Talk:Monty Hell problem

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1 (untitled)
2 Solutions
3 Further discussion
4 Regarding the Harmonic Series
5 Regarding the limit at infinity vs the value at infinity
6 Re: Regarding the Harmonic Series
7 "You can't multiply a zero probability by infinitely many elements" proof attempt
8 Damn it!
9 Gnome mathematicians
10 Possible merge with Ping-pong ball conundrum?
11 Confusion and Bias in Article
12 No comprendo
13 Application of Boole's Inequality
14 Marilyn's name?
15 A comment on the limits
16 Article currently has no citable sources

[edit] (untitled)

The question in the article is unclear to how it relates to the explanation, as it asks solely which banker will have you with the most wealth, making it a much more simple question and having both answers as correct. Can someone please correct the Question so I can more clearly understand this paradox? :)

Perhaps there will be disagreement over the categorization of this problem as falsidical. The classifications that were recently made in the Paradox article won't be changed till any discussion of this question has settled down, if ever. Dandrake 19:39, 15 Oct 2003 (UTC)

I agree with you now. Looking over the explanation, I see that the rec.puzzles explanation had fooled me. Thanks for clearing that up. Perhaps we should de-emphasize the antimony status — the problem was fixed by cleaning up the axioms, right? Paullusmagnus 20:10, 15 Oct 2003 (UTC)

Right. They figured out how to talk about infinite processes without in some sense getting to infinity. And since the antinomy (gotcha -- ain't it hard not to write antimony?) is dead by more than a century, it doesn't deserve much attention, if any, in the article. The article would be better for some editing by whoever first gets around to it. Dandrake 23:56, 15 Oct 2003 (UTC)

The current version of Monty Hell problem says, "Considering both definitions, the solution favoring Marylin uses the first definition (and the use of physical money in the problem favors this interpretation), while the Moloch vs Monty argument uses the second. If only one of the two definitions is used, there is no contradication."

If this is correct, then we have a serious antinomy in probability theory, don't we? A function that is monotonically increasing and everywhere greater than zero (not to mention increasing without limit) gives us zero at infinity when analyzed with probability theory. This not just a little recreational problem; it is important enough that we ought to provide a reference to the published literature on the subject.

"...So it is tempting to assume that a similar failure of intution applies in this case, and that infinitely many probability-zero events might sum to a nonzero expected number of bills. Modern probability theory excludes this assumption, because it takes as an axiom that probabilities are additive even for countably infinite sets of events."

This raises two questions. First, the probability calculation makes no assertion about the amount of money in the bag. If you naively attempt to calculate the expected amount at infinity, you get zero times infinity, which is not zero or anything else.

All right, that's naive, as I said; you really want the infinite sum of (probability of a bill's being there times a bill's value (which is unity)). If the probability of each bill's presence is zero, the answer is obvious. But in fact, at no time is the probability zero for any bill whatever. Zero is a limiting value. Is it axiomatic that these limiting values can be simply summed over an infinite interval? Is there a theorem to that effect?

Perhaps the problem is that the summation of these limiting values is still too naive. What you actually want (sorry to be too hurried to write it in LaTeX) is the limit as t -> infinity of the sum of p(i,t)*v(i) for i = i to M(t). Where p(i,t) is the probability that bill i is in the bag at time t; v(i) is the value of bill i, which is unity; and M(t) is the number of bills in the bag at time t.

This is not a difficult limit to calculate, since it's precisely (identically, in the strict sense) the same as the simpleton's calculation of how much Monty or Marilyn has in the bag.
Dandrake 19:26, 16 Oct 2003 (UTC)

Let us compare the two approaches to calculating Monty's pile. The simpleton's analytic approach says that Monty's money $M (t) = 9 t$ . In the limit this is
$\lim_{t\to\infty} 9 t$ which is, identically, $\sum_{t=1}^\infty 9$ , which is infinite.

The argument from probability is that the expected total in the bag at any time is
$\sum_{i=0}^n p(i)v(i)$ where $p (i)$ is the probability that bill $i$ is in the bag, $v (i)$ is the value of the bill, and $n$ is the number of bills that have ever been added to the bag. The amount in the bag at the end of time is the limit of that quantity.

$v (i)$ is present to emphasize that we are calculating an expectation. As it is a constant, and unity at that, it will be omitted from here on.

So far there is no problem. Even when there are (countably) infinitely many terms, the sum is valid in probability theory. Now, since $p (i)$ is zero in the limit for all $i$ , the conclusion is drawn that in the limit this is a summation of all zero terms, and the sum is zero.

This is a problem. In fact, $p (i)$ is not a constant, but a function of time $p (i, t)$ , which is nowhere zero. (One can formulate the problem with the function having zero value for $t < i / 9$ to represent the fact that the bill is not in the bag before it's placed there, but it is simpler to leave the function undefined for those values and set up the summation accordingly.)

Correctly, the money in the bag at the end of time is represented by
$\lim_{t \to \infty}\sum_{i=1}^{10t} p(i,t)$
The argument for Marilyn's method is, in effect, to replace that formula by
$\sum_{i=1}^\infty \lim_{t \to \infty} p(i,t)$ .
Once this implicit step is brought into the open, it is clearly not valid.

If there is a reduction of the correct formula to a simple closed form, it hasn't been shown yet. Deriving that, and showing that the result is different from the simpleton's analytic formula, is an exercise for those who consider the paradox not to be falsidical. Dandrake 03:19, 17 Oct 2003 (UTC) (Corrected Dandrake 19:40, Oct 17, 2003 (UTC))

Since the reduction of the correct formula hasn't been shown yet, I'll show it: the summation reduces to 9t. Proof is left as an exercise for the reader, as is the computation of the limit. Dandrake 19:40, Oct 17, 2003 (UTC)

I'm pulling out everything below here and replacing it with a new discussion. Please be very wary of putting any of the arguments for the obvious solution back in as factual unless you can back up your intuitive reasoning about the set limit process with formal arguments (this does not mean substituting the numerical limit process and talking about that---everybody here agrees that the number of bills in the bag diverges). Populus 21:58, 17 Oct 2003 (UTC)

[edit] Solutions

One suggested solution is that it is better to bank with Marilyn than with Monty. With Marilyn, no bill that enters the sack ever comes out again, leaving you with infinite wealth upon your eventual release. With Monty, it can be shown that the probability that any particular bill stays in the sack forever is zero, which implies that the probability that there are any bills at all left in the sack when you leave Hell is also zero.

This contradicts the intuitive reasoning that the final total must be greater than any previous day's total because each day sees a net increase in the number of bills, creating a paradox.

An argument against this reasoning considers the bank account of Moloch, who provides the heat at no expense to himself, charging a dollar a day and keeping it all as profit. Every day till eternity, Monty's account is nine times as large as Moloch's. When Moloch becomes infinitely rich, so must Monty, who is nine times as rich.

In Quine's terminology the Monty Hell Problem may be classed as a falsidical paradox, meaning that its reasoning is flawed. (Before the introduction of formal, consistent ways of dealing with the infinite, it would have been an antinomy, pointing out the flaws in some conventional thinking about infinite quantities)

In this particular case, there are two natural ways to define the limit: as a set process in which each bill is tracked separately, and a bill remains "in the limit" only if it is never removed; or as a numerical process in which the total number of bills is tracked, and the limit is defined as the limit of the day-to-day balance. By the second process, Marylin and Monty clearly both add $9 to the sack each day. By the first process, one comes to the conclusion that Monty's bag is empty at the end of time. However, one comes to the same conclusion if one considers a sack where each bill is removed 100 days after it was put in. Clearly, the sack will always contain 100 days worth of bills, even though no bill remains there forever. The intuitive result, that Marylin and Monty are equally favorable choices, is the correct one.

[edit] Further discussion

A key step in the argument for Marilyn's approach is the "which implies". We have the true statement that the probability of any bill's staying in the account forever is zero; or more precisely, it converges to zero in the limit. From this we infer that the probability that anything will be left after infinite time is zero. This inference is invalid. For a simple counterexample, suppose that at time t one puts a bill with serial number t into a sack, and at time t+3 one takes it out again. Every bill eventually comes out of the sack, but that does not mean that the sack will become empty; at time t=100 bills 98, 99, and 100 are in the sack. These bills will all be gone by time t=103, but by that time, new bills will be in the sack, which clearly never becomes empty. Or a more practical example: just because all humans are mortal, we cannot conclude that humanity will eventually become extinct. Everyone dies, but that does not mean that there will be a time when everyone is dead.

Many such perfectly sensible inferences fail, when dealing with infinite processes. For instance, some whole numbers are even and some are not; hence there must be more whole numbers than even numbers; all the more so for perfect squares. But it has been known at least since the time of Galileo that the number of squares is the same as the number of whole numbers; see Galileo's paradox. Or see Hilbert's paradox of the Grand Hotel. So it is tempting to assume that a similar failure of intution applies in this case, and that infinitely many probability-zero events might sum to a nonzero expected number of bills. Modern probability theory excludes this assumption, because it takes as an axiom that probabilities are additive even for countably infinite sets of events.

The above argument may be disputed, however, on theoretical grounds. (As it is in the Talk page).

IMHO this "paradox" should be viewed as the proof that the third Kolmogorov axiom for infinite sets must be wrong, just like Russel's paradox showed that the axiom schema of unrestricted comprehension was wrong. In fact,

A nonprincipal ultrafilter is essentially the same thing as a two-valued probability measure that is finitely additive but not countably additive. [1],

and the existence of a nonprincipal ultrafilter can be proven, at least with the axiom of choice. You can see the relationship of this with AoC looking at the argument at the bottom of this talkpage: without AoC you can't choose "an arbitrary element p in P".

I suppose that somewhere in this paradox hides the axiom of choice, but I can't find it yet. (Yes, Monty does choose a random bill, but the sack is always finite and you don't need AoC to choose an element from a finite set.)--Army1987 21:46, 28 August 2005 (UTC)

[edit] Regarding the Harmonic Series

If a bill is put into the bag at time 1, the probability that it is removed at time t is

${1\over 9t}\prod_{i=1}^{t-1} \big({1 - 1/9i }\big)$

The probability that it is removed eventually is

$P = \sum_{t=1}^\infty {1\over 9t}\prod_{i=1}^{t-1} \big(1 - 1/9i \big)$

If there's any useful way in which this resembles a harmonic series, I don't see it. I also don't know how to show that P=1. I haven't seen any version of the article that successfully argues that each bill does eventually come out of the bag; all the arguments posted so far have been appears to authority or handwaving.

I think this article needs more facts and less argument.

[edit] Regarding the limit at infinity vs the value at infinity

"If we let P be the probability that the bill is never removed, then we have P <= P(t) for any t, so if we can show that the limit of P(t) as t goes to infinity is 0, we will have that P is also 0."

It seems that the distinction between "the probability that the bill is never removed" (P) and "the tendency as the number of days goes to infinity that the bill is not removed by that day" ( $lim_{t \to \infty} P(t)$ ) is the matter of contention of the entire paradox. So in the proof of it, doesn't it seem useless to make that be the step that we gloss over? Without more substantial hand-waving than that found in the You can't multiply a zero probability by infinitely many elements section, I can't imagine anyone being prepared to accept these results. I am not qualified to comment on how to apply the axioms of probability theory, but the proof offered in the appendix does not stand on its own. The biggest thing I learned in my Calculus I class was that the distinction between the limit and the value is essential. (p.s. this is my first wikipedia adventure and I am finding it most enjoyable -- thanks! User_talk:galexander if I'm reopening an old debate that is settled to everyone's satisfaction)

Galexander 22:08, 13 Dec 2003 (UTC)

It's not glossing over (or at least it shouldn't be—if it looks like glossing over we might need to clarify the article further). Let A(t) be the event that the bill is not removed by time t. Let A be the event that it is never removed. Then A is a subset of A(t) for any t, so Pr[A] ≤ Pr[A(t)] for any t. But Pr[A(t)] goes to zero in the limit, so for any x > 0 there is some t for which Pr[A(t)] < x. It follows that Pr[A] is less than any positive x, which means that it's zero.

Populus 00:36, 14 Dec 2003 (UTC)

That looks fairly convincing but I'm having a hard time accepting that being &le ; a function for all values is the same as being equal to the limit in any sense. What you just claimed looks equivalent to: f ≤ 1/x for all x>0 and 1/x goes to 0 at the limit, therefore f = 0. It kind of makes sense but it reminds me of Zeno's paradox. How do you squeeze out of existence those infinitely many values between 1/(really big number) and 0? I can't say it's wrong but it doesn't pass my smell test.

Galexander 19:05, 14 Dec 2003 (UTC)

Maybe it would help to look at it backwards. Suppose I tell you that 0 <= f <= 1/x for all x. Clearly, f = 0 is a solution to this (infinite) set of inequalities. But perhaps there are other solutions? Suppose c > 0 is a solution. Then let x = ceiling(2/c). We then have c <= 1/x <= c/2, a contradiction.

So the answer to "how do you squeeze out the infinitely many values between 1/bi and 0" is that each 1/bigger squeezes out a few more, and that none survive being squeezed out by everybody (which is what we just showed). Populus 02:56, 16 Dec 2003 (UTC)

The c you're describing is an infinitessimal, and I don't think conventional operations are defined for those numbers conjuctive to real numbers. Hence c/2 = NaN, so c <= 1/x <= NaN, which is not strictly contradictory.

Allow me to turn your proof upside down and use it to disprove the existence of transfinite numbers, heavily relied upon by this paradox. Suppose a solution c < infinity to the inequality x <= f <= infinity. Then let x = 2c. Now we have 2c <= x <= c, a contradiction. Or to put it in layman's terms, each "bigger" squeezes out a few more, and none survive....

My point is that if you're willing to accept a number > any number, but not infinity, you have to accept a number < any number, but not 0. Alksub 04:28, 18 August 2006 (UTC)

[edit] Re: Regarding the Harmonic Series

The formula

$P = \sum_{t=1}^\infty {1\over 9t}\prod_{i=1}^{t-1} \big(1 - 1/9i \big)$

can be computed partially with the following QBasic program:

REM MONTY HELL PROBLEM
DIM sum AS DOUBLE
DIM summand AS DOUBLE
DIM factor AS DOUBLE
DIM prod AS DOUBLE
DIM t AS LONG
CLS
sum = 0
prod = 1
FOR t = 1 TO 20971520  'modify this number at will
 IF RND < .0001 THEN PRINT "t="; t, "sum="; sum
 IF t > 1 THEN
  factor = 1 - 1 / 9 / (t - 1)
 ELSE
  factor = 1
 END IF
 prod = prod * factor
 summand = 1 / 9 / t * prod
 sum = sum + summand
NEXT
PRINT sum
END

These are the results:

Number of days in hell	Probability that the first bill will have been removed by this time
5	0.231508
10	0.285019
20	0.336479
40	0.384911
80	0.430155
160	0.472232
320	0.511278
640	0.547470
1280	0.580997
2560	0.61204915
5120	0.64080265
10240	0.66742666
20480	0.69207925
40960	0.71490247
81920	0.73603525
163840	0.75560164
327680	0.77371770
655360	0.79049093
1310720	0.80602084
2621440	0.82039960
5242880	0.83371253
10485760	0.84603863
20971520	0.85745106

When these data points are plotted on a graph, the resulting curve looks like it is going to approach 1 asymptotically. --AugPi 02:40, 5 Apr 2004 (UTC)

Suppose there is a monotonically decreasing series of positive numbers s₁, s₂, s₃, s₄, ... such that

$\sum_{i=1}^\infty s_i = \infty,$

e.g. the harmonic series. Suppose that s_i is the probability that a certain bill is removed on day i. Then the probability that this bill has been removed by day t is

$P_R(t) = \sum_{i=1}^t s_i \prod_{j=1}^{i-1} (1 - s_j)$

and the probability that it has not been removed by day t is

$P_{NR}(t) = \prod_{i=1}^t (1-s_i)$ .

It can be proved by mathematical induction that

P R (t) + P N R (t) = 1

for all t. Notice that

$P_R(t) = \sum_{i=1}^t s_i P_{NR}(i-1)$ .

Induction Basis:

$P_R(1) + P_{NR}(1) = \sum_{i=1}^1 s_i P_{NR}(0) + \prod_{i=1}^1 (1-s_i)$

= s 1 P N R (0) + (1 - s 1) = s 1 + (1 - s 1) = 1.

Notice that

$P_{NR}(0) = \prod_{i=1}^0 (1-s_i) = 1.$

This is because the product of no factors is the multiplicative identity.

Induction Step: Assume that $P R (t) + P N R (t) = 1$ . Then what is the value of $P R (t + 1) + P N R (t + 1)$ ?

$P_R(t+1) + P_{NR}(t+1) = \sum_{i=1}^{t+1} s_i P_{NR}(i-1) + \prod_{i=1}^{t+1} (1 - s_i)$

= s t + 1 P N R (t) + P R (t) + (1 - s t + 1) P N R (t)

= P N R (t)[s t + 1 + (1 - s t + 1)] + P R (t)

= P N R (t) + P R (t) = 1

Therefore $P N R (t) + P R (t) = 1$ for all t, quod erat demonstrandum.

It is true for any x that

$1 + x \le e^x.$

These two functions are tangent to each other at point (0,1). Everywhere else the exponential function is greater than its linear approximation. This can be seen easily by plotting these two functions on a graph, e.g. graphing calculator. Therefore

$1 - x \le e^{-x}$

is also true, for all x.

Then

$P_{NR}(t) = \prod_{i=1}^t (1-s_i) \le \prod_{i=1}^t e^{-s_i} = e^{-\sum_{i=1}^t s_i}$

But

$\sum_{i=1}^\infty s_i = \infty$

as had been assumed earlier for the series s_i, so

$P_{NR}(\infty) \le e^{-\infty} = 0.$

But it was proven above that

P R (t) + P N R (t) = 1

for all t, so that

$P_R(\infty) = 1$ .

The formula

$P = \sum_{t=1}^\infty {1\over 9t}\prod_{i=1}^{t-1} \big(1 - {1 \over 9 i} \big)$

can be expressed as

$P = \sum_{t=1}^\infty s_t \prod_{i=1}^{t-1} (1 - s_i) = P_R(\infty)$

where

$s_i = {1 \over 9 i}$

which is one-ninth of the harmonic series and whose sum is

$\sum_{i=1}^\infty s_i = {1 \over 9} \sum_{i=1}^\infty {1 \over i} = {1 \over 9} \infty = \infty$

This means, given what was proven above earlier, that

$P=1 \$ .

--AugPi 04:18, 5 Apr 2004 (UTC)

Hi all! I stumbled into this page by accident, and I'm just wondering a couple of things:

The question seems to be, that "when you get out of hell, how many bills do you have". At the same time we're talking about infinity - if there is a day you get out, then the total days spent in is finite, and thus equal amout of cash in the sack in both scenarios.
The "not one bill stays in the sack forever"-point makes me wonder, what difference does that make? If you look at the situation at any given point, there will always be bills that have not left the sack. For example, if we suppose the bills get mould after time and must be changed, maybe every 20 years. For each bill in your sack you get a new one. The fact that the bills aren't exactly the same doesn't change anything? You could argue that for each 10 bills you receive each day one "replaces" the one you have given away yesterday.

Ok, if I get answers to those, thank you! --62.237.40.154 14:05, 11 Jun 2004 (UTC)

We formally define a "day at infinity" that occurs after all finite days. See ordinal number.
This is a common confusion caused by thinking of bills as fungible. As explained in the text, if you think of them as individual physical objects, you have to explain how you can have a sack with infinitely many bills in it at the end when every particular bill that might be in the sack has already been removed. The more formal explanation is that it makes a difference whether you take the set limit first and then compute the size or compute the sizes of all the sets and then compute the limit; the problem is phrased to encourage the first interpretation, and the paradox comes from the fact that the two processes give different limits. Populus 18:43, 12 Jun 2004 (UTC)

I'll stick my nose in here and make what seems to me a very straightforward argument, which means that it's most likely either faulty or irrelevant. *grin* On day ω, you get out of hell. You may or may not get paid this day, it's uncertain. Let's assume you didn't, nor was a fee collected. You did get paid the day before, ω-1. You also got paid on ω-2, ω-3, and so on. On day ω-1, one bill was removed at random and you were paid 10 bills, all of which remained in the bag at the end, since no further cash was drawn. On day ω-2, one bill was removed at random and you were paid 10 bills, at least nine of which remained. The first ten have probability 1 of staying. The next ten have probability .9 of staying. If one continues on, one sees that you have 10+10(.9)+10(.9^2)+... bills in the bag. While the first bills had probability 0 of surviving, the bills near day ω had high probability of surviving. Even when you're in the realm of infinity, you can walk a finite distance, it's just that nothing in the finite will notice it since you're still in the realm of infinity. If you somehow step into the finite days, you can apply the finite day logic and then just take the steps back in the realm of infinity. Like I said, it seems straightforward to me, so it's probably either faulty of irrelevant, but I'd be interested in whatever criticisms may come of it. -FunnyMan 07:03, Jul 15, 2004 (UTC)

Short answer: limit ordinals like ω don't have predecessors, so there is no day ω-1. Populus 21:26, 11 Jan 2005 (UTC)

[edit] "You can't multiply a zero probability by infinitely many elements" proof attempt

Let a be a positive integer and P be the set of all positive integers. The probability that an arbitrary element $p \in P$ equals a is: $\frac{1 \mbox{ (a=p)}}{\infty \mbox{ (size of P)}} = 0$

If we sum this probability over P (which is assumed valid since $P \subset \mathcal{Z}$ is countable), we get that the size of $P \cap \{a\}$ is zero.

Yet, we also clearly see $P \cap \{a\}=\{a\} \mbox{ as } \{a\} \subset P$ and thus that the size is one. This is a contradiction. Hence, either the calculation of the probability or the sum of that probability was invalid.

Anybody see something wrong with that? I know I was a little sparse on the probability calculation, but it's pretty simple. -FunnyMan 20:52, Feb 2, 2005 (UTC)

Whoops, I had some broken notation in there. It's fixed now. -FunnyMan 21:37, Feb 2, 2005 (UTC)

This problem hurts my brain, primarily the logic center. I'm thinking of being a math major/stat minor, but not after reading this! Since I can't make up my mind, I'm going to invalidate it in another way: An infinite amount of money is worthless because of inflation, so either way you go, infinite (worthless) or zero (poor), it doesn't really matter! :D DevastatorIIC 19:52, Jun 11, 2005 (UTC)

The paradox relies on the faulty assumption that a countable infinite set may not contain two infinite subsets. Given infinite time, the number of bills taken from the bag is infinite. The number of bills left in the bag is also infinite. Another example of two infinite subsets of an infinite set is the prime numbers and the composite numbers.

I do not see any thing wrong with the mathematics. You can not add infinite zeroes and come up with answer of zero. Look at the problem like this. After n zeroes are added, there will still be infinite zeroes to add. We add multiple numbers in two ways. One way is to add each element to the partial sum. In this way we can never add infinite elements. The second way is to convert addition to multiplication. But we know that zero multiplied by inifinity results in undefined number. kalyan05

[edit] Damn it!

This would probably defeat any objection I can imagine of to this paradox. It is perfectly equivalent to this one, but there's no probability in it. --Army1987 17:26, 3 October 2005 (UTC)

You probability freaks are too concerned with your numbers. Doesn't anyone consider that you can progress towards infinity at different rates? Everyone seems to want to talk about infinity as if it were a single idea, when really there are an infinite number of infinities. As long as you accept that there are two rates of infinity involved, one nine times "faster" than the other, then there is no problem. The problem has nothing to do with probability. Furthermore, since there are infinite number of rates of progression towards infinity, there is no "end" to infinity; and even if you were to hypothesize about it, you would have to accept that there are an infinite number of "ends" to it. If you can appreciate this, all the infinity paradoxes disappear. If you can't, you'll be banging your head against a firy wall forever.24.64.223.203 21:13, 30 October 2005 (UTC)

[edit] Gnome mathematicians

Some gnome mathematicians were sitting in their garden talking about a problem in probability. Gnomes are rather small, and gnome mathematicians are not known to be very bright. The problem was this:

Say you flip a coin. One side is heads, the other is tails. What's the result?

Although some tried to trivialize the problem by making the probabilities zero and one, eventually everyone agreed the real problem, a bit more complicated, could best be simplified if the probability of heads is 1/2, and likewise for tails. But they still couldn't agree on an answer. Some said it was heads, some argued it was tails, until one gnome suggested the problem was incomplete. Either heads or tails was possible depending on which side the coin landed. This seemed like a good solution until one of the "smarter" gnomes pointed out that which side the coin landed on was the very question to begin with. And so the bickering continued.

Of course the problem is incomplete, and it's incredible that gnomes can talk about probability at all considering how little they understand of it. And this is how I come to grips with my own lack of understanding for the Monty Hell Problem. I take it on good faith that either solution is possible. You could have everything, or you could have nothing. It would seem that the problem is stated incompletely, just like Bertrand's paradox. After all, none of us knows what an eternal confinement would be like, much less the end of one. When talking about an infinite length of time, you have to give a bit more information. Do you want to know the amount you've saved or the number of bills you have left? The laws of infinity are independent of the axioms of common sense. Davilla 09:32, 6 February 2006 (UTC)

[edit] Possible merge with Ping-pong ball conundrum?

While this one deals with money, and that deals with Ping pong balls, I think they are fundamentally the same problem. Thoughts? Kirbytime 00:24, 4 April 2006 (UTC)

You're right, they are fundamentally the same problem. The problem with both articles is that they try to be unwontedly even-handed. The "increasing wealth" argument, though superficially appealing, is simply nonsense when you analyze it. The only correct answer is zero. --Trovatore 02:09, 4 April 2006 (UTC)

[edit] Confusion and Bias in Article

This point has been made above, in "Further Discussion," but I will reiterate it here, as it's important: the limit of a probability is not the same as the probability, especially when it's going to be applied to a transfinite set. By definition, a limit is only almost a solution to a function over some x. Imagine an intermediate third banker, Dante, who takes the condemned's pay and adds it to a sack of twenty bills taken from the sinner's main pile. He then selects one to give to the devil. I am suspicious of Appendix 1's proof because Dante's case will be subject to the Monty probabilistic argument for all amounts Dante takes out of the main pile for consideration (even subdivided single bills!) except in the limit case (as the number of bills Dante withdraws approaches 0), which is congruent to Marilyn's case. Revised Alksub 01:35, 30 May 2006 (UTC)

There are two main problems with this article. The first is that the "Attacks" section is biased against the intuitive answer. Its headings have a POV tone and need to be renamed. Also, the probabilistic limit attack illustrated in "Further Discussion" is not represented in the article as far as I can see; the solution section says if you opt for the intuitive answer, you have a bag of bills you have "previously lost forever," based on the questionable proof.

The second problem is that some of the article is written without consideration to what someone at first glace will notice: Marilyn and Monty do exactly the same thing for all intents and purposes. "Carefully" is simply a ruse that insulates Marilyn from complex mathematical discussion. What I am saying is that the article should not attach the paradoxical conclusions to the two figures, but instead synthesize them as two interpretations of one paradox.Alksub 23:59, 29 May 2006 (UTC)

Here's what really turns the trick with this paradox, and it was touched on in the Solutions section. Visualize a row of squares (rather like the tape on your beloved turing machines ;-)) meant for holding bills. They represent every bill given to the sinner. Colour them white if the sinner has lost the bill, or black if the sinner still has the bill. Now, in Marilyn's case, you can imagine all the bills black, except for every tenth bill, which is white. Monty's situation is more complicated. While the last ten squares are most usually black, the first few squares inevitably become white. However, as the transactions continue, a sort of "front" of black squares moves forward on the tape, leaving white squares in its wake. The fallacy, roughly speaking, is to pretend that this front, though huge, doesn't exist, simply because you can't track it all the way to the transfinite shambalaboom. The way posters are doing this is to claim that P a bill is in the sack is 0, and of course that is true for the first n bills, but not for the front.

It's obviously true that Monty's sinner, on the days before he got out of hell, received some bills from Monty whose probability of staying was very high, and hence the sack can't be empty. That's the "front" of bills. The main handwaving that was done was to claim that "transfinite numbers don't have predecessors." I would appreciate it if someone explained to me why this definition of ordinals was established, and how it is valid in the context of this paradox. Moonbat, signing off. Alksub 19:01, 16 June 2006 (UTC)

The definition of ordinal number referred to is quite sound, and omega has no predecessor (it is not correct that "transfinite numbers don't have predecessors" in general). The maths in appendix 1 are correct as far as I can tell. And, I am not sure I understand what Dante does in your example above (maybe you could spell it out). However, it doesn't seem that any of it is really valid in this context. At least not without further assumptions on the nature of a transfinite shambalaboom. Which, I suppose, is the point of the paradox. 192.75.48.150 19:37, 16 June 2006 (UTC)

[the following was moved from "harmonic series"] That may be right, but it begs the question, "Can one get out of hell at all?" Imagine an outside observer who sees the man in hell after he gets out. Ask him the following question, "How much money did you have the day before you got out?" Now by the previous logic against a 'day before' there was no day before, thus he couldnt have gotten out. This is too much like one of zeno's paradoxes.

Ill counter with another paradox i just thought of. Suppose you had an infinite line of people (in both directions.) Each has 1 dollar and every day each person passes the dollar to the right hand person. Thus each day each has 1 dollar. However the probability anyone has any money at all is 0 according to the 'proof' given in the article- even though NO MONEY has left the system. I think that this shows that the argument that there is a problem with the current argument used. Sure it works probabilistically, but that just isnt enough imo.24.36.5.61 00:29, 17 July 2006 (UTC)

No, I think the argument is right... but the assumptions about causality are incoherent (not counter-intuitive, but contradictory). When you use a "non-well-founded" set (e.g. infinite in both directions), it becomes more obvious.

Let me take turn your idea inside-out (from infinite in both directions, to infinite in the middle from both sides, or from ω^*+ω to ω+ω^*): take the situation as written on the page, using Marilyn as the banker (who puts 9 bills in the sack each day and never takes any out). Now on the first day of the afterlife, when God sentenced the prisoner to Hell for his crimes against mathematics, He ordered a countable set of His angels (who had been dancing on the head of a pin) as follows:

He tells the first angel, if the prisoner has not been released by noon on day omega (supposing that day omega starts at midnight), release him, take all the money out of his sack, and give it to him.
He tells the second angel, if the prisoner has not been released by 6 am on day omega, release him, take all the money out of his sack, and give it to him.
He tells the third angel, if the prisoner has not been released by 3 am on day omega, release him, etc.

Each angel has a specific time, halfway between midnight and the previous angel's time, to release you. Now we add our assumption, made explicit:

Naïve causality: If a bill is put in the sack at time T, and not taken out of the sack between times T and U inclusive (U being a later time than T), then it is in the sack at time U. Conversely, if a bill is taken out of the sack at time T, and not put back in the sack between times T and U inclusive, then it is not in the sack at time U.

Obviously, the first angel will take out any remaining bills at 12pm, and nobody touches the sack after that. Therefore, by the assumption, the sack is empty on 12:01 pm on day omega. But clearly, the n'th angel can't take the money out, because the n+1'th angel would already have done so. Therefore, nobody ever takes money out of the sack. But Marilyn put money in, therefore, by the assumption, the sack is not empty on 12:01 pm on day omega. This is not just a counter-intuitive result, this is a contradiction. But the argument is correct, so it must be the naïve causality assumption at fault. But this assumption is obviously valid for our finite experience, so the point of failure for causality must be at an infinity. -Dan 192.75.48.150 14:58, 17 July 2006 (UTC)

Which begs to question whether these situations are logically possible. Also i could have rephrased it to a uni directional problem where odds pass right evens left and 2 to 1. But the other is just easier to see. BTW i love the pot shot in there.24.36.5.61 21:45, 18 July 2006 (UTC)

[edit] No comprendo

I consider myself to be a fairly well-educated person, but the way this is described is completely Greek to me. Starting from the beginning of the section labeled "The Paradox":

Let us start with the obvious explanation why it doesn't make any difference which banker you choose: after t days, both Monty and Marilyn have 9t dollars. Since these quantities both grow without limit, either will give you infinitely many dollars in the end.

Fine so far...

Unfortunately, there is a less obvious explanation that favors Marilyn. This explanation depends on the assumption that the contents of Monty's sack on day ω is a set-theoretic limit of the contents on the preceding days...

I don't know what a "set-theoretic limit" is, and the ariticle on the subject doesn't explain it, either.

where the limit of a sequence of sets A₁, A₂, ...

The what?

is defined to contain some element x only if there is some number N...

Aaargh!

such that x appears in A_n for each $n \geq N$

Stop! I surrender!

This looks like a very interesting article, but I can only guess, because, unlike, say, Monty Hall problem, it's not written in a way that an ordinary person can understand. Is it possible to make it comprehensible to the layman? Thanks -- Mwalcoff 02:03, 28 July 2006 (UTC)

Unless you know basic maths, I'm afraid most likely not. —Nightst a llion (?) 19:53, 31 July 2006 (UTC)

Basic math? I admit I've forgotten my trig and never took calculus, but I'm sure that my knowledge of mathematical concepts is beyond that of a typical American adult. If something is gibberish to someone like me, who has an MA, imagine how incomprehensible it is to the average American adult, who has only a high-school education? Maybe you're all math geniuses in Austria, but I would guess very few Americans -- perhaps 2% -- would understand the sentence described above. -- Mwalcoff 22:53, 31 July 2006 (UTC)

Im not american (canadian), but dont claim to represent americans with your lack of comprehension, ill tell you its sad how in North America professing ignorance isnt something to be ashamed of but is something thta one garners sympathy from.

Ill give a simple explanation. A set is a collection of elements. So a set in this case is the collection of all the individual dollars he has here. But 1 dollar is not the same as another (imagine they all have different serial numbers). At infinity he has none of the individual dollars (by their serial #) that he had at a finite time. Thus he has no money. If you buy the set theoretic limit as the solution.24.36.5.61 02:40, 2 August 2006 (UTC)

Thank you for trying to explain the problem somehow. I still don't understand why the Mariyln or Monty scenarios would be any different.

I realize I cannot speak for all Americans. However, it is clear to me that only a small percentage of Americans would be able to understand this article. In the 2003 U.S. Adult Literacy Survey, only 13% of people got a "proficient" score in "quantitative literacy" (practical mathematics). The example they give of a "proficient" person is someone who can calculate and compare the price per ounce on packages of food. If only 13% of Americans know whether to buy a 6-ounce can of tuna for $1.50 or a 10-ounce can for $2, how many people do you think can understand the sentence above? -- Mwalcoff 03:45, 2 August 2006 (UTC)

What do you mean you don't know what a set-theoretic limit is?! For heaven's sake, it's only taught in every basic university course in set theory, you know. I'm sorry but if you can't be bothered to take an interest in maths that's your own look-out. Apathetic bloody readers, I've no sympathy at all... -Prostetnic Vogon Dan 192.75.48.150 13:57, 2 August 2006 (UTC)

Thank you for making the article easier to understand. I'd be happy to learn what a set-theoretic limit is. Unfortunately, the article set-theoretic limit is completely incomprehensible. Too many Wikipedia editors seem to forget that the point of an encyclopedia is to teach readers things they don't know, not to show off how much you know to your peers in your field. -- Mwalcoff 22:32, 2 August 2006 (UTC)

Oh, good heavens, I suppose now you're going to tell me you don't even know what $\liminf_{i \rightarrow \infty} A_i = \bigcup_i \bigcap_{j \geq i} A_j\mbox{ and }\limsup_{i \rightarrow \infty} A_i = \bigcap_i \bigcup_{j \geq i} A_j$ means? Really! Don't they teach hieroglyphics in American schools anymore? Oh, and of course you shouldn't use Wikipedia to show off to peers in your field. You should show off to people outside your field. Frankly, peers and betters in your own field just get in the way. They tend to show you up and knock over the pet projects and theories that you came to Wikipedia to push in the first place. Try to avoid them as much as possible. 192.75.48.150 14:54, 3 August 2006 (UTC)

OK Mwalcoff Ill explain. Remember the serial number thing? Ok basically they both have the same number of bills. But One of them holds onto the SAME bills with the SAME serial number. The other will end up constantly CHANGING the bills and the serial numbers. In the end that person will NEVER hold onto any INDIVIDUAL bill for infinitely long. Thus he cant have any bill at infinity since he didnt get any at infinity and he never holds on to one for infinitely long. Thus he has no bills. The other person holds on to the same bills with the same serial numbers so this person will have bills at the end (you know which ones too by looking at the serial numbers)

and to be quite frank... there is only so much leaway in dumbing things down, and whoever gave you those numbers is MORON since they are clearly wrong even YOU should be able to see that... if that were true your country wouldnt be able to sustain itself nevermind be as advanced as it is - stop being a sympathy mongerer24.36.5.61 00:23, 4 August 2006 (UTC)

Take it easy, eh? (<-- me being Canadian) Okay, in all seriousness, it probably was more opaque than it needed to be, and if set-theoretic limit was supposed to fill in the blanks, it really didn't, which is why I changed the first section. 192.75.48.150 14:30, 4 August 2006 (UTC)

Thank you again for the explanation. The survey I cited is from the US Department of Education and had 19,000 respondents, so I think it's fairly accurate. I wouldn't think that American's world stature is based on widespread math skills. I'm not looking for sympathy, just trying to get people to understand that others might not be as adept in a particular subject matter as they are. -- Mwalcoff 23:07, 4 August 2006 (UTC)

Still, an encyclopedia can't explain everything to every kind of ignorant person (not meant as an insult). If you don't understand basic math, an encyclopedia won't be able to explain vector theory to you. If you don't get vectors, you'll have a lot of trouble even understanding a few words in an article on tensors. If you know next to nothing about linguistics, why would you *expect* to understand a linguistic article about some unknown language in a remote part of nowhere which details the language as good as possible using accurately defined linguistic termini and clearly defined IPA phonetic symbols? You just can't start to read an article on NP vs P and expect to find an explanation using apples and eggs, simply because a lot of topics in a lot of fields can only be accurately and encyclopedically described to people who know at least something of the field. —Nightst a llion (?) 10:49, 8 August 2006 (UTC)

That may be true in some cases, but I would say that most concepts can be explained (not necessarily thoroughly) in layman's terms. The New York Times often has articles on complex scientific concepts, like the physics of natural-satellite orbits. The user above did an adequate job of explaining the Monty Hell problem. Articles that are by nature too complex for newbies should make clear where the reader can find the background information necessary to understand the article. Part of the problem with the text in question here was that there were not enough links to other articles that could explain it, and the articles that were linked to were themselves not comprehensible.

Here's a metaphor for the argument the article is making. Suppose you have an infinite number of slices of toast (a set of toast), and a jar of jam. You undertake the supertask of spreading the jam evenly over all the pieces of toast. When you're finished, you're dismayed to find that each piece of toast has no jam on it (the set-theoretic limit). Now the article adds up all the zero amounts of jam on the toast and gets 0, and so concludes that you never had any jam in your jar to begin with. But you obviously did. That's the jam. Alksub 07:25, 14 August 2006 (UTC)

At least for me, the hard part to understand is why you end up with no money after an infinite amount of time if you actually make a 9$ profit every day. While it is true that by these calculated limits eventually every 1$ bill will have been removed from the bag, it's also true that every day 9$ are placed in the bag. I understand this might sound silly(I haven't studied complex mathematics), but your gain(9 x an infinite number of days)would be higher than your loss(1 x an infinite number of days), at least in theory...that is I get the feeling that your gain is 'more infinite'than your loss...Sorry if I'm just ranting away, by the way. David88 13:20, 10 September 2006 (UTC)

x * infinity is still infinity. —Nightst a llion (?) 10:11, 15 September 2006 (UTC)

While it is true that x * infinity is infinity for all practical purposes, we have to bend the rules here because an order of infinity is the building block. For example, to examine this problem, we have to say that on day omega the sinner has received (as opposed to making as profit) 10 * infinity bills. We would be getting nowhere if we up and stated that 10 * infinity = infinity because then reasoning would lead to some proof or other that the sinner is being given 1 bill a day, and that's wrong. You can't just pick and choose when you're going to follow the usual axioms of mathematics and when you're going to use definitions that suit a paradox concerning an infinite number of bills. Comment rescinded Alksub 03:29, 5 October 2006 (UTC)

(Unsubstantiated and confrontational comment removed) Alksub 23:38, 6 October 2006 (UTC)

Guess what? Your belief is wrong. You figure out why. --Trovatore 21:22, 4 October 2006 (UTC)

Yes, thank you for that. I concede that the two solutions provided in the article are at least equally valid, but I am still not convinced that the set-theoretic limit is the "better" solution (as the article implies it is). Alksub 02:37, 5 October 2006 (UTC)

OK, that's a little better. I'll grant that the problem needs to be better specified before taking the set-theoretic limit is rigorously justifiable. There are other similarly-worded problems where there's simply no answer (like the one where you turn the light on and off infinitely many times; after that, is it on or off?).

However there's just no justification for claiming the answer should be continuous at infinity. Why should it? At least in the case of the set-theoretic limit, if you specify better some of the unstated assumptions, you come up with the set-theoretic limit as the well-determined answer. There's nothing parallel for continuity at infinity; it's just completely unmotivated. --Trovatore 03:34, 5 October 2006 (UTC)

The number of dollars is "continuous" at omega.
The paths of individual dollars are "continuous" at omega.
The paths of individual dollars do not begin at omega.

All three are "conservation of mass" type intuitions, and any two are compatible in this situation. 192.75.48.150 13:34, 5 October 2006 (UTC)

No, they aren't. There is simply no reason whatsoever to believe the first one. Which dollars do you have at the limit point? Why do you have them? --Trovatore 15:17, 5 October 2006 (UTC)

Well, depends which rules we accept. Let us imagine each dollar bill is marked with the day it was created, and a serial number.

If we take the last two, then we have nothing at all. People (other than you) might not like there being nothing left, but absent rule one, this is permitted.
If we take the first two, then we have an infinite number of dollars, marked with day omega and some serial number. You might not like "day omega" written on a dollar bill, but absent rule three, this is permitted. Consider what would happen if, instead of adding 10 dollars and removing one dollar, the banker added nine dollars and re-labeled one dollar in the sack with the current date. Now what do you think would happen?
If we take the first and third, then we have an infinite number of dollars, each marked with some finite day. You might not like a dollar from a finite day being in the sack after it was removed, but absent rule two, this is permitted.
If we take all three, then we have a contradiction.

In all cases weird stuff happens. 192.75.48.150 16:13, 5 October 2006 (UTC)

The problem is that your rule 1 cannot be motivated by anything in the statement of the problem. It's an external condition that you've added without any justification whatsoever. The number of dollars is a derived quantity, calculable from the dollars that you have. The individual dollars are the fundamental entities. If the problem were stated in terms of your bank balance, without any reference to individual dollars, that would be different; absent any more detailed information, you might choose to assume continuity at infinity, still without any real justification, but just on the basis that you have no contrary information. --Trovatore 16:19, 5 October 2006 (UTC)

Neither was there something in the statement of the problem that says that dollar bills can be marked in an infinite number of different ways, was there. We imposed that as an external condition as well. Even if we accept it, conservation of mass is still a physical rule. Now if the problem were stated in terms of functions, sets, and elements, without any reference to time, sacks, and bills, then maybe that would be different. 192.75.48.150 19:17, 5 October 2006 (UTC)

Yes, as I mentioned, you do have to make some underlying assumptions explicit before you can really justify the answer given by the set-theoretic limit. But they're natural assumptions. The continuity-at-infinity thing just comes out of nowhere. There's no justification for it at all. --Trovatore 20:30, 5 October 2006 (UTC)

I understand your position, and I don't agree with your idea of what's natural and what's not. If anything, conservation of mass is a physical rule whereas "conservation of individuality" is more tenuous. 192.75.48.150 21:01, 9 October 2006 (UTC)

I thought of a simple (and possibly daft) argument. I define a set representing the bills the devil has lost (only to the sinner), and assert it must mirror the bills the sinner has gained. I presume the "loss set" can be managed in any way, so long as a dollar loss enters the set for every bill received by the sinner, and one is removed for every dollar the sinner pays back. Let it be managed in a way that unambiguously indicates it goes to infinity in both the numeric and set limits (i.e. the way Marilyn operates). This appears to contradict the result that the sinner ends up with nothing. A different way of stating this would be to allocate a finite, nonzero number of bills for the devil to give to the sinner, and then to show that this set will become empty as a result of the transactions. An apparent consequence would be bills unaccounted for. Alksub 06:07, 9 October 2006 (UTC)

[edit] Application of Boole's Inequality

Can I assassinate myself and attack the last line of the proof? We have lim P(t) = 0 as t approaches infinity. So Boole's Inequality is a tool used in the proof to limit P of the set to P of a single bill. Which produces this line:

$\sum_{i=1}^{\infty} \Pr[A_i] = \sum_{i=1}^{\infty} 0 = 0.$

But we only had the probability in the limit, so the upper bound from the logical step should be:

$\sum_{i=1}^{\infty} \Pr[A_i] = \lim_{t \to \infty} \sum_{i=1}^{\infty} 0 = \lim_{t \to \infty} 0$

Since we are dealing with competing infinities (the ten-giving and one-taking), they look quite different. Now, reduce the above to absurdity: how did you eliminate the limit? Alksub 04:58, 29 September 2006 (UTC)

Hmm! That each term goes to 0 in the limit doesn't mean their sum goes to 0 in the limit. You're right that there's something missing. 192.75.48.150 14:38, 29 September 2006 (UTC)

I can only assume this is based on the legality of determining that no one bill is in the set (i.e. P -> 0) and concluding that no bills are in the set. While I guess this is tautologous, I'm still unsure of the proof's treatment of P(n). Intuitively, it seems that $\lim_{t \to \infty}$ P(t) = ε and that an infinite sum/product of infinitesimals is ill-defined.

While this may be adequately refuted in the "Just because you have a lot of cadavers doesn't mean there's a rehearsal" section, and may be equivalent to saying "9t," I'll put it here. If you carry the probability-bound limit into the infinite sum, then bring the sum into the limit as a product, you have $\lim_{t \to \infty}$ t[exp(-k(1 + 1/2 + ... + 1/t))] which I think is nonzero. Alksub 07:39, 28 October 2006 (UTC)

[edit] Marilyn's name?

"Marilyn presumably refers to Marilyn vos Savant"

The real Monty Hall's wife is also named Marilyn. Which one is the namesake? -- gparker 04:19, 1 November 2006 (UTC)

[edit] A comment on the limits

I want to point out that the second "limit" necessarily assumes the Axiom of Choice. This is easy to see since we assume that we a dollar bill can be chosen for a countable set of times. This is not strictly in the wording of the problem however. However, having derived 9t as the amount of money after time t, no such assumption is required, the bag will always contain 9t dollars for any time t. Occam's razor would prefer the first solution, which is strictly ZF, to the second, which is ZFC. Antonymous8 02:11, 10 November 2006 (UTC)

[edit] Article currently has no citable sources

FYI, this article has no citable sources. Currently the only reference is that this problem was discussed on an internet forum somewhere. Forums, however, are not considered citable sources.

This article will ultimately need some references from citable sources, such as a published text book or journal. One thing in particular that needs to be verified by an external source is that the name of the paradox is the "Monty Hell problem", and that the description of the paradox is commonly accepted. Terminology and naming are unfortunately not verifiable by logic alone, no are forum posts acceptable verification. Dugwiki 23:08, 14 November 2006 (UTC)

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