Envelope exchange
DcH
number of dollars is greater than or equal to $60 and less than or
equal to $1,000 with all possibilities in between equally likely.
After each person looks at their envelop (and counts the money) they
have the option of keeping the envelope or exchanging the envelop with
the other person. IF both people choose to exchange the envelop with
the other person the envelopes are exchanged. If either person chooses
to keep her envelop then no exchange takes place. In equilibrium, what
is the maximum amount of money a person will choose to exchange?
jonnie303
A person will never choose to exchange unless he holds $60 in his
envelope, assuming both players are rational.
Clearly a person with $1000 will not choose to exchange. Hence the
maximum either person could get from the other is $999. So a person
with $999 will not choose to exchange. Hence the maximum either person
could get from the other is $998. So a person with $998 will not
choose to exchange..... This argument can be repeated right down to
$60.
---------------------
jonnie303
Sevenoaks, uk
James Dow Allen
> On Dec 13, 12:42 am, DcH <ripo...@hotmail.com> wrote:
> > Two individuals receive an envelope filled with dollar bills. The
> > number of dollars is greater than or equal to $60 and less than or
> .
> A person will never choose to exchange unless he holds $60 in his
> envelope, assuming both players are rational.
So, this game was too easy. Let's add some rules.
Move 1: Player A states ("X") how much B needs to pay him to
exchange.
X may be negative, meaning A will pay B for the exchange.
Move 2: Player B can accept. Or make a counteroffer ("Y").
If B accepts A's offer, Mr. Benefactor (the same guy who stuffed
the envelopes in the first place) pays $50 each to A and B.
(For example, A receives (X+50+E_b) altogether
where E_b is original contents of B's envelope.
Move 3. If B makes a counteroffer, A declines or accepts, ending
the game. If he accepts, Mr. Benefactor pays each $50 as before.
James Dow Allen
E Z Peaces
That looks like 941 possible envelopes averaging $530. If the other
person doesn't look in his envelope, you should bet if you don't have
over $529.
That leaves 470 possibilities averaging $294.50. If the other person
has only calculated the $530 average, you should bet if you don't have
more than $294.
If the other player has calculated that, that leaves 235 possibilities
averaging $177, and you should bet if you don't have more than $176.
That leaves 117 possibilities averaging $118. You should bet if you
don't have more than $117.
If the other person has calculated that, it leaves 58 possibilities
averaging $88.50. You should bet if you don't have more than $88.
If the other person has calculated that, that leaves 29 possibilities
averaging $74. You should bet if you don't have more than $73.
If the other person has calculated that, it leaves 14 possibilities
averaging $66.50. You should bet if you don't have more than $66.
If the other person has calculated that, it leaves 7 possibilities
averaging $63. You should bet if you don't have more than $62.
If the other person has calculated that, it leaves 3 possibilities
averaging $61. You should bet if you don't have more than $60.
The recursive logic depends on what you calculated that he calculated
that you calculated that he calculated that he calculated that you
calculated that he calculated that you calculated that he calculated
that you calculated that he calculated that you calculated, etc.
What if you have $63 and he volunteers to swap? How do you know he
wouldn't bet at $64? He'd be risking $4 and might gain far more if you,
too would be willing to break out of the recursion.
Eric Sosman
> DcH wrote:
>> Two individuals receive an envelope filled with dollar bills. The
>> number of dollars is greater than or equal to $60 and less than or
>> equal to $1,000 with all possibilities in between equally likely.
>> After each person looks at their envelop (and counts the money) they
>> have the option of keeping the envelope or exchanging the envelop with
>> the other person. IF both people choose to exchange the envelop with
>> the other person the envelopes are exchanged. If either person chooses
>> to keep her envelop then no exchange takes place. In equilibrium, what
>> is the maximum amount of money a person will choose to exchange?
>
> That looks like 941 possible envelopes averaging $530. If the other
> person doesn't look in his envelope, you should bet if you don't have
> over $529.
I don't think so. Suppose you have $510. Since it's
known that the total amount of money in the game is at
most $1000, the other player has no more than $490. If
you exchange your $510, you are therefore guaranteed to
be less well off.
--
Eric Sosman
eso...@ieee-dot-org.invalid
E Z Peaces
Rich Grise
Are all the bills singles? Am I allowed to feel how thick they are?
Thanks,
Rich
Eric Sosman
> Eric Sosman wrote:
>> On 12/13/2009 7:38 PM, E Z Peaces wrote:
>>> DcH wrote:
>>>> Two individuals receive an envelope filled with dollar bills. The
>>>> number of dollars is greater than or equal to $60 and less than or
>>>> equal to $1,000 with all possibilities in between equally likely.
>>>
>>> That looks like 941 possible envelopes averaging $530. If the other
>>> person doesn't look in his envelope, you should bet if you don't have
>>> over $529.
>>> [...]
>>
>> I don't think so. Suppose you have $510. Since it's
>> known that the total amount of money in the game is at
>>
> The puzzle doesn't mention the total. It describes the contents of an
> envelope filled with dollar bills.
Ah! I see what you mean, but I don't think the O.P. made
clear what he meant by "the number of dollars." I took it to
be the total number of dollars in the game; you took it to be
a description of the contents of each envelope. Different
conclusions follow from different premises, of course.
--
Eric Sosman
eso...@ieee-dot-org.invalid
Eric Sosman
There's some confusion about what "the number of dollars"
means: It might refer to the total amount of money in the game,
or to the contents of each envelope independently. I'll try to
work things out under both assumptions, referring to them as
"the $1K game" and "the $2K game" after the maximum total amount
of money in play.
First Round
-----------
If we assume that Alice and Betty make their swap-or-hold
choices without knowledge of how the other chose, the situation
is pretty simple. Alice looks at her A dollars and computes
Bbar, the expected value of Betty's B dollars. If A is less
than Bbar Alice opts to swap, if A exceeds Bbar she elects to
hold. (If the two are equal she might go either way, which
would make a small difference in subsequent rounds if there
are any.) Betty behaves similarly.
In the $1K game, Alice knows that 0 <= B <= 1000-A. (It's
a little different if A < 60, but we'll get to the same choice
anyhow.) Bbar is then 500-A/2, and Alice will want to swap if
A is less than this, that is, if A < 1000/3. So Alice offers
to swap if 0 <= A <= 333, and holds if 334 <= A <= 1000. The
same goes for Betty.
In the $2K game the values A and B are independent, so the
value of A does not affect Bbar: It's (60+1000)/2 = 530, always.
Alice offers to swap if A < 530, declines to swap if A > 530,
and can choose either if A = 530.
Second Round
------------
The phrase "in equilibrium" suggests some sort of ongoing
process, which I'll suppose means that Alice's and Betty's
initial choices are not irrevocable. So let's suppose that
their choices are announced, and then a Great Voice from the Sky
asks "IS THAT YOUR FINAL ANSWER?" Alice and Betty now have more
information about each other's envelopes, and might use it to
revise their choices.
If Alice offered to swap in the first round while Betty chose
to hold, the situation is stable. Alice now knows that B >= 334
($1K game) or B >= 530 ($2K game), so Alice still wants to swap;
in fact, she's more eager than before. But Betty now knows that
A <= 333 ($1K) or A <= 530 ($2K), so she's now sure that she would
lose on the exchange. If the choices were opposite, the players
will persist in them indefinitely and there will be no exchange.
So let's suppose both players offered to swap. Now Alice
knows that Betty holds no more than $333 ($1K) or $530 ($2K), and
can revise the calculation of Bbar: It's (60+333)/2 = 196.5 ($1K)
or (60+529)/2 = 294.5 ($2K, cautious) or (60+530) = 295 ($2K, bold).
Alice thus lowers her swap threshold, and Betty acts similarly.
Or let's suppose they both decided to hold. Now Alice knows
that Betty holds at least $334 ($1K) or $530 ($2K), and re-estimates
Bbar: (334+1000)/2 = 667 ($1K) or (530+1000)/2 = 765 ($2K). This
causes her to raise her swap threshold for the second round; Betty
does the same.
Later Rounds
------------
If the GVS keeps offering chances to reconsider, the game is
boring. As long as Alice and Betty choose similarly on each round,
each will revise her estimate of the other's holding, will compare
her own holding to the revised estimate, and will re-choose in
light of the new information. This game can only end in two ways:
Either there will come a round where the players choose oppositely
(and this situation persists indefinitely, as discussed before),
or the ranges of possible A and B will "pinch" to certainty, and
a meaningless swap may occur if A = B. So we must suppose some
rule by which the GVS stops offering the chance to change.
One possibility is that the game ends if both Alice and Betty
reaffirm their round K choices in round K+1, that is, if neither
changes her mind. If only one of them changes her mind we are at
the opposite-choices dead end already mentioned, so in fact they
must both change their minds if the game is to continue. But this
leads to the diminishing-range scenario, where the game eventually
ends with opposite choices or with A = B. Boring.
Another possibility is that the GVS offers only a fixed number
of opportunities to reconsider, but this doesn't seem to blend
well with the "in equilibrium" phrase. This short-horizon game
ends when the two choose oppositely, or ends after both have made
the same sequence of swap-or-hold choices and all opportunities
to reconsider have been exhausted. In this case, the amount a
player is willing to swap could lie anywhere between the lowest
and highest possible holdings, with a granularity depending on
the number of reconsiderations the GVS permits. So, no fixed
answer and no "equilibrium" -- I don't think this is the game.
Bluffing
--------
Alice peers in her envelope and finds a paltry $75, so she'd
like to swap. She could offer to swap, but she thinks "Offering
to swap is tantamount to announcing a small holding, which will
make Betty reluctant to swap if she's got a bundle. I'll say I
want to hold, then Betty will think I've got a bundle myself, and
she'll be more willing to swap in a later round. Mwa-ha-haaa!"
Of course, Betty is equally devious and manipulative, and is
thinking along similar lines.
Note that with bluffing, the game does not reach a stalemate
if Alice and Betty choose oppositely. Either (or both) might
change her mind, which she would not have done in the "honest"
games explored above.
Thinking about this variation makes my head hurt. I guess
Alice should proceed by computing a function 0 <= f(A) <= 1
and choosing to swap with probability f(A) or hold with 1-f(A).
(Presumably, f is a decreasing function.) In a multi-round game
she'd use f(A,Bbar[k]) to make her choice for round k+1, drawing
a fresh random number at each round. I don't know where this leads.
--
Eric Sosman
eso...@ieee-dot-org.invalid