switching gives you an expectation of \$x (2+.5)/2 = \$1.25x. It's clear this analysis is faulty because after switching it also says you should switch back!

I'll leave it to another solver to expose the flaw.

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There needs to be some sort of probability distribution of finding any given amount of money in the chests. That is, you ought to be able to say "there's an X% chance that the chest with the smaller amount of money contains between \$Y and \$Z." If you make Y be zero, then you should be able to say "there's an X% chance that the chest with the smaller amount of money contains \$Z or less."

Since the chest with the larger amount of money contains twice as much as the one with the smaller amount of money, you would also be able to say "there's an X% chance that the chest with the larger amount of money contains 2*\$Z or less". And the odds that the chest with the larger amount of money contains \$Z or less is obviously less than the odds that it contains 2*\$Z or less.

So should switch because, regardless of exactly how much money is in the chest, there exists some value for Z that proves that you're more likely to have randomly picked the chest with the smaller amount of money.

Disproof of the above, obviously wrong, hand-waving argument is left as an exercise for the reader.

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I'll give a refutation by symmetry. Mainly because Bayesian formulas are opaque to me. (Read, I tried to understand a priori distributions once.) Your first choice is random. If you switch you end up the the choice you would have made, with 50% probability, without switching. Your expectation for either envelope is \$1.5x, where \$x is the lesser of the two amounts.

OK, yeah, the faulty argument in my post above presumes a uniform probability on the real numbers for \$x. Anyone who thinks \$2 and \$Graham's Number have equal probability needs to immediately make a random deposit into my checking account. (Account number supplied upon request.)

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Here's a refutation based on the impossibility of having chests whose value has uniform probability across the real numbers (or integers.)

Spoiler

Let's look at a simple case where the numbers are random but limited. Say we have two envelopes containing some numbers of dollars in the ratio of 2:1, with no more than 8 dollars in any envelope. Further, let's choose a number at random from {1 2 3 4} for the lower number and get the higher number by doubling the lower. There are four cases, which we can represent by four pairs of envelopes, marked on the outside randomly with A B C D E F G H and containing the following amounts. (For argument's sake let's say it was only by chance that the numbers and letters correlated like this.)

Lower    Higher
A(1)     B(2)
C(2)     D(4)
E(3)     F(6)
G(4)     H(8)

This is the entire sample space. We can therefore do an exact simulation by analyzing each of the eight envelopes that it would be possible to initially choose. We keep track of the payoff from keeping and switching

-----------------------------------------
Choose  Keep(\$)   Switch(\$) Difference(\$)
-----------------------------------------
A       1         2           1
B       2         1          -1
C       2         4           2
D       4         2          -2
E       3         6           3
F       6         3          -3
G       4         8           4
H       8         4          -4
=========================================
Totals   30        30           0
=========================================

Keeping and switching each have an expected payoff of 30/8 = 3.75.

Envelopes A C E G will double if you switch, but they contain an average of \$2.50, which pays on average \$5.00
Envelopes B D F G will halve if you switch, and they contain an average of \$5.00, which pays on average \$2.50

The distribution of dollar values in all 8 cases is {1 2 2 3 4 4 6 8}.
Half of the values (1 2 3) are below the median
In three of four cases (A C E but not B) switching doubles your value.
But you're doubling a small number.

The other half of the values (4 6 8) are above the median.
In only one case (G) is the value doubled. For three cases (D F H) it's halved.

That exposes the fallacy of saying that for an envelope of whatever value, it's equally likely to be halved or doubled. That reasoning holds only if all values are equally likely to be represented. If really valuable envelopes are excluded (as in this case) or made much less probable (as plasmid discusses in his answer) then you're more likely to double a worthless envelope, and to halve a more valuable one.

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