Monty, the Devil and Marilyn
Nichael Cramer
Due to an ill-timed short in the keyboard to your terminal you have just
died rather suddenly and, to your great amazement, find yourself in Hell.
Before you, you see three doors. Standing in front of the doors are Monty
Hall, the Devil and Marilyn Vos Savant.
The Devil approaches you and says the following:
"Greetings. Because you have led a virtuous life and resisted the evils of
C, and because I am a sporting sort, I'm going to give you a chance to buy
your way out of Hell. I shall give you $10 a day and then on our favorite
holiday, AlephNull Day, if you have enough money, you walk away a free being.
"Now there is, needless to say, one slight complication. We don't trust
you to keep track of your own money, so you have to let either Monty or
Marilyn hold the money for you. Don't worry, whatever else they are, they
are totally honest. They do, however, have somewhat different bookkeeping
schemes.
"Monty is somewhat casual in his approach. He will take all of your money
and toss it into his desk drawer. Then, at the end of each business day,
he will reach into the drawer and take out exactly one dollar --as a
service charge for his trouble.
"Marilyn however is much more fastidious and she prefers to do things up
front. Each day, as soon as Marilyn receives your ten dollars, she will
extract her service charge (nine dollars --naturally you would expect to
pay more for such service) and puts the remaining amount snugly away in the
safe behind her desk.
"Now, again, your money is perfectly safe with either one.
"Which do you choose?"
Carl J Lydick
If the safe is more secure than Monty's desk, I'd choose Marilyn. While Monty
and Marilyn might both be honest, we have no guarantee that the devil is. I'd
expect him to steal the money from Monty's desk drawer the day before AlephNull
day. Of course, if AlephNull day is infinitely far in the future, it makes no
difference which of the two you pick.
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CA...@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL
Disclaimer: Hey, I understand VAXes and VMS. That's what I get paid for. My
understanding of astronomy is purely at the amateur level (or below). So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it. If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.
Dave Seaman
>"Monty is somewhat casual in his approach. He will take all of your money
>and toss it into his desk drawer. Then, at the end of each business day,
>he will reach into the drawer and take out exactly one dollar --as a
>service charge for his trouble.
>
>"Marilyn however is much more fastidious and she prefers to do things up
>front. Each day, as soon as Marilyn receives your ten dollars, she will
>extract her service charge (nine dollars --naturally you would expect to
>pay more for such service) and puts the remaining amount snugly away in the
>safe behind her desk.
I choose Marilyn's method. Since she never removes any money from the drawer
after it is put there, I am guaranteed aleph-null dollars on aleph-null day.
With Monty's method it is quite possible to have zero dollars on aleph-null
day, or aleph-null dollars, or any amount in between.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Howard Wachtel
I'll take a shot at this amusing question:
It seems that the point of the problem is to illustrate a familiar
paradox of the infinite. Since Monty reaches into his drawer and pulls
out a dollar "at the end of each business day" from 1 to Aleph Null, for
any positive integer N one can specify the number of the business day at
the end of which dollar number N will be removed from Monty's drawer, which
is another positive integer. Thus, on "Aleph Null Day", all dollars will
have been removed from Monty's drawer and you are stuck in Hell for another
Aleph Null days.
However, with Marilyn's "bookkeeping scheme", she takes out her $9 up
front, so the $1 she puts into her safe is never removed at a later time,
so your nest egg is full on Aleph Null Day.
Should we call this the Monty Hell Problem? :-)
Howard K. Wachtel
University of Illinois at Chicago
U13...@UICVM.CC.UIC.EDU
Phone (312) 975-7970
Bernie Cosell
}In article <67...@bbn.BBN.COM>, ncr...@bbn.com (Nichael Cramer) says:
}>
}>"Monty is somewhat casual in his approach. He will take all of your money
}>and toss it into his desk drawer. Then, at the end of each business day,
}>he will reach into the drawer and take out exactly one dollar --as a
}>service charge for his trouble.
} It seems that the point of the problem is to illustrate a familiar
}paradox of the infinite. Since Monty reaches into his drawer and pulls
}out a dollar "at the end of each business day" from 1 to Aleph Null, for
}any positive integer N one can specify the number of the business day at
}the end of which dollar number N will be removed from Monty's drawer, which
}is another positive integer. Thus, on "Aleph Null Day", all dollars will
}have been removed from Monty's drawer and you are stuck in Hell for another
}Aleph Null days.
Perhaps, but your conclusion isn't immediately obvious. The typical
presentation of the paradox is just as you present it, but Nichael _very_
carefully phrased Monty's procedure so it is not so clear what happens.
The problem is that you have to deal with probabilities and a decreasing
series, and the matterof the limite of the series. Roughly [I tried doing the
real calculation, and it was FAR messier than I was willing to plow through],
Dollar bill "N" has a 1/N chance of being taken on the one day. On the
next day, there are N+9 billsin in the drawer, on the third subsequent day
there are N+18, etc. Now, we can ask "what is the prob that this bill is still
in the drawer on day aleph null? Well, P(surviving one day) = (N-1)/N.
P(surviving second day | survived first day) = (N+8)/(N+9), P(surv 3 | surv 2)
= (N+17) / (N + 18). And in general,
P(surv day K | surv day K-1) = (N + (k-1)*9 -1) / (N + (k-1)*9)
P(surv) = PI[all of the conditional probabilities].
Is that limit always zero? I dunno --- I assume there is some easy
demonstration that it is,but this kindof thing isn't one of my strong
points.. In any event, someone ought to provide the proof and lay the
matter to rest [my intuition is tha tht eprob *is* zero, but it is less
clear to me that if I put in the money faster that it stays zero. For
example, if I put in ten dollars the first day, one hundred the second,
one thousand the third, etc, the prob might well be non-zero]
/Bernie\
Matt Ender
Beg pardon for picking nits, but do you think Omega Day (from the
theory of transfinite numbers) would be more imporant?
Now, not to waste my bandwidth on such a tiny nit, here's the amusing
responses to The Devil, Monty Python [sic], and Marilyn Monroe [sic]
problem. 1) The Devil knows you're a mathematician, and that you
know all about Aleph Null. AlephNull Day has nothing to do with Aleph
Null, but is in fact next Tuesday after lunch. The 'have enough
money' requirement is $9.
2) AlephNull Day is when you think it is. But you will need Aleph One
dollars to get out of Hell...
3) The Devil likes you. You only need enough money on Aleph Null Day
to place a 25 cent call to the diety/ies that will take you out of
hell. Of course, to put a twist on this one, "assume 5% inflation per
annum."
4) The Devil is lying. AlephNull Day is _not_ his/her favorite
holiday. Valentine's day is, and you get to puzzle until AlephNull
Day trying to figure out _why_. (For a hint, read alt.angst)
;)
-- Matt
Matthew T. Russotto
>4) The Devil is lying. AlephNull Day is _not_ his/her favorite
>holiday. Valentine's day is, and you get to puzzle until AlephNull
>Day trying to figure out _why_. (For a hint, read alt.angst)
It's probably the #1 suicide day.
--
Matthew T. Russotto russ...@eng.umd.edu russ...@wam.umd.edu
Your superior intellect is no match for our puny weapons! -- The Simpsons
Just say NO to police searches and seizures. Make them use force.
(not responsible for bodily harm resulting from following above advice)
David Moews
cos...@cosell.bbn.com (Bernie Cosell) writes:
|}In article <67...@bbn.BBN.COM>, ncr...@bbn.com (Nichael Cramer) says:
|}>
|}>"Monty is somewhat casual in his approach. He will take all of your money
|}>and toss it into his desk drawer. Then, at the end of each business day,
|}>he will reach into the drawer and take out exactly one dollar --as a
|}>service charge for his trouble.
[assuming below that Monty removes a randomly chosen dollar on each day]
|... Roughly [I tried doing the
|real calculation, and it was FAR messier than I was willing to plow through],
|Dollar bill "N" has a 1/N chance of being taken on the one day. On the
|there are N+18, etc. Now, we can ask "what is the prob that this bill is still
|in the drawer on day aleph null? Well, P(surviving one day) = (N-1)/N.
|P(surviving second day | survived first day) = (N+8)/(N+9), P(surv 3 | surv 2)
|= (N+17) / (N + 18). And in general,
| P(surv day K | surv day K-1) = (N + (k-1)*9 -1) / (N + (k-1)*9)
|
|P(surv) = PI[all of the conditional probabilities].
|
|Is that limit always zero? I dunno --- I assume there is some easy
|points.. In any event, someone ought to provide the proof and lay the
|matter to rest [my intuition is tha tht eprob *is* zero, but it is less
|clear to me that if I put in the money faster that it stays zero. For
|example, if I put in ten dollars the first day, one hundred the second,
|one thousand the third, etc, the prob might well be non-zero]
Right on both counts. If there are a[j] dollar bills in the drawer on
day j, then the probability of a bill we put in on day k surviving
forever is the product (1 - 1/a[k])(1 - 1/a[k+1])(1 - 1/a[k+2])...,
as you point out. This product is 0 iff the sum of the 1/a[j]'s
for j>=k diverges, i.e., iff the sum of all the 1/a[j]'s diverges.
In our case a[j] grows linearly with j so this sum diverges since
the sum 1+1/2+1/3+... diverges; hence the drawer is empty on
day omega with probability 1. However if we were to put in
10^n (or even just n) bills on day n, the sum of the 1/a[j]'s would
converge and each bill would have a probability exceeding some fixed
positive value of surviving, so we would have an infinite amount
of money on day omega, again with probability 1.
--
David Moews mo...@math.berkeley.edu
Keith Ramsay
>P(surv day K | surv day K-1) = (N + (k-1)*9 -1) / (N + (k-1)*9)
>P(surv) = PI[all of the conditional probabilities].
>Is that limit always zero?
Yes, it is. Take the log of the term (N+9(k-1)-1)/(N+9(k-1))
=1-1/(N+9(k-1)). By Taylor's theorem, for log (1+x)=x-x^2/2+..., this
will be between -1/(N+9(k-1)) and -2/(N+9(k-1)) for large enough k.
The sum of -1/(N+9(k-1)) goes to minus infinity because it is
comparable to a harmonic series, sum of -1/k. The infinite product is
said to "diverge to 0" (i.e., converge to 0 (!), but not by one of the
terms being 0), because the sequence of logs of partial products
diverges to minus infinity. Hence if Monty is being quite random, the
probability of any dollar bill surviving to infinity is zero.
Keith Ramsay
The Twice Upon A Time Video Dood
>and toss it into his desk drawer. Then, at the end of each business day,
>he will reach into the drawer and take out exactly one dollar --as a
>service charge for his trouble.
>
>"Marilyn however is much more fastidious and she prefers to do things up
>front. Each day, as soon as Marilyn receives your ten dollars, she will
>extract her service charge (nine dollars --naturally you would expect to
>pay more for such service) and puts the remaining amount snugly away in the
>safe behind her desk.
>
>"Now, again, your money is perfectly safe with either one.
>
>"Which do you choose?"
At first this question seemed intuitively obvious. I figured that
you should pick Monty, cos that way you get nine dollars a day,
you'll be out of hell that much quicker.
Then I figured there had to be a trick, so I ran through a few days,
just to see what happens. After ten days, here's what you get.
Day: Monty: Total: Marilyn: Total:
1 10 10-1=9 10-9=1 1
2 10 19-1=18 10-9=1 2
3 10 28-1=27 10-9=1 3
4 10 37-1=36 10-9=1 4
5 10 46-1=45 10-9=1 5
6 10 55 10-9=1 6 (this is saturday, monty extracts
no surcharge)
7 10 65 10-9=1 7 (ditto)
8 10 75-1=74 10-9=1 8
9 10 84-1=83 10-9=1 9
10 10 93-1=92 10-9=1 10
Even if Monty takes his dollar out o day siz and seven, you end up
with ninety dollars to Marilyn's ten. Is there something I'm missing?
It seems too easy.
-lightnin
/| Lightning: (n) A Force of Nature consisting of an electric discharge.
/ |_DeityCo: The Major and Minor Deities Placement Agency
/_ /=======================================================================
/ / |Up the long ladder, and down the short rope, |Email:
/ /_ |to Hell with King Billy, and God bless the Pope, | lightnin
/_ / |if that doesn't do, we'll tear him in two, | @wpi.wpi.edu
/ / |and send him to Hell with his red, white, and blue.|__________________
/ / | -Traditional Irish Children's Song
/ /===========================================================================
|/ "Dark is not one of my favorite colors." -Ralph, the All Purpose Animal.
Nichael Cramer
Maybe another way to state the Devil's proposition (which has the
advantages of removing the probabilistic issues and of underscoring the
apparent paradox):
"...Each morning we will add $10 to your pile of money. Then, as the
service charge, either
1] Marilyn will remove $9 from the top of the pile
or
2] Monty will remove $1 from the bottom of the pile..."
N
Matthew T. Russotto
This is a different proposition, and there is no paradox here. In either case,
Joe Damnedsoul ends up with alephnull dollars on alephnull day (assuming
alephnull day is what we all assume it is)
Bernie Cosell
}In article <67...@bbn.BBN.COM> ncr...@labs-n.bbn.com (Nichael Cramer) writes:
}>
}>Maybe another way to state the Devil's proposition (which has the
}>advantages of removing the probabilistic issues and of underscoring the
}>apparent paradox):
}>
}> "...Each morning we will add $10 to your pile of money. Then, as the
}> service charge, either
}>
}> 1] Marilyn will remove $9 from the top of the pile
}>
}> or
}>
}> 2] Monty will remove $1 from the bottom of the pile..."
}This is a different proposition, and there is no paradox here. In either case,
}Joe Damnedsoul ends up with alephnull dollars on alephnull day (assuming
}alephnull day is what we all assume it is)
Close, but no cigar: yes this is different and yes there is no
paradox. But they're not the same: Monty will leave you penniless on
alephnull day.
/Bernie\
Phil Alvin
Help. I've read all the Possible Spoilers and I still don't get it.
I understand limits and probabilities, but it seems to me with
Monty every day you get $10 and every day he takes one. Granted
this may last forever and Monty keeps taking a bill out, but then
you keep giving him more than what he takes out. What am I missing
here?
frustrated,
S
Matthew T. Russotto
Upon further reflection, it looks like on day N, with you have
(10 - 9)n dollars,
whereas with Monty, you have
10n - n dollars.
On alephnull day, you have
lim (10-9)n == alephnull dollars with marylin and
n->alephnull
lim 10n - lim n = alephnull - alephnull = (could be anything)
n->alephnull
How do you get that Monty leaves you penniless?
Bernie Cosell
}In article <khiam0...@cosell.bbn.com> cos...@cosell.bbn.com (Bernie Cosell) writes:
}>russ...@eng.umd.edu (Matthew T. Russotto) writes:
}>
}>}In article <67...@bbn.BBN.COM> ncr...@labs-n.bbn.com (Nichael Cramer) writes:
}>}>
}>}>Maybe another way to state the Devil's proposition (which has the
}>}>advantages of removing the probabilistic issues and of underscoring the
}>}>apparent paradox):
}>}>
}>}> "...Each morning we will add $10 to your pile of money. Then, as the
}>}> service charge,...
}>}>
}>}> 2] Monty will remove $1 from the bottom of the pile..."
}>
}>... But they're not the same: Monty will leave you penniless on
}>alephnull day.
}Upon further reflection, it looks like on day N,
}... with Monty, you have
}10n - n dollars.
}On alephnull day, you have
}lim 10n - lim n = alephnull - alephnull = (could be anything)
}n->alephnull
}How do you get that Monty leaves you penniless?
Because your abstraction into equations isn't proper for the specific problem
at hand. Nichael phrased it very carefully: you put bills in on top and take
them off of the bottom. Let us number the bills. So on day 1 we put in
bills 1-10, on day 2 bills 2-20. On day N we put in bills 10N-9->10N.
In addition, we are told that Monty takes out the bottom bill. So Monty
takes out bill 1 on day 1; bill 2 on day 2; ... bill N on day N.
So now we can ask the question: what bill could possibly be left in the
drawer on day alephnull? The answer is *no* bill: every bill will have
been removed. Hence the drawer must be empty.
/Bernie\
Matt Crawford
That cad! That cur! That montybank!
Rob Hutchings
I think we've all been assuming the standard related paradox, which goes as
follows:
Suppose all the dollar bills are numbered.
Suppose the ones we put in on day n are numbered (10n-9) thru (10n).
Suppose on day n Monty takes out the bill numbered (n).
Now think which bills can possibly be left in after infinity days.
RobH
Rob Hutchings
FALLACY! you have only proved that each individual dollar bill is taken with
probability 1, i.e. survives with probability 0. All that follows from this
is that the expected amount of money in the drawer on day omega is
(the total number of bills put into the drawer) times (the probability that
each bill survives), which is indeterminate --- RobH
Dave Seaman
>FALLACY! you have only proved that each individual dollar bill is taken with
>probability 1, i.e. survives with probability 0. All that follows from this
>is that the expected amount of money in the drawer on day omega is
>(the total number of bills put into the drawer) times (the probability that
>each bill survives), which is indeterminate --- RobH
Probability is countably additive. The probability of a countable union of
null events still has a well-defined probability of exactly zero.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Matthew T. Russotto
Huh? Suppose I have a rational spinner-- that is, a spinner with a pointer
which can point to any rational number between 0 and 1. Certainly the
probability that the spinner will point to any particular rational number
is 0, but the probability that the pointer will point to a rational number
between 0 and 1 is 1.
Dave Seaman
>In article <25...@mentor.cc.purdue.edu> a...@seaman.cc.purdue.edu (Dave Seaman) writes:
>>Probability is countably additive. The probability of a countable union of
>>null events still has a well-defined probability of exactly zero.
>
>Huh? Suppose I have a rational spinner-- that is, a spinner with a pointer
>which can point to any rational number between 0 and 1. Certainly the
>probability that the spinner will point to any particular rational number
>is 0, but the probability that the pointer will point to a rational number
>between 0 and 1 is 1.
And then you can use a bijection between the integers and the rationals in
[0,1] to establish a uniform probability on the integers, such that all
integers are equally likely to be chosen.
Right.
Probability theory is derived from measure theory, and measure is countably
additive. The measure of the rationals in [0,1] is zero, not one. There is no
way to normalize the measure so that you get one.
--
Dave Seaman
a...@seaman.cc.purdue.edu
David G. Caraballo
>>
>>The Devil approaches you and says the following:
>>
>>"Greetings. Because you have led a virtuous life and resisted the evils of
>>C, and because I am a sporting sort, I'm going to give you a chance to buy
>>your way out of Hell. I shall give you $10 a day and then on our favorite
>>holiday, AlephNull Day, if you have enough money, you walk away a free being.
The Devil WOULD say something like this! We can count days with positive
integers: 1,2,3,4,5,...
To say that we ever reach Aleph_nul Day is tantamount to saying that Aleph_nul
is a positive integer. (We were counting, adding one for each day, and this
process ended in finite time.) This is already a problem, since, given any
positive integer N, the integer N+1 is strictly greater than N, whereas
Aleph_nul + 1 = Aleph_nul, and so Aleph_nul cannot be an integer. Hence, we
can never reach Aleph_nul day, and we see where the Devil gets his
reputation. :)
If, however, the Devil specified a finite amount of money, then either
Marilyn's scheme or Monty's scheme will work, but Monty's would be nine times
as fast. No doubt, though, the Devil wants Aleph_nul dollars.
With either scheme, the amount of money saved by day d, namely 9d or d,
depending on the bookkeeper, will approach Aleph_nul as d-> Aleph_nul, but it
will never reach Aleph_nul, since d can never reach Aleph_nul. It looks like
it's time to make another deal with the Devil...
Here goes...
Mr. Devil, let's assume that I will have enough money on Aleph_nul day to pay
you Aleph_nul dollars. I can save at most $9 per day, so that on the day
before Aleph_nul day, I'd have at least Aleph_nul - 9 dollars, but this equals
Aleph_nul, so I should be released no later than the day before Aleph_nul day.
The day before that, I had at least Aleph_nul - 9 = Aleph_nul dollars, so I
should be released 2 days before Aleph_nul day. Proceeding in this fashion,
it is clear that you should release me right away.
Devil: Oh dear, I hadn't thought of that... (and promptly vanishes in a puff
of illogic) (ala Doug Adams...)
David
David G. Caraballo
>and toss it into his desk drawer. Then, at the end of each business day,
>he will reach into the drawer and take out exactly one dollar --as a
>service charge for his trouble.
I tried, but I couldn't resist responding to this one. With either
bookkeeper, you'll have enough money at Aleph_nul day -- not that it will do
you any good since one can never reach that day! There seems to be no doubt
that Marilyn's scheme of saving $1 of each of your $10 will give you enough
money eventually. Monty's scheme can be seen as a paradox, but, as written,
it is not a true paradox.
Let f(t) be the number of dollars Monty has saved for you by day t. Clearly,
f(t) = 9t, since he puts aside nine dollars each day. Now, one can look at
the problem differently. One can define functions in(t) and out(t),
representing the amount of money put into the drawer and taken out, resp.
Again, it is clear that in(t) = 10t, and out(t) = t. Here's the problem.
(All limits below are for t--> Aleph_nul )
Clearly, lim f(t) exists.
IF lim in(t) and lim out(t) exist, and if lim in(t) - lim out(t) is not an
indeterminate form (such as infinity minus infinity), then
lim f(t) = lim in(t) - lim out(t). However, lim in(t) - lim out(t) IS an
indeterminate form, to wit infinity minus infinity, and so this limit is
meaningless (as opposed to zero).
Fortunately, lim f(t) exists. It equals lim 9t = infinity. The apparent
paradox results from (incorrectly) equating lim f(t) with lim in(t)- lim out(t)
and claiming that the latter limit is zero. When subtracting infinity, one
can make statements like "given any particular dollar, it will eventually be
removed...", since it seems that anything minus infinity should be zero, but
this is not the case.
Here's a different way to view the problem. Start numbering the dollars, and
let T(t) be the set of numbers of our dollars at day t. (f(t) equals the
number of elements in T(t), by construction.)
T(1) = {2,3,4,5,6,7,8,9,10}, since a bill, say #1, was removed on day 1.
T(2) = {3,4,5,6,...,19,20}, since a second bill, say #2, was removed on day 2.
T(3) = {4,5,6,7,...,29,30}, since a third bill, say #3, was removed on day 3.
and so on. Clearly, T will not ever decrease in size, and since it increases
in a continuous way, it can never become an empty set.
NOTE: The numberings on the removed bills were arbitrary.Whichever bill
happens to be removed first, we can call it #1 and label the others 2 - 10.
Whichever is removed second, we can call it #2 and label the remaining bills
3 to 20, and so on.
>"Now, again, your money is perfectly safe with either one.
>
>"Which do you choose?"
Either one will work, but Monty seems to be more honest than Marilyn, so I'd
choose him.
David
Rob Hutchings
The argument can, however, be restored in the following way.
Let the dollar bills be numbered in the order they are put into the drawer,
and let X at any time be the number of the lowest numbered bill in the drawer.
On day n, the probability of Monty choosing the lowest numbered bill is
1/(9n+1) and therefore the expected value of X increases by at least that
amount (possibly more., but that doesn't matter). Hence the expected value of X
on day omega is at least the sum of the series sum(1/9n+1) which is divergent.
I.e. X increases without limit and Monty leaves you penniless. ---RobH
Joe Keane
Nothing. These people are confused, even David Moews. The result of having
nothing left at the end is not only counter-intuitive; it's just plain wrong.
The last time this came up, it was numbered balls in an urn. I'd say that we
can take it as an axiom that if you remove a ball, it doesn't matter *which*
one you remove, as far as the number of balls is concerned. So it makes no
sense to say that there are an infinite number left if you always take out the
highest-numbered one, but there are none left if you always take out the
lowest-numbered one.
Or, look at the last ball that you put in. Certainly it's still there, or if
you took it out last, then another one you put in at the same time must still
be there. So there must be at least one ball left, and in fact nine out of
every ten that you put in are still there.
So where have these people gone wrong? Simple. Let's look at that last ball.
What number does it have on it? Well, it obviously must be infinite. But uh
oh, they haven't been considering that case. What they're talking about is
whether there are any balls left at the end *with a finite number on them*.
It's interesting to see whether this is likely or not under various
conditions. But you shouldn't claim that there aren't *any* left.
--
Joe Keane, amateur mathematician
j...@osc.com (uunet!amdcad!osc!jgk)
Keith Ramsay
>The Devil WOULD say something like this! We can count days with positive
>integers: 1,2,3,4,5,...
>
>To say that we ever reach Aleph_nul Day is tantamount to saying that
>Aleph_nul is a positive integer.
This is assuming the usual (plausible) assumptions about the nature of
time, which are being (implicitly) dropped for the sake of the puzzle.
There's no *mathematical* reason why there couldn't be an infinite
sequence of days, followed by another day afterward (forever and a
day!), however strange this may seem.
In article <51...@osc.COM> Joe Keane <j...@osc.com> writes:
>Nothing. These people are confused, even David Moews. The result of having
>nothing left at the end is not only counter-intuitive; it's just plain wrong.
I beg to differ. If we assume that any sequence of events requires
only finitely many days, then naturally the whole scenario is
impossible. If we don't assume that, then your reasoning doesn't hold
up.
>The last time this came up, it was numbered balls in an urn. I'd say that we
>can take it as an axiom that if you remove a ball, it doesn't matter *which*
>one you remove, as far as the number of balls is concerned.
If X and Y are sets with the same cardinality, x a member of X, y a
member of Y, then X-{x} and Y-{y} have the same cardinality, yes. This
shows that if two urns have the same number of balls on day n, and
if between day n and day n+1 one adds 10 balls to both, and takes 1
from each, that they will have the same number of balls on day n+1 as
well.
*By induction*, this shows that two such urns which have the same
number of balls added and taken away each day, and which started with
the same number of balls on day 0, will have the same number of balls
after any *finite* number n of days have passed. As you know, this
form of mathematical *induction* applies to finite sequences only. (It
is, in effect, a definition of "finite".)
>So it makes no
>sense to say that there are an infinite number left if you always take out the
>highest-numbered one, but there are none left if you always take out the
>lowest-numbered one.
It certainly does make sense. The assumptions are that balls (bills)
are individuals, that a ball (bill) will be present on day aleph-0 if
it was placed there on a finite day, and never removed on any later
finite day, and that a ball will be absent on day aleph-0 if it was
absent on some finite day, and never added again on a later day.
People also have trouble with Zeno's paradox, until it occurs to them
that there isn't necessarily any problem with there being infinitely
many events in between two events. I for one don't see any inherent
problem with there being an infinite sequence of events, followed by
more action later. Some kind of continuity assumption is required,
however, if we're to be able to tell what's going to happen.
Our solution is implicitly assuming a kind of "physics" for a world
("Hell") in which time doesn't behave the way we expect it to. Given
that it's a puzzle, these seem reasonable assumptions. Formulating a
complete "physics" for our "hell" would be harder- the assumptions
I've mentioned don't tell us what would happen if on alternate days a
ball is added and taken out of an urn, for example. This is perhaps
why we don't expect there to be any times infinitely far into the
future; our ideas of cause and effect run counter to such ideas.
But let's not get too serious after all; it's a puzzle!
>Or, look at the last ball that you put in.
There's no last ball that was put in. In a finite ordered collection,
there's always a last element, but not necessarily in an infinite
ordered collection.
>So where have these people gone wrong? Simple. Let's look at that last ball.
No, there isn't one!
>What number does it have on it? Well, it obviously must be infinite. But uh
>oh, they haven't been considering that case.
Because there is no such case!
Keith Ramsay
David Moews
|In article <61...@ut-emx.uucp> s...@ut-emx.uucp (Phil Alvin) writes:
||Help. I've read all the Possible Spoilers and I still don't get it.
||I understand limits and probabilities, but it seems to me with
||Monty every day you get $10 and every day he takes one. Granted
||this may last forever and Monty keeps taking a bill out, but then
||you keep giving him more than what he takes out. What am I missing
||here?
|
|Nothing. These people are confused, even David Moews. The result of having
|nothing left at the end is not only counter-intuitive; it's just plain wrong.
|
|The last time this came up, it was numbered balls in an urn. I'd say that we
|can take it as an axiom that if you remove a ball, it doesn't matter *which*
Hmmmm. Let me state my thoughts on this:
In the problem with Marilyn and Monty, and in this urn problem (in which,
let's say, the urn starts out empty and on day n+1 we put in balls #10n
through #10n+9 and take out ball #n,) we are given the state of the world on
day 0 (W[0]), and a procedure for deriving W[n+1] from W[n] for all finite n.
Then an inquiry is made as to some property of W[omega]. (Omega is the
number that you count just after you've counted all the natural numbers,
in case that isn't clear by now.) Now W[omega] is not strictly defined by
the conditions of the problem. Clearly it must be obtained by some
limiting process. Let N[i] be the number of balls on day i. We might
assume that our limiting process makes N continuous. Then since, for finite i,
N[i] = 9i, N[omega] must be infinite. This is the argument for having
an infinite number of balls on day omega. However, suppose we look at things
with a finer grain, and consider B[i], the set of balls in the urn on day i.
If the limiting process makes B continuous, in the sense that if a ball is
in B[i] for all big finite i it is in B[omega] and if a ball is absent
from B[i] for all big finite i it is absent from B[omega], then B[omega]
is empty and N[omega] = 0.
Evidently we have to give up continuity for either N or B. I think that
one should make the finer-grained function, namely B, continuous. (So
it does matter which ball you remove.) This is because we can easily
compute N from B but not vice versa: if B[omega] is empty, then
N[omega] is definitely 0, but if we assume N continuous and
hence make B[omega] an infinite set, it is difficult to construct B[omega] in
a non-arbitrary way. I think there might be a construction of an element
of B[omega] here:
|...Or, look at the last ball that you put in. Certainly it's still there, or
|if you took it out last, then another one you put in at the same time must
|still be there. So there must be at least one ball left, and in fact nine out
|of every ten that you put in are still there.
|
|So where have these people gone wrong? Simple. Let's look at that last ball.
|What number does it have on it? Well, it obviously must be infinite....
but it seems rather complicated. Is the idea that we take out ball #3 but
put in #30, take out #30 but put in #300, and so on? If so we would seem to
get a continuum of balls on day omega (one for balls #3, #31, #314, #3141, ...,
one for balls #2, #27, #271, #2718, ..., and many more,) but that's surely too
many. Anyway, why make trouble for yourself? Just set B[omega] to be empty.
--
David Moews mo...@math.berkeley.edu
Howard Wachtel
(David G. Caraballo) says:
>
>>In article <67...@bbn.BBN.COM>, ncr...@bbn.com (Nichael Cramer) says:
>>>
>>>The Devil approaches you and says the following:
>>>
>
>The Devil WOULD say something like this! We can count days with positive
>integers: 1,2,3,4,5,...
>
>To say that we ever reach Aleph_nul Day is tantamount to saying that Aleph_nul
>is a positive integer. (We were counting, adding one for each day, and this
>process ended in finite time.) This is already a problem, since, given any
>positive integer N, the integer N+1 is strictly greater than N, whereas
>Aleph_nul + 1 = Aleph_nul, and so Aleph_nul cannot be an integer. Hence, we
>can never reach Aleph_nul day, and we see where the Devil gets his
>reputation. :)
Let's fix this by reformulating the problem
in the following way: Suppose we fix Aleph_Null Day at a
point in the future, and suppose that the Devil gives you $10 at 1/n
of a day prior to Aleph_Null Day, for all positive integers n (and
then Monty or Marilyn disposes of the money simultaneously
in the same fashion as stated in the problem originally). Then
we don't have to worry about never reaching Aleph_Null Day.
>It looks like
>it's time to make another deal with the Devil...
>
>Here goes...
>
>Mr. Devil, let's assume that I will have enough money on Aleph_nul day to pay
>you Aleph_nul dollars. I can save at most $9 per day, so that on the day
>before Aleph_nul day, I'd have at least Aleph_nul - 9 dollars,
Wait a minute--there's no such thing as "the day before Aleph_Nul Day"!
>but this equals
>Aleph_nul, so I should be released no later than the day before Aleph_nul day.
>The day before that, I had at least Aleph_nul - 9 = Aleph_nul dollars, so I
>should be released 2 days before Aleph_nul day. Proceeding in this fashion,
>it is clear that you should release me right away.
I don't think that the inducation works in reverse like that--it would
only prove that for all positive integers n, the Devil could let you
out on day Aleph_Nul - n (if there were such a day!), but not "right
"away", as surely day 0 is not day Aleph_Nul - n for any finite n.
Matthew T. Russotto
>>In article <67...@bbn.BBN.COM>, ncr...@bbn.com (Nichael Cramer) says:
>>>
>>>The Devil approaches you and says the following:
>>>
>>>"Greetings. Because you have led a virtuous life and resisted the evils of
>>>C, and because I am a sporting sort, I'm going to give you a chance to buy
>>>your way out of Hell. I shall give you $10 a day and then on our favorite
>>>holiday, AlephNull Day, if you have enough money, you walk away a free being.
>
>
>you Aleph_nul dollars. I can save at most $9 per day, so that on the day
>before Aleph_nul day, I'd have at least Aleph_nul - 9 dollars, but this equals
>Aleph_nul, so I should be released no later than the day before Aleph_nul day.
>The day before that, I had at least Aleph_nul - 9 = Aleph_nul dollars, so I
>should be released 2 days before Aleph_nul day. Proceeding in this fashion,
>it is clear that you should release me right away.
>
>Devil: Oh dear, I hadn't thought of that... (and promptly vanishes in a puff
> of illogic) (ala Doug Adams...)
Naa, the Devil merely says "Aleph_null dollars? Who said anything about
Aleph_null dollars? I want Aleph_1 dollars."
Ian Collier
[I'm sure everyone knows the question by now, so I've deleted it]
>Upon further reflection, it looks like on day N, with you have
>(10 - 9)n dollars,
>whereas with Monty, you have
>10n - n dollars.
>On alephnull day, you have
>lim (10-9)n == alephnull dollars with marylin and
>n->alephnull
>lim 10n - lim n = alephnull - alephnull = (could be anything)
>n->alephnull
OK, I think I have now worked out the answer...
Firstly, the second limit above is not correct. If on day n you have 10n-n
dollars, then on day oo [that's short for AlephNull, by the way] you have
lim (10n-n) and NOT: lim 10n - lim n
n->oo n->oo n->oo
Distributing the limit through the subtraction is not mathematically correct
except under certain circumstances, which are not met here. Hence on day oo
you must have oo dollars.
And secondly, we can express the problem in terms of an infinite summation,
rather than a limit, and roughly the same thing happens. In general, we seem
to have no problem in deciding that the final sum of money you get from
Marilyn is
oo oo
\sum (10-9) = \sum 1 = oo
n=1 n=1
Now the problem with Monty is that people don't seem to want to write
oo
10 - 1 + 10 - 1 + 10 - 1 + ... = \sum 9 = oo
n=1
because Monty takes the dollar from the drawer instead of from the money
which is deposited. (I don't think this makes a difference). However note that
the infinite summation is not "absolutely convergent". It is an unfortunate
fact of mathematics that a summation which is not absolutely convergent may
be rearranged in order to give a whole lot of different answers, including
(in this case) +oo, -oo, 0, and so on. For example, since there are
infinitely many of the numbers 10 and -1,
10 - 1 + 10 - 1 + ... = 10-1-1-1-1-1-1-1-1-1-1 + 10-1-1-1-1-1-1-1-1-1-1 ...
equals zero. Or,
10 - 1 + 10 - 1 + ... = 10+10+10+10+... + -1-1-1-1-1-... = oo - oo
equals undefined.
But the fact is, that since we have a clearly defined order in which to add
up the summation (that is, add 10, take 1, repeat), the correct answer for
the problem is +oo.
Ian Collier
Ian.C...@prg.ox.ac.uk | i...@ecs.ox.ac.uk
Dave Seaman
>Now the problem with Monty is that people don't seem to want to write
>
> oo
>10 - 1 + 10 - 1 + 10 - 1 + ... = \sum 9 = oo
> n=1
>
>because Monty takes the dollar from the drawer instead of from the money
>which is deposited. (I don't think this makes a difference). However note that
>the infinite summation is not "absolutely convergent". It is an unfortunate
>fact of mathematics that a summation which is not absolutely convergent may
>be rearranged in order to give a whole lot of different answers, including
>(in this case) +oo, -oo, 0, and so on. For example, since there are
>infinitely many of the numbers 10 and -1,
>
>10 - 1 + 10 - 1 + ... = 10-1-1-1-1-1-1-1-1-1-1 + 10-1-1-1-1-1-1-1-1-1-1 ...
>
>equals zero. Or,
>
>10 - 1 + 10 - 1 + ... = 10+10+10+10+... + -1-1-1-1-1-... = oo - oo
>
>equals undefined.
>
>But the fact is, that since we have a clearly defined order in which to add
>up the summation (that is, add 10, take 1, repeat), the correct answer for
>the problem is +oo.
Conditional convergence and rearrangement of terms have nothing to
do with it. The reason Monty's method leaves an indeterminate
amount of money in the drawer is that it is not specified exactly
which bills he removes. Here are three possible scenarios, each
consistent with the original problem statement, giving three
different answers.
(1) On day 1, bills 1-10 go into the drawer and Monty
takes bill 1. On day 2, bills 11-20 go into the drawer
and Monty takes bill 2. On day n, bills (10n-9)
through (10n) go into the drawer and Monty takes bill
n. On aleph-null day the drawer is empty. Proof:
Suppose not. Then there must be a least number n such
that bill n is still in the drawer. But this bill
was removed on day n and never returned to the drawer,
which is a contradiction. The limit analysis fails
because it does not make use of all the available
information, and also because it depends on an incorrect
assumption, namely, that f(aleph-null) necessarily
equals lim f(n). It does not, as this example shows.
n -> oo
(2) On day 1, bills 1-10 go into the drawer and Monty
takes bill 2. On day 2, bills 11-20 go into the drawer
and Monty takes bill 3. On day n, bills (10n-9)
through (10n) go into the drawer and Monty takes bill
(n+1). On aleph-null day the drawer contains one
dollar (namely, bill number 1). All the other bills
have been removed, as explained in (1). Once again,
the limit analysis fails because it does not make use
of all the available information and because it makes
an incorrect assumption.
(3) On day 1, bills 1-10 go into the drawer and Monty
takes bill 2. On day 2, bills 11-20 go into the drawer
and Monty takes bill 4. On day n, bills (10n-9)
through (10n) go into the drawer and Monty takes bill
(2n). On aleph-null day the drawer contains aleph-null
dollars (all the odd-numbered bills). All the
even-numbered bills have been removed. Once again,
the limit analysis fails because it does not make use
of all the available information. The assumption
about limits happens to work out in this case (but
only because I chose my assumptions in such a way as
to make it happen).
--
Dave Seaman
a...@seaman.cc.purdue.edu
D. J. McCarthy ~
> Conditional convergence and rearrangement of terms have nothing to
> do with it. The reason Monty's method leaves an indeterminate
> amount of money in the drawer is that it is not specified exactly
> which bills he removes.
There's still something... I don't know... unbelievable about
determining how much money is left in the drawer solely on which dollar
is taken out every day. Yes, there are proofs, and I can follow them
and agree with the outcome, but it still leaves a bad taste in my mouth
- and I don't trust them as much as I feel they need to be.
How about a scenario that you don't mention - namely, that
Monty takes a dollar at random? On the first day the probability of
each dollar remaining is 9/10. On the second day, each of the first
ten dollars have a staying probability of, what, (9/10) * (18/19)
[this is undoubtedly incorrect] and each of the second ten dollars has
a probability of staying of 18/19. The rest of the days are too
strenuous for my flu-modified brain to figure out at the moment.
Anyone want to take this train of thought to a depot?
--
| D. J. McCarthy Qui nos rodunt confundantur
| dmc...@swtec1.intel.com et cum iustis non scribantur.
| ...!intelhf!mipos3!modl01!dmccart Io!
David R. Morrison
> The reason Monty's method leaves an indeterminate
> amount of money in the drawer is that it is not specified exactly
> which bills he removes. Here are three possible scenarios, each
> consistent with the original problem statement, giving three
> different answers.
> [proof that removing that the bottom bill => f(aleph-null) = 0]
> [proof that removing that the next to bottom bill => f(aleph-null) = 1]
> [proof that removing that the bottom even bill => f(aleph-null) = oo]
Just a few points to ponder:
Suppose there was a gremlin that came in every night and reordered the
bills. As far as Monty and you proof know, nothing has changed. But
now I can transfer from one of your proofs to the other.
If this gremlin doesn't change the problem, does it matter which bill
Monty removes?
If this gremlin does change the problem, how does it? The total number
of bills is unaffected (don't tell me it messes up your accounting...the
way you do you accounting has no affect on how many are actually there).
Does removing an infinite amount of something mean there's nothing left?
Would this Monty Hell argument work with the IRS, if you got taxed on
aleph-null day?
Joe Keane
Moews) writes:
>Hmmmm. Let me state my thoughts on this:
>Evidently we have to give up continuity for either N or B.
But continuity gives two different answers, so why do you trust it at all? We
know it works fine for problems stated with finite numbers. I think that the
reason continuity fails in this problem is that it explicitly names an
infinite ordinal, omega. So you get new interactions which you have to take
into account.
>I think there might be a construction of an element of B[omega] here [Joe's
>stuff deleted] but it seems rather complicated. Is the idea that we take out
>ball #3 but put in #30, take out #30 but put in #300, and so on?
Something like that; let me make it explicit. Let the `prefix' of a ball be
its number in decimal, but with any trailing zeros deleted. Then on each day
N we add prefixes 10*N+1 through 10*N+9, and in no case does a prefix ever go
away. In fact, if we just write the prefixes on the balls, then we can
cleverly recycle them. On each day we take a ball out, put it right back in,
and then add nine others. I figure that you can also skip the first two steps
without changing much. Now by the same continuity argument, by day omega all
finite prefixes will be represented in the urn. But, there are no balls in
it? This is the contradiction we have to resolve.
I have a counter-puzzle. On day zero i take a ball and write the letter `X'
on it. I put it in an urn and pour tons of cement over them to prevent
tampering. Then on a blackboard i write `X = 0'. Every day i come in and
increment the value for `X' i have written, but i never go near the urn. Now
on day omega, is the ball still in the urn? Have fun guys.
>Anyway, why make trouble for yourself? Just set B[omega] to be empty.
Sure.
David G. Caraballo
>In article <67...@bbn.BBN.COM> ncr...@labs-n.bbn.com (Nichael Cramer) writes:
>>
>> "...Each morning we will add $10 to your pile of money. Then, as the
>> service charge, either
>> 1] Marilyn will remove $9 from the top of the pile or
>> 2] Monty will remove $1 from the bottom of the pile..."
>This is a different proposition, and there is no paradox here. In either
>case, Joe Damnedsoul ends up with alephnull dollars on alephnull day
>(assuming alephnull day is what we all assume it is)
If we assume that Joe can reach Aleph_nul Day, then there does seem to be a
paradox in Monty's scheme. On the one hand, the number of dollars increases
with each day. On the other hand, no dollar bill can stay in the pile until
Aleph_nul Day, so that Joe cannot have any money on Aleph_nul Day. One can
deduce the statement "If Joe lives to see Aleph_nul Day then he will have no
money" (assuming he uses Monty's scheme). The problem is to reconcile this
statement with the fact that he's growing nine dollars richer per day. The
resolution is that Joe never lives to see Aleph_nul Day. (This is a form of
the result that the statement "If A then B" is automatically true whenever A
does not occur/is false.)
Dave Seaman
> How about a scenario that you don't mention - namely, that
>Monty takes a dollar at random?
In fact, I have mentioned that scenario -- several times (but probably only
once in the current incarnation of this discussion). It is fairly easy to
show that if each dollar in the drawer is equally likely to be chosen for
removal, then any given dollar has probability zero of remaining in the drawer
on aleph-null day. (Hint: express the probability as an infinite product,
and compare the logarithm of the infinite product with the sum of the harmonic
series.) By countable additivity of probability, it follows that with
probability one, the drawer is empty on aleph-null day.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Trent Tobler
> in the following way: Suppose we fix Aleph_Null Day at a
> point in the future, and suppose that the Devil gives you $10 at 1/n
> of a day prior to Aleph_Null Day, for all positive integers n (and
> then Monty or Marilyn disposes of the money simultaneously
> in the same fashion as stated in the problem originally). Then
> we don't have to worry about never reaching Aleph_Null Day.
last I checked, 1/1 + 1/2 + 1/3 + 1/4 + ... + 1/n + ... diverges to infinity.
Same problem.
> Wait a minute--there's no such thing as "the day before Aleph_Nul Day"!
Sure there is:
Aleph_Nul - 1 ( = Aleph_Nul, so Aleph_Nul is it's own successor and
predicessor)
> I don't think that the inducation works in reverse like that--it would
> only prove that for all positive integers n, the Devil could let you
> out on day Aleph_Nul - n (if there were such a day!), but not "right
> "away", as surely day 0 is not day Aleph_Nul - n for any finite n.
Ah, but we are saying, let n grow beyond any bounds, and we get
Aleph_Nul - Aleph_Nul = ????
--
Trent Tobler - tto...@csulx.weber.edu
Dave Seaman
>Suppose there was a gremlin that came in every night and reordered the
>bills. As far as Monty and you proof know, nothing has changed. But
>now I can transfer from one of your proofs to the other.
Assuming that the bills are reordered at random, this is equivalent
to the version in which the bills are unnumbered and one is randomly
removed each day. As I have explained elsewhere, this means that
with probability one, the drawer is empty on aleph-null day. Yes,
the reordering makes a difference.
>If this gremlin doesn't change the problem, does it matter which bill
>Monty removes?
The gremlin does change the problem, and yes, it matters which bill
Monty removes.
>If this gremlin does change the problem, how does it? The total number
>of bills is unaffected (don't tell me it messes up your accounting...the
>way you do you accounting has no affect on how many are actually there).
The total number of bills is unaffected on any finite day. To find
the number of bills on aleph-null day, we need some way to determine,
for any given bill, whether it is in the drawer on aleph-null day.
Then, we merely count the ones remaining. This is actually just
the continuity principle, but operating one level lower than most
people thought of applying it. For each bill b and day d, let
f(b,d) = 1 if bill b is in the drawer at the end of day d, and 0
otherwise. Then f(b,oo) = lim f(b,d) = 1 if the bill is in the
b->oo
drawer on aleph-null day, and 0 otherwise. The number we seek is
the sum of f(b,oo) for all b.
Someone else suggested a scenario in which bills that are removed
from the drawer may be recycled. If the same bill is added to the
drawer on every odd-numbered day and removed again on every
even-numbered day, then we cannot decide where that particular bill
is on aleph-null day. The value of f(b,oo) = lim f(b,d) is
b->oo
undefined. If there are bills whose position on aleph-null day is
unknown, then we cannot determine how many bills remain in the
drawer.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Herman Rubin
.....................
> How about a scenario that you don't mention - namely, that
> Monty takes a dollar at random? On the first day the probability of
> each dollar remaining is 9/10. On the second day, each of the first
> ten dollars have a staying probability of, what, (9/10) * (18/19)
> [this is undoubtedly incorrect] and each of the second ten dollars has
> a probability of staying of 18/19. The rest of the days are too
> strenuous for my flu-modified brain to figure out at the moment.
This is much easier than it looks. On the n-th day, there are <10n
dollars in the drawer (except = for n=1) before Monte takes one out.
Now the sum of 1/10n is infinite, and therefore the probability that
any particular dollar will survive is 0. Since there are only a
countable number of dollars, the probability that any will survive
is 0.
--
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
Phone: (317)494-6054
hru...@pop.stat.purdue.edu (Internet, bitnet)
{purdue,pur-ee}!pop.stat!hrubin(UUCP)
Joe Miller
around again and again with total disregard for my sanity.
My turn:
It seems to me that there will be aleph-null dollars on aleph-null days whether
you choose Marilyn or Monty. The big sticking point with Monty's accounting
method seems to be the following fact.
- For day n, and lowest bill b: As n --> oo, b --> oo.
This simply means that if the days go on without bound then the lowest numbered
bill gets higher without bound. I think we can all accept that. It is from
this seemingly harmless statement that the supporters of the Monty Leaves You
Penniless Paradox conclude:
- Because, for any finite bill n there is some day d beyond which the lowest
bill left is greater then n, therefore there can be no bills left on aleph-null
day.
This would seem to be a wonderful conclusion except that it only takes into
account the bills with finite numbers. Pandora's box has been opened. If we
assume that we can reach aleph-null day in the first place then we must ask
ourselves what numbers the last bills we added had on them. It is clear that
the numbers on the bills must themselves be infinite. My conclusion:
- On aleph-null day Monty's infamous drawer will have an infinite number of
bills all numbered with something equivalent to oo. True, the lowest bill left
will be infinite but we will have a range from oo to oo which in this case
contains aleph-null bills.
In article <51...@osc.COM> j...@osc.COM (Joe Keane) writes:
> I have a counter-puzzle. On day zero i take a ball and write the letter `X'
> on it. I put it in an urn and pour tons of cement over them to prevent
> tampering. Then on a blackboard i write `X = 0'. Every day i come in and
> increment the value for `X' i have written, but i never go near the urn. Now
> on day omega, is the ball still in the urn? Have fun guys.
I think that Joe is making a related point. On day omega the ball is still in
the urn but it has a value (symbolically) of oo.
Conclusion:
We must be very careful when we futz with infinity.
Joe aka (fa...@wam.umd.edu)
Danil Patrick Suits
way. Rather than sitting back and being smug, I thought I would
put my own thoughts up to the flame test.
Suppose that at 6 am we deposit our $10 with either banker, and
Monty ends his day at 6 pm (when he pulls a bill)
let _a_ be a sequence determined by
a(t) = the amount of money held by marilyn at 6am of day j if t=2j-1
= the amount of money held by marilyn at 6pm of day j if t=2j
similarly define _b_ by
b(t) = the amount of money held by monty at 6 am of day j if t = 2j-1
= the amount of money held by monty at 6 pm of day j if t = 2j
Now we know what _a_ looks like
1,1,2,2,3,3,4,4,5,5,6....
and we have an idea how _b_ starts at least
10,9,19,18,28,27....
Now _b_ is strictly greater than _a_ for all t. Also _a_ is a non
decreasing sequence which diverges. since b(t) > a(t) for all t
it follows that lim b(t) = lim a(t) = oo
I haven't checked my copy of Howard Anton's calculus text for a while
but I seem to remember that relation holds.
Neat problem tho.
--
A 9 5 2 --- Danil ---
K - Q 8 6 4
- J ^ da...@rice.edu
*
m...@vax5.cit.cornell.edu
(Data is taken at the end of the day after Monty takes his dollar)
bill #'s bill #'s bill #'s
Day issued (total) in drawer (total) Monty has (total)
1 1 to 10 $10 2 to 10 $ 9 1 $1
2 11 to 20 $20 3 to 20 $18 1 to 2 $2
3 21 to 30 $30 4 to 30 $27 1 to 3 $3
4 31 to 40 $40 5 to 40 $36 1 to 4 $4
5 41 to 50 $50 6 to 50 $45 1 to 5 $5
6 51 to 60 $60 7 to 60 $54 1 to 6 $6
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
n 10n-9 to 10n $10n n+1 to 10n $9n 1 to n $n
These figures will hold for any finite n.
On any given day n (n here is finite),
- Amount issued $ 10n. (bill #s 1 to 10n)
- Monty has $ n. (bill #s 1 to n)
- Drawer has $ 9n. (bill #s n+1 to 10n)
=$ 10n-n. (bill #s 1 to 10n) less (bills 1 to n)
Clearly, the amount of money in the drawer grows linearly as n increases
(from one finite number to the next.)
Limiting: what happens as n->oo? (please pardon the sloppy notation below)
- Amount issued $ 10oo. (bill #s 1 to 10oo)
- Monty has $ oo. (bill #s 1 to oo)
- Drawer has $ 9oo. (bill #s oo+1 to 10oo)
=$10oo-oo. (bill #s 1 to 10oo) less (bills 1 to oo)
This is _not_ the case:
x Amount issued $ oo. (bill #s 1 to oo)
x Monty has $ oo. (bill #s 1 to oo)
x Drawer has $ 0. (bill #s -- )
=$ oo-oo. (bill #s 1 to oo) less (bills 1 to oo)
Consider the following fallacious proof:
lim (10-1)n = (lim 10n) - (lim 1n) | $$ in drawer = $$ issued - $$ monty
n->oo n->oo n->oo |
|
oo = oo - oo | oo = oo - oo
oo = 0 | oo = 0
It _seems_ that this is duplicated by the proponents of the Marilyn theory.
O.K. That is about where I have it I still can't see why you lose with Monty.
Where have I gone wrong???
-Greg
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David R. Morrison
>>> [Argument that removing lowest numbered bill => no bills left]
>>> [Argument that removing next lowest numbered bill => 1 bill left]
>>> [Argument that removing lowest even numbered bill => oo bills left]
>>
>>Suppose there was a gremlin that came in every night and reordered the
>>bills.
>
>>Monty removes?
>
>The gremlin does change the problem, and yes, it matters which bill
>Monty removes.
Could resolve the following variation?
You have decided you don't quite trust Monty, so you seal each bill
along with a secret device in each envelope. You number these envelopes
and bills sequentially.
Now this secret device inside the envelope renumbers the bills. When
Monty removes a envelope from the drawer, the bill inside the envelope
gets renumbered to contain the lowest even number in the current lot.
The bill which was the lowest even number before Monty removed one (if
it wasn't the one removed), gets the number of the one Monty removed.
By your first argument, there will be no envelopes left.
By your third argument, there will be an infinite number of bills.
Keith Ramsay
>O.K. That is about where I have it I still can't see why you lose with Monty.
> Where have I gone wrong???
You have shown again that the amount of money goes to infinity as the
day number n goes to infinity. This is so. This does not mean that
there are any individual bills which survive, however.
This puzzle didn't explicitly contain all of the information which
would be necessary to give a rigorous solution to it. So there are
alternative "models" one could give of "hell" which would perhaps
allow for different answers.
I've been assuming dollar bills are individuals, and that the sequence
of days goes 1,2,3,..., followed by aleph-0. I've also assumed that a
bill which is either in or out of the desk on day n, and isn't either
added or taken away afterward, is still where it was put on day
aleph-0. If you think a different solution is more natural, try to
find an alternative model for how dollar bills behave in "hell".
The puzzle is very reminiscent of questions which arise in
mathematics. The monty puzzle is a lot like the "hump rolling out to
infinity" counterexample in analysis. Suppose f_n are non-negative
functions, that the integral of f_n goes to infinity as n->infinity,
and that for each x, f_n(x) tends to f(x). One might think that the
"limiting region" under the curve y=f(x) would have to have infinite
area if the areas under the curves y=f_n(x) tends to infinity, but
this reasoning is fallacious. It is possible for the f_n to tend to
zero at each x, making the area under the limiting curve y=f(x) be
zero. The area under the curve "rolls off into infinity" as it were.
The reason your reasoning doesn't apply to my model of Monty's "hell"
is the same reason it doesn't apply to limits of integrals in
mathematics.
Keith Ramsay
Dave Seaman
>It seems to me that there will be aleph-null dollars on aleph-null days whether
>you choose Marilyn or Monty. The big sticking point with Monty's accounting
>method seems to be the following fact.
>
>- For day n, and lowest bill b: As n --> oo, b --> oo.
I don't recall anyone even mentioning this fact before. The
reasoning has been that for any bill b, there is a day n such that
bill b is removed from the drawer and never returns. If you are
going to criticize an argument, you should at least try to begin
by stating the argument correctly.
[lengthy critique of the wrong argument omitted...]
>This would seem to be a wonderful conclusion except that it only takes into
>account the bills with finite numbers. Pandora's box has been opened. If we
>assume that we can reach aleph-null day in the first place then we must ask
>ourselves what numbers the last bills we added had on them. It is clear that
>the numbers on the bills must themselves be infinite.
It's not clear to me. Would you like to explain how you deduced
that? There are infinitely many finite numbers, you know. Therefore
we can have infinitely many bills, each bearing a finite number,
with no two numbers equal. Why must there be transfinite numbers
on the bills? Mind you, I am not saying it is impossible for a
bill to have a transfinite number on it. At this very moment, I
have in my pocket a bill with a transfinite number on it; I wrote
it there myself. Even if we assume, however, that there are some
bills around with transfinite numbers, the question is when would
they be added to the drawer? Every predecessor of aleph-null is
finite, and therefore every bill added before aleph-null day has
a finite number.
>My conclusion:
>- On aleph-null day Monty's infamous drawer will have an infinite number of
>bills all numbered with something equivalent to oo. True, the lowest bill left
>will be infinite but we will have a range from oo to oo which in this case
>contains aleph-null bills.
I defy you to name even ONE bill with a transfinite number that is
in the drawer on aleph-null day. What is the number? When was it
added to the drawer?
>In article <51...@osc.COM> j...@osc.COM (Joe Keane) writes:
>> I have a counter-puzzle. On day zero i take a ball and write the letter `X'
>> on it. I put it in an urn and pour tons of cement over them to prevent
>> tampering. Then on a blackboard i write `X = 0'. Every day i come in and
>> increment the value for `X' i have written, but i never go near the urn. Now
>> on day omega, is the ball still in the urn? Have fun guys.
>I think that Joe is making a related point. On day omega the ball is still in
>the urn but it has a value (symbolically) of oo.
Joe Keane's example strongly hints, but never comes right out and
says, that the ball is never removed from the urn. Assuming this
to be the case, it follows that the ball is still there on aleph-null
day. What I don't understand is what any of this has to do with
anything, unless you actually believe your own argument (as stated
above), which seems to be of the form "Any problem that mentions
infinity must have infinity as the answer." (Perhaps this is not
a fair statement of your argument, but I find that I am unable to
state your argument in a form that actually makes sense. Corrections
are gladly accepted.)
>Conclusion:
> We must be very careful when we futz with infinity.
The defense rests.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Joe Miller
[lengthy critique of my lengthy critique of the wrong argument omitted...]
>> ...Pandora's box has been opened. If we
>>assume that we can reach aleph-null day in the first place then we must ask
>>ourselves what numbers the last bills we added had on them. It is clear that
>>the numbers on the bills must themselves be infinite.
>
> It's not clear to me. Would you like to explain how you deduced
> that? There are infinitely many finite numbers, you know. Therefore
> we can have infinitely many bills, each bearing a finite number,
> with no two numbers equal. Why must there be transfinite numbers
> on the bills? Mind you, I am not saying it is impossible for a
> bill to have a transfinite number on it. At this very moment, I
> have in my pocket a bill with a transfinite number on it; I wrote
> it there myself. Even if we assume, however, that there are some
> bills around with transfinite numbers, the question is when would
> they be added to the drawer? Every predecessor of aleph-null is
> finite, and therefore every bill added before aleph-null day has
> a finite number.
Hmmm... Every bill has a finite number? How about this.
X=the highest bill ever added to the drawer (which Dave tells us is finite and
Dave is an honorable mathematition.)
But bill X was added to the drawer on day D=Trunc(X/10)+1. Right?
This leads to two possible conclusions:
1) Day D+1 is aleph-null day. This goes against the general assumption
that the predecessor of a finite number is finite.
2) Day D+1 is a finite day and the Devil gives Joe ten more bills with
numbers greater then X.
Both are contradictions. Therefore the highest bill ever added to the drawer
cannot have a finite value. The same can be said of the 2nd highest, the 3rd
highest, the 4th highest ... the 987,445,345th highest ... etc.
>>My conclusion:
>>- On aleph-null day Monty's infamous drawer will have an infinite number of
>>bills all numbered with something equivalent to oo. True, the lowest bill
>>left will be infinite but we will have a range from oo to oo which in this
>>case contains aleph-null bills.
[absurd demands deleted]
[Joe Keane's example deleted, yet still appreciated]
[Blah, blah blah. Oh, we've reached the end]
I am still of the unshaken belief that Monty's drawer will contain aleph-null
bills on aleph-null day. It appears however that I am in the minority. Oh
well, that's nothing new.
Joseph Miller aka (fa...@wam.umd.edu)
David Moews
| mo...@math.berkeley.edu (David Moews) writes:
|#Hmmmm. Let me state my thoughts on this:...
|#Evidently we have to give up continuity for either N or B.
|
|But continuity gives two different answers, so why do you trust it at all? We
|know it works fine for problems stated with finite numbers. I think that the
|reason continuity fails in this problem is that it explicitly names an
|infinite ordinal, omega. So you get new interactions which you have to take
|into account.
I don't understand. Continuity gives no restriction for problems that only
use, as times, the finite ordinals (= natural numbers), since in this case the
time topology is discrete. So it can only be used when an infinite ordinal
is present, which is the case when you say it fails. Anyway, isn't continuity
of N the basis of your conviction that N[omega] is infinite?
|...Let the `prefix' of a ball be
|its number in decimal, but with any trailing zeros deleted. Then on each day
|N we add prefixes 10*N+1 through 10*N+9, and in no case does a prefix ever go
|away. In fact, if we just write the prefixes on the balls, then we can
|cleverly recycle them. On each day we take a ball out, put it right back in,
|and then add nine others. I figure that you can also skip the first two steps
|without changing much. Now by the same continuity argument, by day omega all
|finite prefixes will be represented in the urn. But, there are no balls in
|it? This is the contradiction we have to resolve.
It is important which balls you take out and add, but unimportant what
the balls are labelled; it is not important if you rewrite the labels,
either (to answer one poster's hypothesis of a bill renumberer.) After all,
the balls *could* all be blank, and Monty could take one out at random, and
the urn would probably still be empty on day omega. So, if you take a ball
out and put another one in with the same label, it's not the same as if you
took the ball out and put the same ball back in. In the first case, the
urn will be empty; in the second case, the urn might be full, although the
fact that each ball is outside the urn infinitely many times makes me
uncertain.
There may be a second argument in the text I included above, to the effect that
if we define a function Pre such that Pre[M,D] is the number of balls with
prefix M in the urn on day D, and if we add and remove balls as we originally
did, then, for each M, Pre[M,D] always becomes 1 for all big enough D, so, by
continuity, Pre[M,omega] should be 1 for all M. I don't like this argument
for the same reason I didn't like the argument from the continuity of N. Since
Pre is essentially defined in terms of B, which is more specific and,
physically, more relevant, continuity of B is more important than continuity
of Pre.
|... I have a counter-puzzle. On day zero i take a ball and write the letter
|`X' on it. I put it in an urn and pour tons of cement over them to prevent
|tampering. Then on a blackboard i write `X = 0'. Every day i come in and
|increment the value for `X' i have written, but i never go near the urn. Now
|on day omega, is the ball still in the urn? Have fun guys.
Sure, it's there. The label doesn't matter.
--
David Moews mo...@math.berkeley.edu
Edward C. Hook
>In article <25...@mentor.cc.purdue.edu> a...@seaman.cc.purdue.edu (Dave Seaman)
>writes:
>> finite, and therefore every bill added before aleph-null day has
>> a finite number.
>
>Hmmm... Every bill has a finite number? How about this.
>X=the highest bill ever added to the drawer (which Dave tells us is finite and
>Dave is an honorable mathematition.)
... Then, he probably wouldn't make THIS mistake. Why do you think that this
'X' of which you speak exists ??? ( Hint: It need not. The integers are
well-ordered, so every non-empty set of integers has a SMALLEST element,
but there certainly are such sets which have NO largest element. The
remainder of the argument vanishes, in a wispy little puff of smoke ...8^)
===============================================================================
Ed Hook | "I have yet to see any problem,
Computer Sciences Corporation | however complicated, which, when
NAS | you looked at it in the right
NASA/Ames Research Center | way, did not become still more
Mountain View, CA 94039 | complicated." - Poul Anderson
+1 415 604 1272 |-----------------------------------
Internet: ho...@nas.nasa.gov | They're MY opinions ... ALL mine!!
===============================================================================
David Seal
>This simply means that if the days go on without bound then the lowest numbered
>bill gets higher without bound. I think we can all accept that. It is from
>this seemingly harmless statement that the supporters of the Monty Leaves You
>Penniless Paradox conclude:
>
>- Because, for any finite bill n there is some day d beyond which the lowest
>bill left is greater then n, therefore there can be no bills left on aleph-null
>day.
>
>This would seem to be a wonderful conclusion except that it only takes into
>account the bills with finite numbers. Pandora's box has been opened. If we
>assume that we can reach aleph-null day in the first place then we must ask
>ourselves what numbers the last bills we added had on them. It is clear that
>the numbers on the bills must themselves be infinite. My conclusion:
>
>- On aleph-null day Monty's infamous drawer will have an infinite number of
>bills all numbered with something equivalent to oo. True, the lowest bill left
>will be infinite but we will have a range from oo to oo which in this case
>contains aleph-null bills.
No, this won't do. The reason we conclude that there is no finitely-numbered
bill left is that every one of them has been put into the drawer on a
finitely-numbered day, but removed again on a subsequent finitely-numbered
day.
But we can also conclude that there are no infinitely-numbered bills in the
drawer, for an even simpler reason: they cannot have been put into the
drawer in the first place! We've added money to the drawer and let Monty
take his cut on every finitely-numbered day. The bills added on day n are
numbered 10n-9 to 10n; if n is finite, so are all these numbers. So we've
only added finitely-numbered bills to the drawer on finitely-numbered days.
The only possibility left is that we've added infinitely-numbered bills to
the drawer on infinitely-numbered days. But we know that we stop at day
AlephNull, which is the first infinitely-numbered day. So we *didn't* add
bills to the drawer on any infinitely-numbered day...
By the way, "day AlephNull" should really be called "day Omega" - these are
ordinal numbers we're talking about, not cardinal numbers. And remember that
subtraction is not always defined on the ordinals: "day Omega-1" is not a
meaningful concept.
Note that your question about what the numbers on the last bills we added
are is also not meaningful: there are no such "last bills". Asking to find
them is like asking to be shown the largest positive integer, or the largest
real number that is strictly less than 2: whatever candidate you mention, a
larger/later one can be found.
>Conclusion:
> We must be very careful when we futz with infinity.
Agreed :-) In particular, almost any intuitive idea of "infinity" is likely
to be wrong, in the sense that it can quickly be made to contradict itself.
If you want to deal with the idea of "infinity" safely, you really have to
use mathematics. And it's worth bearing in mind the fact that mathematicians
took a long time to come up with safe theories of the infinite - Zeno's
paradox shows that it was a recognised problem in Greek times, and major
aspects of the theory were still being worked out this century. Indeed, they
are still being worked out, and still lead to heated debate among
mathematicians - for instance, some mathematicians find the results so
absurd that they reject the idea that "infinity" has any sensible meaning at
all...
David Seal
ds...@armltd.co.uk
All opinions are mine only...
Dave Seaman
>X=the highest bill ever added to the drawer (which Dave tells us is finite and
>Dave is an honorable mathematition.)
>But bill X was added to the drawer on day D=Trunc(X/10)+1. Right?
>This leads to two possible conclusions:
> 1) Day D+1 is aleph-null day. This goes against the general assumption
> that the predecessor of a finite number is finite.
> 2) Day D+1 is a finite day and the Devil gives Joe ten more bills with
> numbers greater then X.
>Both are contradictions. Therefore the highest bill ever added to the drawer
>cannot have a finite value. The same can be said of the 2nd highest, the 3rd
>highest, the 4th highest ... the 987,445,345th highest ... etc.
Actually, your argument was entirely correct right up to your
conclusion. The correct conclusion is:
Both are contradictions. Therefore there cannot be any
such thing as a "highest bill ever added to the drawer."
Also, I find it highly interesting that when I challenged you to
actually name a bill that is left in the drawer,
>>I defy you to name even ONE bill with a transfinite number that is
>>in the drawer on aleph-null day. What is the number? When was it
>>added to the drawer?
you dismissed this with
>[absurd demands deleted]
As the lion said to the elephant, "Just because you can't answer
the question is no reason to get angry." Transfinite numbers do
have names, you know. Here is one: omega+10. Unfortunately, that
bill does not get added to the drawer until the day AFTER aleph-null
day (which really should be omega day). The bills omega through
omega+9 would be added on omega day, but only AFTER the devil comes
around to collect and finds an empty drawer.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Joe Keane
>drawer is that it is not specified exactly which bills he removes.
Bleah.
>On day 1, bills 1-10 go into the drawer and Monty takes bill 1. On day 2,
>bills 11-20 go into the drawer and Monty takes bill 2. On day n, bills
>(10n-9) through (10n) go into the drawer and Monty takes bill n. On
>aleph-null day the drawer is empty. Proof: Suppose not. Then there must be
>a least number n such that bill n is still in the drawer. But this bill was
>removed on day n and never returned to the drawer, which is a contradiction.
I agree that there is a least number n, but i don't think that it's finite.
Prove that n is finite and i'll accept your argument.
Dave Seaman
[Re: proof that drawer is empty on aleph-null day]
>I agree that there is a least number n, but i don't think that it's finite.
>Prove that n is finite and i'll accept your argument.
My assumptions, based on the problem statement, are:
1. No bill can be in the drawer at the beginning of aleph-null day
unless it was placed there at some time before aleph-null day.
2. Only finite-numbered bills are placed in the drawer on
finite-numbered days.
I also need one fact (definition, actually) from set theory:
Aleph-null is the smallest transfinite cardinal number.
Therefore, all days before aleph-null day are finite-numbered days.
I'm sure you can take it from there.
Really, we should be talking about omega (the smallest transfinite
ordinal number) instead of aleph-null, but the distinction is not
particularly important in the present context.
--
Dave Seaman
a...@seaman.cc.purdue.edu
Merlyn LeRoy
This puzzle is (a bit) easier to understand when you realize that
the set of numbers (1, 2, 3, ...) is the same size as the set
(10, 20, 30, ...). Monty is saying he will give you as many bills as there
are elements in set 1 and only removing as many bills as there are elements
in set 2. Marilyn is removing the fee first, and only giving you as many
bill as there are in set 2.
But both sets are the same size (countably infinite = aleph null).
---
Merlyn LeRoy