You have three cards in a hat. One is white on both sides, one is red on both sides, the third is white on one side and red on the other. You pull a card out of the hat and lay it down. The face is white. What are the chances that the other side of the card is also white?
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card problem 9
- Thread starter BobRodes
- Start date
>It strikes me as a more general formulation of the problem, of which my example is a specific instance.
Look at guitarzan's and traingamer's examples (and even my extreme example involving billions of extra x/x cards). In ALL of those the answer to the original problem remains 2/3 - in other words your problem is NOT a special case.
However, Olaf's solution would differ (and be different for each specific quantity of extra cards that we introduce) because he is not answering the question that was asked.
Although the following is not what Olaf has suggested, it is akin to someone spotting that dividing 2 by 3 results in an answer of 2/3, which matches the correct answer and is therefore somehow a legitimate restatement of the question - which of course it is not, despite the fact the the answers happen to be the same
Look at guitarzan's and traingamer's examples (and even my extreme example involving billions of extra x/x cards). In ALL of those the answer to the original problem remains 2/3 - in other words your problem is NOT a special case.
However, Olaf's solution would differ (and be different for each specific quantity of extra cards that we introduce) because he is not answering the question that was asked.
Although the following is not what Olaf has suggested, it is akin to someone spotting that dividing 2 by 3 results in an answer of 2/3, which matches the correct answer and is therefore somehow a legitimate restatement of the question - which of course it is not, despite the fact the the answers happen to be the same
@ BobRodes re: That's like insisting that the question of how many legs a tarantula has is a completely different question from the number of legs a spider has, and that the fact that the answers are the same is meaningless.
No its not like spider legs. It like restating the question to be "What percentage of cards can satisfy the premise of the question" Only 2 of 3 cards can be white side up so the answer is 2/3. Same answer, different question.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
No its not like spider legs. It like restating the question to be "What percentage of cards can satisfy the premise of the question" Only 2 of 3 cards can be white side up so the answer is 2/3. Same answer, different question.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
Olaf Doschke
Programmer
Bob Rodes,
strongm,
it's not true that I'm rephrasing the problem as you see it. But even if I'd not argue that, this reformulation would not hold true if you add a blue/blue card, because that new card would break the symmetry of the problem in a way tat my equivalent transformation of the problem would not hold true.
Therefore arguing that applying your view of my new question on the extended situation computes to a wrong result is no proof at all, that my result is only coincidally right. You invalidate a result by applying the transformed problem on a transformed outset, which of course gives the wrong result, but only because you broke the nature of the problem.
My result is the same as in the original problem, because I did a correct equivalent transformation of the problem. Even though the different problem can also be solved by counting the mono colored cards like you say it, or the number of two colored cards like I did. This doesn't mean this is just true by coincidence. It's just astonishing that the problem get's so simple and different by the transformation, which is as Bob Rodes puts it a generallisation of the problem.
This generalisation does not hold true when adding the blue card therefore you can't apply the transformed question then to proof it's wrong. In fact in a first step I'd then say the blue/blue card does nothing to the problem and can be disregarded.
The red/red card can also be disregarded in the original problem, and if you rephrase the problem with the red color the white/white card can be disregarded. In my different view on the problem the white/red card can be disregarded. In my rephrasing of the problem to a "totally different" one ,you don't have to think in terms of the single faces to get to the result. In that way it is simpler.
I agree after all this discussion, that this kind of simplyfication is not at all simple to understand and agree on.
Bye, Olaf.
strongm,
it's not true that I'm rephrasing the problem as you see it. But even if I'd not argue that, this reformulation would not hold true if you add a blue/blue card, because that new card would break the symmetry of the problem in a way tat my equivalent transformation of the problem would not hold true.
Therefore arguing that applying your view of my new question on the extended situation computes to a wrong result is no proof at all, that my result is only coincidally right. You invalidate a result by applying the transformed problem on a transformed outset, which of course gives the wrong result, but only because you broke the nature of the problem.
My result is the same as in the original problem, because I did a correct equivalent transformation of the problem. Even though the different problem can also be solved by counting the mono colored cards like you say it, or the number of two colored cards like I did. This doesn't mean this is just true by coincidence. It's just astonishing that the problem get's so simple and different by the transformation, which is as Bob Rodes puts it a generallisation of the problem.
This generalisation does not hold true when adding the blue card therefore you can't apply the transformed question then to proof it's wrong. In fact in a first step I'd then say the blue/blue card does nothing to the problem and can be disregarded.
The red/red card can also be disregarded in the original problem, and if you rephrase the problem with the red color the white/white card can be disregarded. In my different view on the problem the white/red card can be disregarded. In my rephrasing of the problem to a "totally different" one ,you don't have to think in terms of the single faces to get to the result. In that way it is simpler.
I agree after all this discussion, that this kind of simplyfication is not at all simple to understand and agree on.
Bye, Olaf.
Olaf Doschke
Programmer
I'll simply take on the 4 card problem and show you how that changes arguing:
these three formulations are NOT all equivalent in that card set situation:
1. white face up, what's the probability of the other face being white?
2. red face up, what's the probability of the other face being red?
3. blue face up, what's the probability of the other face being blue?
It's easy to see that the anser to the 3. problem is 1 or 100%, as there only is the blue/blue card you know the other face is blue too. The first two problems do have different options and while their probability is not that easy to see it's easy to see that the rsul is not 100%.
You'd need to add a blue/red and blue/white card to reastablish the symmetry again and to be able to repharse it to:
4. any face up, what's the probabilty of the other face being the same color?
But only then.
Bye, Olaf.
these three formulations are NOT all equivalent in that card set situation:
1. white face up, what's the probability of the other face being white?
2. red face up, what's the probability of the other face being red?
3. blue face up, what's the probability of the other face being blue?
It's easy to see that the anser to the 3. problem is 1 or 100%, as there only is the blue/blue card you know the other face is blue too. The first two problems do have different options and while their probability is not that easy to see it's easy to see that the rsul is not 100%.
You'd need to add a blue/red and blue/white card to reastablish the symmetry again and to be able to repharse it to:
4. any face up, what's the probabilty of the other face being the same color?
But only then.
Bye, Olaf.
Olaf Doschke
Programmer
Bob Rodes,
what I wanted to write to you: Thanks, you're exactly seeing my point.
Bye, Olaf.
what I wanted to write to you: Thanks, you're exactly seeing my point.
Bye, Olaf.
>You invalidate a result by applying the transformed problem on a transformed outset, which of course gives the wrong result, but only because you broke the nature of the problem
Not at all. I only invalidate your solution. The problem remains EXACTLY the same as it was before.
If you read my solution to this problem, you will see that I have repeatedly stated that all cards apart from the white/white and the white/red are completely irrelevant to the problem, they are red herrings, specifically designed to confuse the issue. The problem, when simplified, boils down to:
I have a red/white card, and a white/white card. I'm showing a white face, what are the odds that the other side is white?
(or, if you prefer, I have a red/white card, and a red/red card. I'm showing a red face, what are the odds that the other side is red?)
Not at all. I only invalidate your solution. The problem remains EXACTLY the same as it was before.
If you read my solution to this problem, you will see that I have repeatedly stated that all cards apart from the white/white and the white/red are completely irrelevant to the problem, they are red herrings, specifically designed to confuse the issue. The problem, when simplified, boils down to:
I have a red/white card, and a white/white card. I'm showing a white face, what are the odds that the other side is white?
(or, if you prefer, I have a red/white card, and a red/red card. I'm showing a red face, what are the odds that the other side is red?)
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2
- #127
This interesting puzzle is more than 100 years old.
Olaf,
your point 4 in your post above concurs with my simplification of the problem (also above). The symmetry of this I happily accept - but that isn't how you originally stated your solution, and it is that original solution that I have a problem with.
your point 4 in your post above concurs with my simplification of the problem (also above). The symmetry of this I happily accept - but that isn't how you originally stated your solution, and it is that original solution that I have a problem with.
lionelhill
Technical User
Olaf's solution isn't coincidence, it works for valid arithmetical reasons based on the three cards stated in the original problem; so does the 6-face approach to solving the problem. The fact that it doesn't deal with blue-blue cards is irrelevant since they aren't in the original problem.
Think in Excel terms: there may be two formulae that produce the right answer from all the current expected input, but they might do it in different ways, and could differ in how they would handle novel, different input. Which is right?
Think in Excel terms: there may be two formulae that produce the right answer from all the current expected input, but they might do it in different ways, and could differ in how they would handle novel, different input. Which is right?
Olaf Doschke
Programmer
strongm,
surely you did interpret differently how I initially stated my solution, but still I meant it from the beginning.
It's true that either adding the blue/blue card or taking away the red/red card doesn't change the problem as it is stated, as it is stated for the white face up. It was in fact not you but guitarzan that stated my solution would compute to 3/4 in the changed situation, which is complete nonsense, as my solution couldn't be applied to the changed problem.
Of course if you limit it to the only cards that matter, you can also solve it by computing the odds for the four faces problem, it's simple, true.
But having the symmetry and using it is not wrong, too. And certainly not just happens to be right by coincidence, as you said it to.
Bye, Olaf.
surely you did interpret differently how I initially stated my solution, but still I meant it from the beginning.
It's true that either adding the blue/blue card or taking away the red/red card doesn't change the problem as it is stated, as it is stated for the white face up. It was in fact not you but guitarzan that stated my solution would compute to 3/4 in the changed situation, which is complete nonsense, as my solution couldn't be applied to the changed problem.
Of course if you limit it to the only cards that matter, you can also solve it by computing the odds for the four faces problem, it's simple, true.
But having the symmetry and using it is not wrong, too. And certainly not just happens to be right by coincidence, as you said it to.
Bye, Olaf.
Olaf,
Your "simple" solution is reposted below:
Kindly explain to me that the "2" represents and what the "3" represents in your solution.
Because in strongm's (correct) solution, I know exactly what they represent. The "2" represents side A of the white/white card and side B of the white/white card. And the "3" represents the total of sides that are white (the two listed above, plus the white face of the white/red card).
But I want to hear your justification of the numbers. What does the "2" and the "3" mean in your solution? Two what? out of three what?
Your "simple" solution is reposted below:
So your answer matches the correct answer, which is 2/3 (2 out of 3).OlafDoschke said:And this whole thread could've been much short with this very simple explanation:
With no outset given the chance to pick the red/white card is 1/3. So the chance, the downside face of the card has the different color is 1/3 and therefore the chance the downside face is the same color is 2/3. That's true for either outset of a white or red face showing.
Did I make you blush
Kindly explain to me that the "2" represents and what the "3" represents in your solution.
Because in strongm's (correct) solution, I know exactly what they represent. The "2" represents side A of the white/white card and side B of the white/white card. And the "3" represents the total of sides that are white (the two listed above, plus the white face of the white/red card).
But I want to hear your justification of the numbers. What does the "2" and the "3" mean in your solution? Two what? out of three what?
Olaf: I can see that what you are doing and I can see why it yields the same answer. I can see the symmetry and I can see that the equation balances.
You have calculated that "any face up, the probabilty of the other face being the same color = 2/3"
The question asks: "The face is white. What are the chances that the other side of the card is also white?"
You must PROVE equivelency. You can't simply state that it exists and if I can't see it then you give up trying to explain it.
I'm willing to bet that the method you need to use to prove equivelency resembles the solutions put forth earlier by many people including myself.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
You have calculated that "any face up, the probabilty of the other face being the same color = 2/3"
The question asks: "The face is white. What are the chances that the other side of the card is also white?"
You must PROVE equivelency. You can't simply state that it exists and if I can't see it then you give up trying to explain it.
I'm willing to bet that the method you need to use to prove equivelency resembles the solutions put forth earlier by many people including myself.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
Given the symmetry of the RR/WW card distribution, you can (and Olaf did) legitimately rephrase the original question from:
'The face is white. What are the chances that the other side of the card is also white?'
to
'What are the chances of drawing a card with the same colour on both sides'
...at which point you're just counting cards.
This solution scales so long as the distribution of RR/WW cards is identical.
ie. given 3 RW, 5 RR, and 5 WW cards (13 total), the correct answer to both the original question and the 'rephrased question' is now 10/13.
Equivalency proof?
R = Number of Red/Red Cards
W = Number of White/White Cards
M = Number of Red/White Cards
1) given white, 2nd face also white. p(wg,w) = 2W/(2W+M)
2) given white, 2nd face red. p(wg,r) = M/(2W+M)
3) given red, 2nd face also red. p(rg,r) = 2R/(2R+M)
4) given red, 2nd face white. p(rg,w) = M/(2R+M)
The chances of seeing a white face in the first side are : (2R+M)/(2R+2W+2M)
For R=W, this simplifies to (2R+M)/(4R+2M) = 1/2
(and likewise for red)
p(w,w) = 2W/(4W+2M)
p(w,r) = M/(4W+2M)
p(r,r) = 2W/(4W+2M)
p(r,w) = M/(4W+2M)
5) p(same) = p(r,r)+p(w,w) = 4W/(4W+2M) = 2W/(2W+M)
6) p(diff) = p(r,w)+p(w,r) = 2M/(4W+2M) = M/(2W+M)
The terms 1) and 5) are identical for R=W
'The face is white. What are the chances that the other side of the card is also white?'
to
'What are the chances of drawing a card with the same colour on both sides'
...at which point you're just counting cards.
This solution scales so long as the distribution of RR/WW cards is identical.
ie. given 3 RW, 5 RR, and 5 WW cards (13 total), the correct answer to both the original question and the 'rephrased question' is now 10/13.
Equivalency proof?
R = Number of Red/Red Cards
W = Number of White/White Cards
M = Number of Red/White Cards
1) given white, 2nd face also white. p(wg,w) = 2W/(2W+M)
2) given white, 2nd face red. p(wg,r) = M/(2W+M)
3) given red, 2nd face also red. p(rg,r) = 2R/(2R+M)
4) given red, 2nd face white. p(rg,w) = M/(2R+M)
The chances of seeing a white face in the first side are : (2R+M)/(2R+2W+2M)
For R=W, this simplifies to (2R+M)/(4R+2M) = 1/2
(and likewise for red)
p(w,w) = 2W/(4W+2M)
p(w,r) = M/(4W+2M)
p(r,r) = 2W/(4W+2M)
p(r,w) = M/(4W+2M)
5) p(same) = p(r,r)+p(w,w) = 4W/(4W+2M) = 2W/(2W+M)
6) p(diff) = p(r,w)+p(w,r) = 2M/(4W+2M) = M/(2W+M)
The terms 1) and 5) are identical for R=W
@ Brigmar. (Stepping in for Olaf)
Now compare your proof to the original solutions to the problem. (The most detailed being Guitarzan @ 20 May 10 11:48 )
1- Is is simpler?
2- Is is just an expansion?
3- Is it legitimate to offer a solution that does not answer the question asked without proving equivelency?
The reason I ask these questions is that there are many people responding to this post that don't accept the mathmatical proofs already given. Olaf's solution obviously does not adhere to the premise of the question. It is very easy to dismiss based on this varience and as such requires additional work to prove equivelence.
Is it the right answer - Yes
Is it based on the same mathematical probabilities - Yes
Does it answer the question as asked - No
Is it simpler - No (considering the extra proofs involved)
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
Now compare your proof to the original solutions to the problem. (The most detailed being Guitarzan @ 20 May 10 11:48 )
1- Is is simpler?
2- Is is just an expansion?
3- Is it legitimate to offer a solution that does not answer the question asked without proving equivelency?
The reason I ask these questions is that there are many people responding to this post that don't accept the mathmatical proofs already given. Olaf's solution obviously does not adhere to the premise of the question. It is very easy to dismiss based on this varience and as such requires additional work to prove equivelence.
Is it the right answer - Yes
Is it based on the same mathematical probabilities - Yes
Does it answer the question as asked - No
Is it simpler - No (considering the extra proofs involved)
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
Olaf Doschke
Programmer
kwbMitel,
I don't see what kind of extra proofs you need. There is a symmetry in the distribution of colors that way: If you exchange all red faces with white faces and all white faces with red faces, the cardset is the same again.
Or if you keep the same cardset and rephrase the problem to red face up and asking for probabilities of the other face being red too, you're getting the exact same result for reason of the symmetry.
So in fact the name of the initial color does not really matter, what yields the correct result is asking for the probability of the other side being the same color.
I already admitted it might not be that easy to get it, but in general finding and using symmetries in physics is the basis of many physical laws and models. I wanted just to stress out the symmetry. Unfortunately many don't seem to get that.
Bye, Olaf.
I don't see what kind of extra proofs you need. There is a symmetry in the distribution of colors that way: If you exchange all red faces with white faces and all white faces with red faces, the cardset is the same again.
Or if you keep the same cardset and rephrase the problem to red face up and asking for probabilities of the other face being red too, you're getting the exact same result for reason of the symmetry.
So in fact the name of the initial color does not really matter, what yields the correct result is asking for the probability of the other side being the same color.
I already admitted it might not be that easy to get it, but in general finding and using symmetries in physics is the basis of many physical laws and models. I wanted just to stress out the symmetry. Unfortunately many don't seem to get that.
Bye, Olaf.
Olaf - My mistake earlier was in agreeing with others that your solution was only correct by coincedence. I see that is not the case. Most recently I stated that I agree with your answer and agree that the method is based on the same probabilities.
The main difference is that by including all possibilities instead of the ones specified by the question you must subsequently prove that they are equivelent to answer the question.
In proving the equivelency, you must use the solution to the problem provided by others much earlier in this thread.
This is not the simpler way.
My issue with your solution has mainly been that it does not answer the question as posed. The fact that it is correct is beside the point. If I were convinced that the answer was 50% as many others in this thread believe, your answer would not sway me as it obviously includes variables intentionally excluded by the puzzle.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
The main difference is that by including all possibilities instead of the ones specified by the question you must subsequently prove that they are equivelent to answer the question.
In proving the equivelency, you must use the solution to the problem provided by others much earlier in this thread.
This is not the simpler way.
My issue with your solution has mainly been that it does not answer the question as posed. The fact that it is correct is beside the point. If I were convinced that the answer was 50% as many others in this thread believe, your answer would not sway me as it obviously includes variables intentionally excluded by the puzzle.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
lionelhill
Technical User
Can I make a tentative suggestion that if anyone wishes to continue this thread, they start a new "continuation thread" with a similar subject title? This one has reached such a length that explorer on my PC cannot display this text I am typing half as quickly as I can type it...
Thanks all for an interesting debate
Thanks all for an interesting debate
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