Talk:Bertrand's box paradox

message: 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)
replies: text: 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)

text: 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)
thread_title: False article?

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

text: 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)
thread_title: What's the answer (2/3)

message: 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)
replies: text: I found that Martin Gardner has written about it under the name of "The three cards" .  This should give more leads.  --C S (Talk) 08:49, 12 January 2006 (UTC)
replies: text: 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)
thread_title: From Heraclesprogeny to Arthur Rubin

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

message: 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)
thread_title:  Colors 

message: 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)
replies: text: 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)
replies: text: 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)

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

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)

text: 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)
thread_title:  This problem and Lewis Carroll's two coins problem 

message: 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)
thread_title: And the rest?

message: 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)
thread_title:  Minor clarification suggestion 

message: 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)
thread_title: Why three cards?

message: 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)
replies: text: Agreed, I'll merge the articles. Melchoir (talk) 00:47, 30 March 2008 (UTC)
thread_title:  Add a history section? 

message: 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)
replies: text: 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)

text: 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)
thread_title: Talk archive of [[Bertrand's box paradox]]

message: 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)
thread_title:  Merged [[Three cards problem]] into [[Bertrand's box paradox]] 

message: I would like to change the sentence "However, this reasoning fails to exploit all of your information; you know not only that the card on the table has a black face, but also that one of its black faces is facing you." to "However, this reasoning fails to exploit all of your information; you know not only that the card on the table has at least one black face, but also that it has a black face which is facing you." I realize this may seem redundant, as the only way to know the first part of the statement is by knowing the second part. But anyway prefer this to the current sentence, which strongly suggests the card has two black faces. Comments? RomaC (talk) 23:02, 19 June 2009 (UTC)
replies: text: I agree that the wording strongly suggested the card has two black faces.  I was about to make your clarifying changes, but then I realized the whole sentence was dubious.  So I have gotten rid of the "face which is facing you" bit altogether.  More complete explanations are in the section below it.   Open4D (talk) 20:41, 16 September 2009 (UTC)
thread_title: Clarity

message: I was introduced to this problem with the three coins in a bag version.   However, the difference is that here in this article the problem is stated as, paraphrased, "If heads is drawn, what is the probability the opposite side is heads" compared to, "No matter what side is drawn, what is the probability that the opposite side is heads."  Answer to the latter problem is 1/3.  The idea being that the coin "game" is a carnival game where the odds/probability are stacked in favor of the house at 2/3.  Apologies if I goofed something. Ij3n (talk) 00:31, 28 August 2009 (UTC)
thread_title: Minor Concern How the Problem Is Stated

message: I am about to remove this addition.  Why?  Let X represent 100 silver coins in a drawer.  If "each silver coin is replaced by 100 silver coins", then the problem simply changes from GG/GS/SS to GG/GX/XX and is otherwise unchanged.  I imagine this is not what the author intended; if so can I suggest a more detailed explanation?  This would be better placed in new section titled something like "alternative explanations if you still don't get it", which could also have many of the discussions in these talk pages written up and added to it.  (Although, if you believe in WP:NOTTEXTBOOK, then I suppose that section I have just described probably belongs in Wikibooks.) Open4D (talk) 10:10, 16 September 2009 (UTC)
thread_title:  Removing the "100 silver coins" bit 

message: The article currently explains the "box" problem, and then restarts from scratch with the "card" problem, explaining much of the same probability stuff all over again, with different wording and some extra bits thrown in. Everything that is generic should be explained once only, under whichever variant is covered first (which presumably should be the box, since it is called the "box paradox"). The "card" explanation then just needs to demonstrate how it is exactly the same problem (which is pretty obvious really). 109.151.57.10 (talk) 18:40, 4 December 2011 (UTC)
thread_title:  Poor organisation 

message: I made a minor change in the informal description to reflect more specifically the probability distinction for the casual reader who may still assume even probability distribution.  Flying Hamster (talk) 17:13, 4 March 2012 (UTC)
replies: text: Well, it already said as much; and the change you made is not quite logically consistent since it states the conclusion - that the two remaining possibilities are not equally likely - as a reason for the conclusion. But I'm going to try to clean it up a little to address your intended point. JeffJor (talk) 21:29, 5 March 2012 (UTC)
thread_title:  Wording 

message: The last sentence of the above section reads "It can be resolved only by recognizing how observing a gold coin changes the probability that the box is GS or GG, but not GG." Now I'm not an expert here, but it seems to me this is a typo and should in fact read "It can be resolved only by recognizing how observing a gold coin changes the probability that the box is GS or GG, but not SS". ???? Starfiend (talk) 16:38, 10 April 2012 (UTC)
replies: text: Then I guess I worded it poorly, since you misunderstood. No probability actually changes; but the combination of a specific observation with each event in {GG, GS, SS} is different than the probability of just each event itself. For example, if "OG" is the event "observe a gold coin," the probability is 1/3 for each of the events {GG, GS, SS}. But the probability is 1/3, 1/6, and 0, respectively, for {GG&OG, GS&OG, SS&OG}. This is the change I meant, and GG is the one that doesn't change in combination with OG. Then, the conditional probability is found by normalizing these probabilities with the factor 1/2, the sum of those probabilities. So P(GG|OG)=2/3, P(GS|OG)=1/3, and P(SS|OG)=0. Let me think about a way to word it better. JeffJor (talk) 21:16, 10 April 2012 (UTC)
thread_title:  Typo in "The paradox as stated by Bertrand" 

message: I added the Bayes' rule version of the solution by Bayes theorem, and filled in the formal details of a solution by symmetry. I also corrected the wording and the notation in the first solution (refering to the original silver and gold coins version of the problem) . The probabilities which are defined there are various conditional probabilities and the solution is actually the Bayes' rule solution. Richard Gill (talk) 12:10, 19 April 2013 (UTC)
thread_title:  Bayes rule, symmetry arguments 

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thread_title:  External links modified 

message: I've been looking for a translation of the original text to base a major re-write upon (as opposed to the poorly-worded section I provided in 2014), and hadn't found one. But I did just (re-)locate the source that cites it and is what I was recalling then. Eugene Northrop, Riddles in Mathematics, starting on page 165.
The issue is, that the "paradox" which Joseph Bertrand refers to is not the problem itself, which appears to be a veridical paradox to some. It is the way that Bertrand demonstrated that the naive solution, which concludes that the probability is 1/2, could not be right. The problem itself was stated like this:
:Three boxes are identical in external appearance. The first box contains two gold coins, the second contains two silver coins,
and the third contains a coin of each kind, one gold and one silver. A box is chosen at random. What is the probability that
it contains the unlike coins?''
The answer is trivial: 1/3. But what if you remove one coin from the chosen box, without looking at it? The naive solution says that if you were to take one coin out and look at it, the answer would change to 1/2 no matter what you saw. So you don't actually need to look at it - simply taking a coin out changes the answer to Bertrand's actual question. That absurd conclusion is the paradox Bertrand refers to.
What I intend, is to change the problem statement to what Bertrand actually said, and highlight the differences between how subsequent authors have re-interpreted it. It may take a bit of time to do so, so I'll save drafts until I'm done. And I'm open to comments and suggestions. JeffJor (talk) 12:09, 26 July 2017 (UTC)
replies: text: I believe you mean he said "if you take one coin out, look at it, and see that it's gold", which would appear to lead to the 1/2 conclusion and the paradox. I think this is related to an effect I can't find a reference to. In one example of it, subjects in an experiment were slower to correctly identify an object that gradually became less blurred if they initially saw it in a blurred state and misidentified it. In another example, possibly related to but not mentioned on Phonemic restoration effect, people watching a video or possibly a silent video insist a phoneme or word is what they expect it to be; possibly lip-reading a P or B sound. The point is that the "naive" solution of 1/2 is hard to understand for someone who has already agreed to the answer of 1/3. For someone asking "what is the color/material of the second coin", they won't see any reason to think the probability is 1/2. That only happens if you ignore the previous answer of 1/3 and don't think about the unseen coin, and just ask yourself "which boxes have a coin like the one I just removed". I do think the current description could be "revisionist" and not reflect the original statement. 23.121.191.18 (talk) 03:22, 15 October 2017 (UTC)
thread_title:  The current article is based on revisionist history 

message: The similarly-named Bertrand paradox (probability) illustrates the difficulty of defining a random probability. I think what is happening here is we are primed to think of 'fate': we didn't select the box with silver coins. It is not hard to go from there to saying that if the 'random' box was the one with mixed coins, it was 'fate' that we selected the gold one. The word "if" means we are speculating about a subset of hypothetical scenarios, and it's easy to think we should give the hypothetical scenario of selecting the mixed box and then the gold coin the same probability or weight as selecting the box with gold coins. (For example, maybe the gold coin is closer to the opening to the box.) This suggests the range of answers could vary with a practical experiment compared to asking about the hypothetical situation. Instead of repeatedly trying until they get a gold coin, the experimenter could just ask about whatever type of coin they happen to randomly pick out.
I think this is more accurate than saying that people have been trained to think abstractly about similar objects, and that this causes them to 'merge' the two gold coins in the same box, and the scenarios of selecting each of those coins, in their mind. I mention this because just like most psychology majors, my initial answer was 1/2. 23.121.191.18 (talk) 03:22, 15 October 2017 (UTC)
thread_title:  Consolation for psychology majors 

message: Hello. I am not familiar with this paradox and it seems to me that a sentence in the introduction is poorly written to people in my situation. The 'paradox' is in the probability, after choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, of the next coin drawn from the same box also being a gold coin. I don't understand if the first coin mentioned is withdrawn from the box – itself randomly chosen – or from any of the three boxes. I also do not understand the sequence of events, as it is not clear. Should we pick a coin, look at it and put it back in its original box would it turn out to be a silver coin? All this is extremely poorly worded, making the reading of the rest of the article tedious because we also have to be guessing what the rules were meant to be from the results, which is almost impossible to achieve in an article that is about things being counterintuitive. Can someone fix that? Thierry Caro (talk) 09:52, 9 August 2020 (UTC)
thread_title:  The introduction is poorly written 

message: It would appear even experts in this particular field don't know anything about this field.
This doesn't bode well for the rest of us either.
I will let my work stand for itself. Feel free to run it and look for errors.
   <script >   
   function randomIntFromInterval(min, max) { // min and max included 
   return Math.floor(Math.random() * (max - min + 1) + min)
    }
   function Ball() {
    var gold = 0;
    var winner = 0;
    var i = 0;
    var rndInt = 0;
    var NumberofRounds = 100000;
    while (i < NumberofRounds) {
        rndInt = randomIntFromInterval(1, 3);
        //if 1, then box contains 2 gold balls, if 2, then 1, if 3, then 0.
        if (rndInt != 3) //you drew a gold ball, congrats
        {
            gold += 1; //increment the game counter
            if (rndInt != 2) //you are in the box with 2 gold balls
            {
                winner += 1; //winner winner chicken dinner
            }
        }
        i++;
    }
    odds = Math.round((winner / gold) * 100);
    //alert("the odds of drawing any gold is " + gold.toString()  + " in " +  NumberofRounds.toString()  + ". The odds of drawing two is : " +  winner.toString() +  " out of " +  NumberofRounds.toString() + " rounds in total, of which "+ gold.toString() + " contain valid gold. The thought experiment is your odds of drawing gold again. For that, your odds are " + winner.toString() + "/" + gold.toString() + ". The answer, therefore, is : " + odds.toString() + " %."); // show popup with the ratio
    return odds;
   } 
   </script>
:Your error is in your main conditional part.
   if (rndInt != 3) //You chose a box with at least 1 gold ball
   {
      gold += 1; //, but have not yet drawn a ball from it.
:It should be
   if (rndInt != 3) //you still have a chance of getting gold
   {
      ball = randomIntFromInterval(1, rndInt); //draw a ball from the box
      //if 1 then gold, if 2 then silver
      if (ball == 1)
      {
         gold += 1; //increment the game counter
         if (rndInt != 2) //you are in the box with 2 gold balls
         {
             winner += 1; //winner winner chicken dinner
         }
      }
   }
:, where
   var ball = 0;
:was declared along with the other declarations.  JumpDiscont (talk) 05:08, 22 October 2021 (UTC)
thread_title:  The answer is 1/2 

message: Due to Vaidman, Aranohov et al.  should it be mentioned?  cheers, Michael C. Price talk 19:12, 3 July 2021 (UTC)
thread_title:  Quantum three box paradox 

message: https://github.com/isghe/Paradosso_delle_tre_carte
Isidoro Ghezzi. 2.39.50.17 (talk) 21:16, 13 November 2022 (UTC)
thread_title:  It looks we need just TWO cards 

message: In the first draw, you pick 1 out of 3 boxes. This can be written as:
\frac{1}{3}+\frac{1}{3}+\frac{1}{3}
However since 1 box doesn't contain any gold coins, that possibility is discarded, so it's 1 out of 2, which would be:
\frac{1}{2}+\frac{1}{2}
After that, in both cases, 1 out of 2 coins is chosen. The result depends on the box chosen, so it looks like this:
\frac{1}{2}\times\left(\frac{1}{2}+\frac{1}{2}\right)+\frac{1}{2}\times\left(\frac{1}{2}+\frac{1}{2}\right)
But again, one of those possibilities (silver coin being chosen) has to be discarded, so it's actually like this:
\frac{1}{2}\times\left(\frac{1}{2}+\frac{1}{2}\right)+\frac{1}{2}\times\left(\frac{1}{1}\right)
As a result, the boxes now look like either of those 3 variants:
# [G] [GS] [SS]
# [G] [GS] [SS]
# [GG] [S] [SS]
Now it's time for the 2nd draw, including all 3 boxes. The draw will be different based on the result of the previous draw, so we have this:
\frac{1}{2}\times\left(\frac{1}{2}\times\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)+\frac{1}{2}\times\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)\right)+\frac{1}{2}\times\left(\frac{1}{1}\times\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)\right)
Finally the second coin is chosen based on the box chosen and the result from the previous draw. The probability distribution of pulling the gold coin from each box looks like this:
# [G] [GS] [SS] (1/1); (1/2); (0/2)
# [G] [GS] [SS] (1/1); (1/2); (0/2)
# [GG] [S] [SS] (2/2; (0/1); (0/2)
Which means the final calculation looks like this:
\frac{1}{2}\times\left(\frac{1}{2}\times\left(\frac{1}{3}\times\frac{1}{1}+\frac{1}{3}\times\frac{1}{2}+\frac{1}{3}\times\frac{0}{2}\right)+\frac{1}{2}\times\left(\frac{1}{3}\times\frac{1}{1}+\frac{1}{3}\times\frac{1}{2}+\frac{1}{3}\times\frac{0}{2}\right)\right)+\frac{1}{2}\times\left(\frac{1}{1}\times\left(\frac{1}{3}\times\frac{2}{2}+\frac{1}{3}\times\frac{0}{1}+\frac{1}{3}\times\frac{0}{2}\right)\right)
= \frac{5}{12}
= 0,41(6)
 
Archimedes5000 (talk) 19:26, 18 March 2023 (UTC)
thread_title:  The solution in the article makes no sense