Talk:Bertrand's box paradox

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This article was nominated for deletion on January 8, 2006. The result of the discussion was keep. An archived record of this discussion can be found here.

1 Talk archive of Three cards problem
2 Talk archive of Bertrand's box paradox
3 Merged Three cards problem into Bertrand's box paradox

[edit] Talk archive of Three cards problem

[edit] False article?

Think. What are the odds that the other side is also black? So, the card what is just picked up, can't be white/white, which means you don't need white/white card at all to perform this "problem". Then probability that you just picked up white/black or black/black, is of course 50/50! I agree only one thing in this article; try it!!! I did try, probability was 50/50. 195.163.176.146 06:28, 14 December 2006 (UTC)

See if this helps. Let us assume that you have only the 2 cards as you say. If you draw one and do not look at either side, then there is a 50/50 chance it can be either. But there is more to this puzzle. Now I let you look one side of it. This is more information and changes your "bet". If you see white, there is certainty it is the white/black card. If you see black (as the problem states) you should still change your bet. This new information changes the odds. It is now 2/3 as this article explains - since you see a black side, 2 out of the three black sides are on the B/B card, so this card is more likely. From a common sense point of view, since you change your bet when you get to look and see white, you should also change your bet if you get to look and see black. More information changes the odds.Obina 13:12, 14 December 2006 (UTC)

Read the problem description carefully! "The side facing up is black. What are the odds that the other side is also black?" There is only one card that is black from other side too. From two cards (you don't have to count white/white as I explain earlier) probability to get B/B is 1/2. What I meant by "False article?" is, are this article trying to describe real paradox, or is it just attempt to terrorize wikipedia? We don't have to argue about this, just try it!← 212.16.102.117 16:29, 14 December 2006 (UTC)

I told people we should have retained the incorrect argument which the anons give above and explain why it's incorrect. Please check the earliest revisions of the article for that argument. As for the anon, all the arguments given in the article are correct, even if we don't explain why the argument proposed above is wrong. — Arthur Rubin | (talk) 17:06, 14 December 2006 (UTC)

How do you explain that the real world result are 50/50? PLEASE TRY IT! 212.16.102.117 18:33, 14 December 2006 (UTC)

Ok, I was wrong, I've done error, sorry to all. Result is 66,666...% = 2/3. And for future idiots, here is Vb code to simulate problem (just make form and command button to the right edge):

Private Type Card
 Upface As Boolean
 Downface As Boolean
End Type

Private Sub Command1_Click()
Dim ThreeCards(1 To 3) As Card
Dim Pullout As Long
Dim n As Long
Dim ii As Long
Dim BothBlack As Long

Randomize Timer

'First we define cards...
'Lets say that True means black.

'Card one, both sides are black.
ThreeCards(1).Upface = True
ThreeCards(1).Downface = True

'Card two, both sides are white.
ThreeCards(2).Upface = False
ThreeCards(2).Downface = False

'Card three have two positions, so it will be defined later.


For ii = 0 To 1000000
Pullout = Int(Rnd * 3) + 1

If Pullout = 3 Then
'Card three, upface could be either one, black or white.
 If CInt(Rnd) = 1 Then
 ThreeCards(3).Upface = True
 ThreeCards(3).Downface = False
 Else
 ThreeCards(3).Upface = False
 ThreeCards(3).Downface = True
 End If
End If

If ThreeCards(Pullout).Upface = True Then
n = n + 1
 If ThreeCards(Pullout).Downface = True Then
 BothBlack = BothBlack + 1
 End If
End If
Next ii


Print "Probability is " & Str((BothBlack / n) * 100) & "%"
End Sub

You can use this code for free as you like. 212.16.102.117 19:49, 14 December 2006 (UTC)

[edit] What's the answer (2/3)

The key sentence in this analysis I think is

'Thus, there are only two possible cards, double green-faced or purple/green-faced, and each has an equally likely probability of being the one you chose.'

I think this is not true. The history of where the cards came from is key. How do we know there is an equal chance of those 2 cards?

When assessing the cards on the table, you must assess where they came from, the cards in the hat. Just like guessing whether a person on a team is a man or a women, it is important to know whether this is a rugby team or not.

There are 2 single colour cards in the hat, and only one dual colour card. The chance of a single colour cards will be 2/3, regardless of what colour you see.Obina 12:26, 11 January 2006 (UTC)

Agreed. - Haukur 12:28, 11 January 2006 (UTC)

Here is another way to consider the problem using the colour and side rather than the card. We see a green card so we know the double purple is gone. So there are only 2 cards in the running. How many permutations and combinations are there so we can assess the probability? Of these 2 cards, if we look at one side, we could at look at the top of card A, the bottom of card A, the top of B or the bottom of B. Of these, 3 are green and one is purple. But we look at see that it is not purple. So we could be looking at one of 3 sides. Top of A, Bottom of A, or Top of B. So there is a 2/3 chance we are looking at card A, and thus 2/3 chance that the bottom is green.

Anyone who still has doubts - this is a very easy experiment to try. Try it 20 times with, say, 3 business cards with P or G written on each side. Record how many times the colour under matches the top. Obina 18:20, 11 January 2006 (UTC)

The article cites a poll of 53 students, of which 35 guess 1/2 and only 3 guessed 2/3. That leaves 15 unaccounted for. I find it hard to believe that so many guessed neither the right answer nor the naive wrong answer. Does anybody have more detail on that? 171.64.71.123 04:54, 30 August 2006 (UTC)

As a former professor, I can answer this; it's easy. 35 students thought the question was a no-brainer. 3 students had heard this problem before, and 15 students didn't want to look like idiots, so they didn't raise their hands. Rklawton 15:23, 27 February 2007 (UTC)

[edit] From Heraclesprogeny to Arthur Rubin

I think I fixed the article, leaving the naive 1/2 in, but showing the correct number is 2/3. Arthur Rubin | (talk) 00:04, 12 January 2006 (UTC)

Improvement well done! Fine to leave and explain how 1/2 can seem like the quick answer, but even qual approach leads to 2/3 if probability is considered right. As said on talk page we must find a reference or this page will not survive as original research.Obina 00:13, 12 January 2006 (UTC)

I found that Martin Gardner has written about it under the name of "The three cards" [1]. This should give more leads. --C S (Talk) 08:49, 12 January 2006 (UTC)

Good find! Indeed it does give leads; see question and answer by John Schuyler. The pdf you found is a copy, probably illegal, of "Science Puzzlers", first published by Scholastic in 1960, and republished by Dover in 1981 as "Entertaining Science Experiments with Everyday Objects". Melchoir 09:08, 12 January 2006 (UTC)

[edit] Approaches

I've taken out the sections on the "qualitative approach" because they basically make no sense. Melchoir 07:50, 12 January 2006 (UTC)

[edit] Article name?

What is the puzzle called in the sources you're using? It'd be nice to have a name for the article which gets at least *some* Google hits :) - Haukur 12:24, 12 January 2006 (UTC)

Duh, actually reading the above shows that a move to a title like The Three Cards might be desirable since that seems to be Gardner's title. I'm not sure about the definite article or the capitalization, though, and maybe the name is ambiguous.

Stupid Gardnerian puzzlecruft ;) - Haukur 16:23, 12 January 2006 (UTC)

I like The Three Cards. Sure, it's ambiguous, but I can't see anybody typing in "The Three Cards" and being disappointed by the result. Of course, that's because I can't see anybody typing it in at all... Melchoir 16:39, 12 January 2006 (UTC)

How about The three cards (probability). No caps per Wiki norm - is not a proper name! And put a link to it on the Probability page.Obina 20:06, 12 January 2006 (UTC)

I contacted my statistics professor from CalTech. He couldn't find a name or a reference, but he used it in Ma 2 (sophomore math). (And he uses red/white cards, with an odd reference to three coins, with colors gold and silver.) Not helpful, I'm afraid, but I thought I'd report progress (or lack thereof). Arthur Rubin | (talk) 01:35, 13 January 2006 (UTC)

One more possible name The Three Cards problem to make it like [Monty Hall problem]]. I still prefer The three cards (probabilty). I'll change to one of these in a day or 2 and add a few links to here as mentioned.Obina 14:13, 14 January 2006 (UTC)

I'd still prefer Gardner's title but either of those will do too. We could also use something like Three cards problem or The three cards puzzle. - Haukur 14:29, 14 January 2006 (UTC)

The "odd reference" sounds like he is referencing Bertrand's Box Paradox, which should really be merged here. Bertrand's problem, though, involved six coins contained in three boxes, rather than three coins. -- Antaeus Feldspar 18:50, 2 February 2007 (UTC)

It has also been called the Three Card Swindle. This name is applied when the swindler tries to get the mark to bet on the outcome on the theory that the relevant probability is 1/2 rather than 1/3. But this name doesn't find many pages on Google. Jim 14159 03:44, 5 November 2007 (UTC)

[edit] Colors

This is a trivial issue, but it's easiest to fix trivial issues. The article currently uses

green/purple.

By listing green first I mean that green is the color shown after the draw. The external links are divided between

black/red,
red/white,
black/white,
red/white.

Schuyler and Gardner, listed on this talk page, use

black/white,
black/white.

I don't know if the apparent consensus on the ordering (black > red > white) tells us anything about psychology, but I'm going to do a find-and-replace to follow Gardner. Yes, I know I have too much time on my hands! Melchoir 16:25, 12 January 2006 (UTC)

[edit] Expert tag

I think this article still needs a tag at the top indicating that it's not "done", whatever that means, so I'm adding Template:Expert. I think it's mainly "Formal approach" that needs logical cleanup, and we also need to add a correct, informal explanation. Melchoir 18:18, 14 January 2006 (UTC)

Why exactally do you feel an expert is needed? This is a very very simple probability question that prays on people ignoring the first part of the information so that they pick the 'obvious' answer of 1/2. I don't seen how an expert can provide more through an answer than the long winded one already shown. There are three cards, BB WW BW. If one card is showing B then its either the BB card one way up, the BB card the other way up or the BW card with B up. Thus 2/3 of those possible situations have B on the under side. Very very very simple, User_talk:Dacium|(talk)]] --Dacium 08:41, 13 February 2006 (UTC)

As I said before, the "Formal approach" section still needs work; historically, it's an outgrowth of an incorrect solution, and it should probably be scrapped entirely. As for the simple explanation, by all means, add it! The expert tag is not meant to discourage non-experts. Heck, I'm not an expert in probability myself, and it doesn't stop me! Melchoir 09:07, 13 February 2006 (UTC)

Eh, I did it myself. Melchoir 09:28, 13 February 2006 (UTC)

[edit] This problem and Lewis Carroll's two coins problem

There is something puzzling for me in the way this problem is presented; it differs from how I've seen the same problem presented before, in a way that does not change the answer but does actually duplicate a second, related problem from Lewis Carroll.

Carroll's problem was as follows: You have a bag, and in this bag are one regular coin with a heads and a tails, and one double-headed coin. You shake the bag, reach in and pull out a coin and look at only one side. The side you look at is heads; what is the chance that the other side is also heads?

Obviously, this is the same puzzle, in all functional respects, as the three cards problem as currently described in the article: simply substitute "cards" for coins", "heads" for "black" and "tails" for "white". Which highlights the puzzling part about the current presentation of the problem: even though it's called the "three cards problem", one of those three cards is absolutely irrelevant. Because we specify that the card face we're looking at is black, the card that's white on both sides can never be the one we're looking at.

I suggest that we might describe the problem in what I believe is closer to its original form: instead of specifying that we see a card face that's black, and asking the chance that the other side is black, we simply ask what the chances are that the other side is the same color as the side we're looking at, whichever that is. The answer remains the same: the chances are 2/3 that we're looking at one of the faces of a card with the same color on both sides. (This form of the puzzle also has a literary reference that can be mentioned -- it was used in a Leslie Charteris story about Simon Templar.) -- Antaeus Feldspar 15:47, 1 March 2006 (UTC)

I agree with you on mathematical grounds, but I disagree with your proposal because in all the references I've seen, the problem is given as it currently appears here. I think it would be better to expand "Symmetry" into a top-level section and mention Simon Templar there. It would contain the alternate statement of the problem, as well as the story of the scammer, if you know what I'm talking about. Melchoir 21:11, 1 March 2006 (UTC)

I'm afraid I don't know the story; fill me in? As far as how the problem appears in the references, Schuyler credits Gardner as his source, but Schuyler specifies that the card face looked at is black, which Gardner does not. I can't check the Nickerson or Bar-Hiller and Falk references right now, but I do find it rather hard to believe that the majority of commentators discussing "the three cards problem" are actually using a form of it where the existence of the third card is absolutely irrelevant. -- Antaeus Feldspar 00:19, 2 March 2006 (UTC)

The 2 coin problem is similar - perhaps identical mathmatically by some solutions. I think, though, the third card adds to the common sense wrong answer. And considering pulling one card out of a bag of 3 helps illustrate the value of considering the sample from which a items is drawn. This is very important in population probabilities, used by drug developers, marketing execs, and polititians. A card could be either one colour or two. But if one pulls it from a bag with 2/3 of the card having one colour, there is a 2/3 chance the card is monochromatic. As expressed here the problem helps one move to the same problem, say, where there are 5 cards in a bag. If we are told one is all white, one is all black, and the other 3 are W/B, we can solve this directly.Obina 11:50, 4 March 2006 (UTC)

Sorry, I forgot to respond earlier. The scam story I was talking about goes something like this: I draw one of the three cards from a hat. No matter what it shows, I invite you to bet money that the other side is the opposite color. Of course, regardless of the color, if you bet you have a 2/3 chance of losing. Mathematically speaking, the underlying problem is exactly what you described in your original comment: what matters is the probability of getting a match, not a particular color. However, the scam (hypothetically) works because I have shown you a color, and I ask an intentionally misleading question about a color. I am hoping that you don't think about generalities, but you are confused by the situation at hand and assume the probability is 1/2. Melchoir 19:28, 4 March 2006 (UTC)

[edit] And the rest?

In a survey of 53 Psychology freshmen taking an introductory probability course, 35 incorrectly responded 1/2; only 3 students correctly responded 2/3. What did the other 15 psychology students suggest? Anyone knows? Just curious. INic 02:40, 10 October 2006 (UTC)

[edit] Minor clarification suggestion

Hi - It might be just me, but I feel that in the "Solutions/Intuition" section, where it says:

"Two of the 3 black faces belong to the same card. The chance of choosing one of those 2 faces is 2/3."

it would be clearer if it said something like:

"Two of the 3 black faces belong to the same card. Given that you have choosen a black face, the chance of choosing one of those 2 faces is 2/3."

Without that, I at least was still thinking in terms of the overall cards...

Comments? Gwynevans 12:43, 2 February 2007 (UTC)

[edit] I think there is a logical error involved here somewhere (Or not)

Okay, if you're dealing with three cards, the probability of drawing a card with 2 of the same faces is 2/3. That much we know. But drawing from the three cards, you actually, without knowing any information about the card you've drawn, have a 1/3 or 33.33...% chance of drawing any single card. Now, if you look at one side of the card, you are left with 2 possible conclusions 1) The other side of your card has the same color as the side you see 2) The other side of your card has the another color

There are no thirds to consider, there's a 50% chance that the color will come up either way.

Looking at it as a matter of faces is an error, because that would mean you could draw each face randomly, and then you would divide the problem into 6ths, with a 2/5 chance of drawing another black face, and a 3/5 chance of drawing a white face. However, if we remove 2 white faces, IE eliminating the white/white card, we're left with a 2/3 chance to draw one of the two black faces, and a 1/3 chance to draw the white face, and that's where the error lies. Because we do not draw the other side of the card randomly, in essence, the way the problem is worded would give a 50% chance of either because the white/white card is extraneous. The card can be one of 2 options. Try doing the experiment without the white/white card, since that's what the wording in this problem suggests, and the trend will tend towards 50/50 unless you have a fluke.

Now, I do agree that the probability of drawing a card with the same color on either side is 2/3, or 66.666%, but if the problem was presented as shown in this article, then the kids who said 50% were correct. —The preceding unsigned comment was added by 68.227.203.149 (talk) 01:14, 14 February 2007 (UTC).

Added: I'm signing this because I just made an account.--Maveric Gamer 01:18, 14 February 2007 (UTC)

I'm afraid you're wrong. Just look at the symmetry argument. If the probability that you draw a card with the same color on both sides is 2/3, then that probability cannot magically become 1/2 if you specify the color you observe on the top. — Arthur Rubin | (talk) 01:45, 14 February 2007 (UTC)

Yes, but if there is a card in front of you that is black, there are only 2 cards it could be, thus there's a 50% chance that it is the black/black card, and a 50% chance it's the black/white card. Before the draw, there was a 1/3 chance that you would draw the black/black card, a 1/3 chance of drawing the black/white card, and a 1/3 chance of drawing the white/white card. After drawing a card that you don't see, that chance is still 33% either way, as there are 3 possibilities. After looking at the table with the card on it, and seeing the black card, one card gets eliminated as a possibility, so it can only be one of two cards. As point of fact, any information we can gain about this card will only serve to either make the odds 100% (if we were told that the sides are different) or 50% (both sides are the same). If we only know one side, then we eliminate one of three options, and since two options are left to us, it's still 50/50.

Even if we look at the sides. How many sides are white? 3. How many sides are black? 3. So what are the odds of drawing a card with the white side up? 50%. With the black side up? 50%. With the white side down? 50%. Black side down? Yep, still 50%.

Maveric Gamer 02:56, 14 February 2007 (UTC)

Okay, I think I get it now. Basically, what's happening, is that when there is a black side up, there are 3 possible scenarios 1) Side 1 of the B/B card is up 2) Side 2 of the B/B card is up 3) Side 3 (Black side of B/W) is up

In cases 1 and 2, the other side will also be black. I would delete everything I've said, but the discussion may help someone else comprehend this better. 68.227.203.149 07:06, 14 February 2007 (UTC) (Sorry about the IP, forgot to re-log in. Me again Maveric Gamer 15:29, 14 February 2007 (UTC))

I'm glad 68.227.203.149 got it. Here's the safe approach to this type of problem: List all outcomes of the fundamental experiment in a way where it's clear that there is symmetry, i.e. that they all have the same probability. E.g., flipping two identical dice simultaneously, you can only distinguish three outcomes (no heads, one head, or two heads), but to have symmetry you must list four outcomes (HH, HT, TH, and TT, where H=head and T=tail, and pretending you can distinguish HT from TH).

In our case, there are three cards of equal probability, and each card has two sides of equal probability, so even though the two sides of the B/B card are indistinguishable (and likewise the two sides of the W/W card), we list $3\times2=6$ outcomes of equal probability:

One side of B/B
Other side of B/B
One side of W/W
other side of W/W
B side of B/W
W side of B/W

These are all equally probable outcomes of the fundamental experiment, and as there are no other possibilities, each has an initial probability of 1/6.

Now, we are told we are watching a black side, which means we can discard possibilities 3, 4 and 6 in my list, leaving us with just the three possibilities listed by 68.227.203.149 above. There's no information making any of these possibilities more or less probable than the other two, so we still have symmetry, with a probability of 1/3 for each.

With this approach, try your hand at e.g. the Boy or Girl paradox (easy), or maybe even the Monty Hall problem (harder)!--Niels Ø (noe) 11:33, 14 February 2007 (UTC)

Ok, now I need some further help from you smart people. It seems that the above argument assumes that we are drawing sides, not cards. If we are drawing cards, there are three possibilities:

We draw a card, and find a black side showing. This eliminates W/W as a possibility. Only now the question is asked, "What is the chance that the other face is Black?". With only B/B and B/W as possibilities, is the probability not 1/2? It seems false to imply the "probability of seeing a black face" has anything to do with the problem, as the black face has already been seen, akin to saying that a coin flip has a 50% chance of being tails after the coin has already been flipped and tails is showing. Any help here will be appreciated.

Adl116 20:11, 14 September 2007 (UTC)

The problem with your reasoning, Adl116, is that we have more information than "one of the sides is black", when we take the card out of the bag and place it on the table, we're labelling the sides on that card, we are indeed drawing sides, and we know that "this side is black".

If, on the other hand, someone else, which we couldn't see, pulled out the card, looked at both sides, and said "(at least) one of the sides is black", your reasoning would be correct, and the chances would be 50-50.

This is also an interesting point. If someone else pulls the card, and says "one of the sides is black", we need to know if said person has looked on both sides, (or at least know the probability that he did,) before we can accurately estimate the probability that both sides are black.

Bogfjellmo 21:02, 11 October 2007 (UTC)

[edit] Why three cards?

What does the white/white card have to do with anything? It's not clear to me why it's included in the problem at all. Historic reasons?

Actually, there's a nice explanation of the correct answer using the third card! After somebody answers you 50/50, just ask them what's the probability, if you take a random card and place it on the table, that the hidden color is the same as the shown color... With three cards, the answer is obviously 2/3. Now you can explain that the fact the shown color is black doesn't change the probability. Ratfox (talk) 22:04, 28 November 2007 (UTC)

[edit] Add a history section?

I don't know anything about this problem other than what I've read on Wikipedia and the discussion pages. But obviously the Bertrand's Box Paradox is the exact same problem, so I think that these two articles should be consolidated. In addition, some Wikipedia users discussed deleting the Three cards article in January 2006 because they thought it was original research. Therefore, including Lewis Carroll (as discussed above) and Joseph Bertrand or anyone else in this article under something like a "History of the problem" section could be worthwhile for the article's credibility and interest. —Rafi Neal 04:22, 30 March 2007 (UTC)

Agreed, I'll merge the articles. Melchoir (talk) 00:47, 30 March 2008 (UTC)

[edit] Ambiguity in the Problem

The question boils down to: Are we picking a black side that goes with a card, or are we picking a card that happens to be black on one side?

If we are picking a black side, then there are 3 possible black sides, 2 of which belong to the all-black card, rendering the probability of choosing the all-black card 2/3.

If we are picking a card that happens to be black on one side, then there are 2 possible cards that have at least one black side. Therefore, choosing the all-black card has 1/2 probability.

This problem is based on the ambiguity of whether we know that we are picking a black side or not, not on its counterintuitive nature. —Preceding unsigned comment added by Aznthird (talk • contribs) 02:54, 14 November 2007 (UTC)

It is tempting to think that the confusion is due to an ambiguous statement of the problem. But this is not the case. The current article text is

"You put all of the cards in a hat, pull one out at random, and place it on a table. The side facing up is black."

Since we are observing only the side facing up, which turns out to be black, we have picked a black side as you put it. Picking a black card requires a much more elaborate setup, such as:

You put all of the cards in a hat, pull one out at random, and blindly pass it to an assistant, with the instruction that he should place it on the table with a black side facing up, if possible. The assistant takes the card, looks at both sides, and places it on the table. The side facing up is black."

This is clearly not the problem under consideration. Melchoir (talk) 00:45, 30 March 2008 (UTC)

[edit] Hmm

Ok first off: "What are the odds that the other side is also black?" I recommend learning the difference between odds and probability. Refer to wikipedia. The odds are 2

Now if your asking what is the probability that the other side is black that is 2/3rds. To those down below that think it is 1/2: the "trick" to this is labeling the faces 1 through 6. With 1,2 being the black card, and 3 being the black face of the black/white card. You have been told one of {1,2,3} is showing with equal probability. What is the probability that the face underneath is one of {1,2,3}. That probability is 2/3rds.

Probability: 1/3*1+1/3*1 +1/3*0 = 2/3

The odds are (2/3) / (1/3) = 2.

I would edit the thread but its just to darn much fun as is. Jeremiahrounds (talk) 21:59, 29 March 2008 (UTC)

The odds aren't "2" but "2 to 1". Since such odds aren't stated as simple numbers, there is no danger of confusion between the two formats, which are logically equivalent. And the difference between them is not as interesting as the paradox under discussion.

I like the labeling trick; fortunately it's already in the article. Melchoir (talk) 01:24, 30 March 2008 (UTC)

[edit] Talk archive of Bertrand's box paradox

This article should probably be consolidated with Three cards problem, since they are exactly the same problem, only with cards substituted for boxes, sides and their colors substituted for drawers and their coins. -- Antaeus Feldspar 05:22, 10 January 2007 (UTC)

I agree. Also, there was some discussion on the Three cards problem's talk page about a similar Lewis Carroll problem. Also, apparently the three cards article was nominated for deletion in January 2006 on the grounds of it being original research—so adding Bertrand and Carroll to the article, perhaps in a "History of the problem" section, would give the article more credibility. I'll add this comment to the Three cards problem's talk page. —Rafi Neal 03:57, 30 March 2007 (UTC)

I came to the discussion page to make exactly this same suggestion. I tagged the article for merge. — brighterorange (talk) 00:43, 24 April 2007 (UTC)

[edit] Merged Three cards problem into Bertrand's box paradox

I've just merged Three cards problem into Bertrand's box paradox. I think this is consistent with the comments in the above pre-merge talk page sections. Melchoir (talk) 01:17, 30 March 2008 (UTC)