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From: Brent Meeker <meekerdb.domain.name.hidden>

Date: Mon, 4 Oct 2004 22:19:42 -0000

*>-----Original Message-----
*

*>Norman Samish:
*

*>
*

*>>The "Flip-Flop" game described by Stathis Papaioannou
*

*>strikes me as a
*

*>>version of the old Two-Envelope Paradox.
*

*>>
*

*>>Assume an eccentric millionaire offers you your choice
*

*>of either of two
*

*>>sealed envelopes, A or B, both containing money. One
*

*>envelope contains
*

*>>twice as much as the other. After you choose an
*

*>envelope you will have the
*

*>>option of trading it for the other envelope.
*

*>>
*

*>>Suppose you pick envelope A. You open it and see that
*

*>it contains $100.
*

*>>Now you have to decide if you will keep the $100, or
*

*>will you trade it for
*

*>>whatever is in envelope B?
*

*>>
*

*>>You might reason as follows: since one envelope has
*

*>twice what the other
*

*>>one
*

*>>has, envelope B either has 200 dollars or 50 dollars, with equal
*

*>>probability. If you switch, you stand to either win
*

*>$100 or to lose $50.
*

*>>Since you stand to win more than you stand to lose, you
*

*>should switch.
*

*>>
*

*>>But just before you tell the eccentric millionaire that
*

*>you would like to
*

*>>switch, another thought might occur to you. If you had
*

*>picked envelope B,
*

*>>you would have come to exactly the same conclusion. So
*

*>if the above
*

*>>argument is valid, you should switch no matter which
*

*>envelope you choose.
*

*>>
*

*>>Therefore the argument for always switching is NOT
*

*>valid - but I am unable,
*

*>>at the moment, to tell you why!
*

Of course in the real world you have some idea about how much

money is in play so if you see a very large amount you infer it's

probably the larger amount. But even without this assumption of

realism it's an interesting problem and taken as stated there's

still no paradox. I saw this problem several years ago and here's

my solution. It takes the problem as stated, but I do make one

small additional restrictive assumption:

Let: s = envelope with smaller amount is selected.

l = envelope with larger amount is selected.

m = the amount in the selected envelope.

Since any valid resolution of the paradox would have to work for

ratios of money other than two, also define:

r = the ratio of the larger amount to the smaller.

Now here comes the restrictive assumption, which can be thought of

as a restrictive rule about how the amounts are chosen which I

hope to generalize away later. Expressed as a rule, it is this:

The person putting in the money selects, at random (not

necessarily uniformly), the smaller amount from a range (x1, x2)

such that x2 < r*x1. In other words, the range of possible

amounts is such that the larger and smaller amount do not overlap.

Then, for any interval of the range (x,x+dx) for the smaller

amount with probability p, there is a corresponding interval (r*x,

r*x+r*dx) with probability p for the larger amount. Since the

latter interval is longer by a factor of r

P(l|m)/P(s|m) = r ,

In other words, no matter what m is, it is r-times more likely to

fall in a large-amount interval than in a small-amount interval.

But since l and s are the only possibilities (and here's where I

need the non-overlap),

P(1|m) + P(s|m) = 1

which implies,

P(s|m) = 1/(1+r) and P(1|m) = r/(1+r) .

Then the rest is straightforward algebra. The expected values are:

E(don't switch) = m

E(switch) = P(s|m)rm + P(l|m)m/r

= [1/(1+r)]rm + [r/(1+r)]m/r

= m

and no paradox.

Brent Meeker

Received on Tue Oct 05 2004 - 02:16:15 PDT

Date: Mon, 4 Oct 2004 22:19:42 -0000

Of course in the real world you have some idea about how much

money is in play so if you see a very large amount you infer it's

probably the larger amount. But even without this assumption of

realism it's an interesting problem and taken as stated there's

still no paradox. I saw this problem several years ago and here's

my solution. It takes the problem as stated, but I do make one

small additional restrictive assumption:

Let: s = envelope with smaller amount is selected.

l = envelope with larger amount is selected.

m = the amount in the selected envelope.

Since any valid resolution of the paradox would have to work for

ratios of money other than two, also define:

r = the ratio of the larger amount to the smaller.

Now here comes the restrictive assumption, which can be thought of

as a restrictive rule about how the amounts are chosen which I

hope to generalize away later. Expressed as a rule, it is this:

The person putting in the money selects, at random (not

necessarily uniformly), the smaller amount from a range (x1, x2)

such that x2 < r*x1. In other words, the range of possible

amounts is such that the larger and smaller amount do not overlap.

Then, for any interval of the range (x,x+dx) for the smaller

amount with probability p, there is a corresponding interval (r*x,

r*x+r*dx) with probability p for the larger amount. Since the

latter interval is longer by a factor of r

P(l|m)/P(s|m) = r ,

In other words, no matter what m is, it is r-times more likely to

fall in a large-amount interval than in a small-amount interval.

But since l and s are the only possibilities (and here's where I

need the non-overlap),

P(1|m) + P(s|m) = 1

which implies,

P(s|m) = 1/(1+r) and P(1|m) = r/(1+r) .

Then the rest is straightforward algebra. The expected values are:

E(don't switch) = m

E(switch) = P(s|m)rm + P(l|m)m/r

= [1/(1+r)]rm + [r/(1+r)]m/r

= m

and no paradox.

Brent Meeker

Received on Tue Oct 05 2004 - 02:16:15 PDT

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