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From: Ross A. Finlayson on 2 Oct 2009 02:22
On Oct 1, 9:41 pm, Tim Little <t...(a)little-possums.net> wrote:
> On 2009-10-01, Len <lwapn...(a)gmail.com> wrote:
>
> > Can someone offer a resolution?
>
> There are many resolutions, with different resolutions more or less
> acceptable to different people.
>
> > That is, one should be willing to pay any finite amount of money in
> > exchange for the unknown content of the envelope.  (I wish to ignore
> > utility considerations.  Or, simply assume the contents are utils,
> > as opposed to dollars.)
>
> You do have to assume that utility is unbounded above and that a
> linear weighted average is the only rational model for determining the
> utility of a probabilistic mix of utilities.  Both appear to be very
> poor assumptions.
>
> For example, they imply that there exists a positive outcome such that
> it is perfectly rational to enter a game in which you cause horrible
> painful deaths for everyone you care about, except for a 10^-100
> chance of winning the prize.
>
> - Tim

Of course it depends on risk tolerance in terms of the currency of
utils but if the envelope has eight dollars the unconditional
probability that the other envelope has less is 75%, or the sum of
1/2^-n, that being 1/(2^n) for less than three, there is a function
that shows the probability of getting a lesser envelope for each value
in the envelope. There's an assumed goal of gaining utils in their
interchange as units.

For the sum of the envelope it is known that the result of the ergodic
sequence was this sequence of positive in true/false. In the general
case of the n-sided die, it is that alphabet in the ergodic, where the
ergodic is for example the normal sequence or a sequence where the
elements in the sequences have a fixed alphabet and probability of
symbol occurrence unrelated to previous and following signals. So,
log 2 of the contents of the envelope, is the number of heads tossed
in a row.

The player gets the first envelope and reads the contents, he knows
the expected value of losing, and the probability of losing, and the
"expected" value of winning, and the probability of winning each of
those values. The player knows that tossing two heads in a row is
normal but three is an event.

Consider then where the user sees there were four heads in a row. To
do better, he would have to think whether he would eventually win or
lose and his various goals in the game. Not thinking instead
following a plan, the player sees that it's 1/2^five, 1/32,
probability of doing better, where doing better is doing at least
twice as better, with 31/32 of doing worse, with that being
partitioned into how bad.

Then in consideration of basic risk tolerance then if there's any cost
then the user could get a number that is irrational to their risk
tolerance, in terms of that for some small number of players in this
non-fundable game they would, if for example 2, or 2x2, ... people
played, where "utils" are valued linearly: someone would get so many
as to devalue them, besides busting the banker.

Now if the user gets 2 or 1 head in a row, then they almost certainly
should switch. That's because they're normal, the sequences of
events. Three heads in a row is unlikely. The user would notice it
if they were observing the sequences, that besides two heads in a row
which are unconnected, the three heads in a row mean that the user can
expand the alphabet and still represent them as separate events. That
is compared to where otherwise in the normal sequence, their wouldn't
need to be at least all the combinations of three instead of all the
combinations of two, where at each step of the event translation the
user could maintain in the record that it was a normal sequence as
opposed to some huge unexpected sequence, here Poisson. Then, it's an
indicator about how much space the user has to maintain information
about sequences in the past with the constant parameters of the
distributions. If the user was maintaining a cache on the counting,
it would only need to cache events for three or larger because for two
in a row it's no event that there would be a change in the resources
of algorithms that decide plays, where playing is a limited
resource.

The user is interested as to whether a particular sequence the gamer
drew was long enough that in a normal sequence where algorithms are
determined in their use of resources in at least a linear form but
maybe even sublinearly, that the sequence was long enough that to
collect a long enough (bit still linear) continuing sequence to ensure
it's normality towards ergodicity would take more than the product of
the linear algorithms would take in area that is their use of the
algorithm, the idea here is to indicate number-theoretic facts about
the continuous and discrete with simple explanations of non-analytic
behavior in terms of mathematical primitives like area.

Here then the notion is to determine what fundamental operations that
comprise what i.i.d. heads are enough in the envelope to oblige the
player in his decision to keep it. He's already playing the game, he
can effectively discard the known or keep it, the contents of the
other envelope is a secondary consequence.

So, if the user found that maintaining the decision on the play,
keeping up the data structures to balance the known against an unknown
(if not necessarily the already generated contents of the other
envelope). The player knows that given a fundamentally cheap decision
or flipping a coin (little investment in the decision), he's got a a
50% chance of getting an expected value. When the player then gets
anything outside the range of those values by the cheapest decisions,
then it becomes more worthwhile to either invest more in a decision or
to accept default decisions based on that expecting to make certain
decisions that have linear input find that the linear input in the
transition from the normal to the ergodic sees that being exponential
as well.

Basically playing costs the players and gamers besides the rules,
besides gaining them. Also that's quantifiable.

Regards,

Ross F.


From: Herman Rubin on 2 Oct 2009 13:21
In article <00fdef0f-b72c-409f-bc22-b804e90c5ccb(a)p10g2000prm.googlegroups.com>,
Len <lwapner2(a)gmail.com> wrote:
>Can someone offer a resolution?

>The content of each of two identical envelopes is determined by a
>=93St. Petersburg procedure=94. A fair coin is tossed repeatedly until
>heads first appears. If heads appears on the first toss, $2 will be
>placed in one envelope. If heads does not appear until the second
>toss, $4 will be placed in that envelope. If three tosses are
>required, $8 goes into the envelope, and so on, doubling the amount
>for each additional required toss of the coin. So, in general, if n
>tosses are required, $2^n goes into the envelope. The envelope is
>sealed and the expected value of the content is infinite. That is,
>one should be willing to pay any finite amount of money in exchange
>for the unknown content of the envelope. (I wish to ignore utility
>considerations. Or, simply assume the contents are utils, as opposed
>to dollars.) A similar procedure is independently employed to fill
>the second envelope. It too is sealed.

<You randomly select an envelope and observe its content. You may keep
<the content or exchange it for the other, unopened envelope. Should
<you be willing to do so?

<One could argue =93yes=94, by the above discussion. But if this were the
<case, one should always be willing to make the exchange. Why open the
<selected envelope? Just switch! By symmetry this seems ludicrous.

<Resolution?

This problem is unavoidable with infinite expected utility.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin(a)stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
From: LauLuna on 2 Oct 2009 19:45
On Oct 2, 7:21 pm, hru...(a)odds.stat.purdue.edu (Herman Rubin) wrote:
>
> <Resolution?
>
> This problem is unavoidable with infinite expected utility.

So the expected utility cannot be infinite? Are you meaning a
reductio?
From: Herman Rubin on 2 Oct 2009 20:21
In article <10074bf8-ccc5-4773-8e30-e7882fcce3ac(a)m38g2000yqd.googlegroups.com>,
LauLuna <laureanoluna(a)yahoo.es> wrote:
>On Oct 2, 7:21=A0pm, hru...(a)odds.stat.purdue.edu (Herman Rubin) wrote:

>> <Resolution?

>> This problem is unavoidable with infinite expected utility.

>So the expected utility cannot be infinite? Are you meaning a
>reductio?

This is correct. One can get all sorts of paradoxes.

Going to finitely additive measures does not help.
The Radon-Nikodym Theorem does not hold there, and
it can be shown quite easily that attempting to
approximate purely finitely additive measures by
countably additive can give essentially any result.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin(a)stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
From: jillbones on 6 Oct 2009 16:32
On Oct 1, 11:13 am, Len <lwapn...(a)gmail.com> wrote:
> Can someone offer a resolution?
>
> The content of each of two identical envelopes is determined by a
> “St. Petersburg procedure”.  A fair coin is tossed repeatedly until
> heads first appears.  If heads appears on the first toss, $2 will be
> placed in one envelope.  If heads does not appear until the second
> toss, $4 will be placed in that envelope.  If three tosses are
> required, $8 goes into the envelope, and so on, doubling the amount
> for each additional required toss of the coin.  So, in general, if n
> tosses are required, $2^n goes into the envelope.  The envelope is
> sealed and the expected value of the content is infinite.  That is,
> one should be willing to pay any finite amount of money in exchange
> for the unknown content of the envelope.  (I wish to ignore utility
> considerations.  Or, simply assume the contents are utils, as opposed
> to dollars.)   A similar procedure is independently employed to fill
> the second envelope.  It too is sealed.
>
> You randomly select an envelope and observe its content.  You may keep
> the content or exchange it for the other, unopened envelope.  Should
> you be willing to do so?
>
> One could argue “yes”, by the above discussion.  But if this were the
> case, one should always be willing to make the exchange.  Why open the
> selected envelope?  Just switch!  By symmetry this seems ludicrous.
>
> Resolution?
>
> Thanks,
>
> Len

Resolution: Open the selected envelope! If you don't
like what you find, you can always switch. If you
open the second envelope, you can't switch back to
the first.

regards, Bill J
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