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?
Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations Rhinorhino on being selected by the Tek-Tips community for having the most helpful posts in the forums last week. Way to Go!
card problem 9
- Thread starter BobRodes
- Start date
lionelhill
Technical User
This thread has suddenly become the world of programing, in miniature:
Olaf restated the original problem in a different way, and as a result was able to find the probabilities very quickly and easily. The objection raised to his reasoning is that it wouldn't apply if you add an extra card (but surely that is moving the goal-posts?) His way of thinking doesn't scale to solve the problem correctly if you change the problem.
In programing terms, Olaf has done what we used to in the good old days: He's noticed it is a special case and exploited it to make things faster and easier. The rest of us are doing what we do nowadays: keep everything totally generalised and don't worry if it makes everything big, slow, and difficult to understand.
Give alternative approaches a chance... Three cheers for a bit of sideways thinking!
Olaf restated the original problem in a different way, and as a result was able to find the probabilities very quickly and easily. The objection raised to his reasoning is that it wouldn't apply if you add an extra card (but surely that is moving the goal-posts?) His way of thinking doesn't scale to solve the problem correctly if you change the problem.
In programing terms, Olaf has done what we used to in the good old days: He's noticed it is a special case and exploited it to make things faster and easier. The rest of us are doing what we do nowadays: keep everything totally generalised and don't worry if it makes everything big, slow, and difficult to understand.
Give alternative approaches a chance... Three cheers for a bit of sideways thinking!
Lionhill, As Guitarzan says, Olafs "Solution" is coincidental. Olafs solution solves the question "What are the chances that the reverse side is the same as the top side". Yes that is a very simple question to answer. 2 cards have the same sides and 1 does not. 2/3 = answer.
Unfortunately, this is not the question asked by the puzzle.
Alternative maybe, sideways maybe, but completely wrong none the less.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
Unfortunately, this is not the question asked by the puzzle.
Alternative maybe, sideways maybe, but completely wrong none the less.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
>was able to find the probabilities very quickly and easily
As others have already said it was a coincidence only. It is not a special case. It is not a solution to the problem posed. And, in my opinion, it isn't simpler than a couple of the correct solutions already put forward in the thread.
As others have already said it was a coincidence only. It is not a special case. It is not a solution to the problem posed. And, in my opinion, it isn't simpler than a couple of the correct solutions already put forward in the thread.
Olaf Doschke
Programmer
Sorry, if you don't get my point. So if you insist the outset of a white face (or red face up) makes a difference, simply open your eyes and see it's a symmetrical problem no matter what outset you give. Adding a blue/blue card is unfair arguing, even though it still will keep it a symmetrical problem it introduces a third class of color for which the situation is different than for the other two colors, so it breaks the symmetry in regard to the three colors, blue has a different distribution.
And it even stays symmetric with a fourth blue/blue card, because the 1/4 chance to pick the red/white card would then not be relevant and the result would not change to 3/4. If you'd "interpolate" my argueing that way you haven't understand it. Besides that the situation is as it is.
It's a very important thing to see the symmetry in problems, but you don't seem to see that and argue the simplicity of the solution by adding asymmetry.
Bye, Olaf.
And it even stays symmetric with a fourth blue/blue card, because the 1/4 chance to pick the red/white card would then not be relevant and the result would not change to 3/4. If you'd "interpolate" my argueing that way you haven't understand it. Besides that the situation is as it is.
It's a very important thing to see the symmetry in problems, but you don't seem to see that and argue the simplicity of the solution by adding asymmetry.
Bye, Olaf.
traingamer
Programmer
Having just stumbled onto this thread, I was in the 50/50 camp (it has to be one or the other), until I got to strongm's post at: 20 May 10 12:45 and the succeeding ones from him and cc.
You obviously physically have one of two cards (appearance of 50/50), but you simply do not have equal chances of having those two cards - you have twice as many chances of having the white/white one - giving you the 2/3 answer.
Greg
People demand freedom of speech as a compensation for the freedom of thought which they seldom use. Kierkegaard
You obviously physically have one of two cards (appearance of 50/50), but you simply do not have equal chances of having those two cards - you have twice as many chances of having the white/white one - giving you the 2/3 answer.
Greg
People demand freedom of speech as a compensation for the freedom of thought which they seldom use. Kierkegaard
Olaf Doschke
Programmer
Disregard the second paragraph of my previous posting. The point is, that a blue/blue card would really break the symmetry of the problem, even though there'd still be kind of a symmetry in that eg mirroring all cards you'd get the same card set, but blue has a different distribution than red and white.
Bye, Olaf.
Bye, Olaf.
Olaf, Your solution does not answer the question that the puzzle asks. I get your point, don't assume I don't. Yes I do insist the outset of a white face, the puzzle clearly states it. Your solution ignores the puzzle and answers a completely different question. The fact that the answers are the same is meaningless.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
Olaf Doschke
Programmer
traingamer, yes. Perhaps the simplest way to look at it is to see it as 6 faces instead of 3 cards. the chance to have a white face up are 1/2, as three faces are white. That's the outset, so that probability does not matter. But in one of these 3 cases you have the white face of the white/red card and in 2 of the three cases you have one white face of the white/white card.
Bye, Olaf.
Bye, Olaf.
Olaf Doschke
Programmer
kwbMitel,
you still don't get me: I don't disregard the outset. I just take a step back and see the symmetry of the problem.
Let's say the outset was the opposite: a red face up, and the question would be, what is the probability of the other side being red? That would be the exact same problem. And that's what I used to simplify the problem.
Bye, Olaf.
you still don't get me: I don't disregard the outset. I just take a step back and see the symmetry of the problem.
Let's say the outset was the opposite: a red face up, and the question would be, what is the probability of the other side being red? That would be the exact same problem. And that's what I used to simplify the problem.
Bye, Olaf.
Olaf, Your last post finally answers the correct question. Welcome to the club. Looking back on your "simple explaination" do you now see how it differs from answering the puzzle as stated.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
Olaf
The blue/blue thing was only to illustrate that the "simple" explanation you gave had nothing to do with the original question. You basically stated that the answer is 2/3 because you have a 2/3 chance of having drawn a card where both faces are the same. This is NOT the original question.
However, the explanation you just gave above is absolutely correct.
The blue/blue thing was only to illustrate that the "simple" explanation you gave had nothing to do with the original question. You basically stated that the answer is 2/3 because you have a 2/3 chance of having drawn a card where both faces are the same. This is NOT the original question.
However, the explanation you just gave above is absolutely correct.
Olaf Doschke
Programmer
Yes, I see that it differs, but the difference doesn't matter. I still think of my first solution as more elegant, even though it might not be simler to understand. We can agree on that.
See, I'm a physicist and not just a programmer. That gives a really different view on mayn things. I just transformed the problem to a more general one in two steps:
1. outset white face, probability of white other face.
That's the problem as we know it.
2. outset red face, probability of red other face.
That's exact the same problem in this kind of outset.
3. outside any colored face, probability of same other face.
Because 1 and 2 are equivalent problems, this is also an equivalent problem. And it's an easier view on the problem if you consider my initial arguing.
Bye, Olaf.
See, I'm a physicist and not just a programmer. That gives a really different view on mayn things. I just transformed the problem to a more general one in two steps:
1. outset white face, probability of white other face.
That's the problem as we know it.
2. outset red face, probability of red other face.
That's exact the same problem in this kind of outset.
3. outside any colored face, probability of same other face.
Because 1 and 2 are equivalent problems, this is also an equivalent problem. And it's an easier view on the problem if you consider my initial arguing.
Bye, Olaf.
Olaf,
Your re-statement of the premise of the puzzle is equivelent to:
Given the 3 cards how many have the same face front and back.
I do not agree that this differnce does not matter.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
Your re-statement of the premise of the puzzle is equivelent to:
Given the 3 cards how many have the same face front and back.
I do not agree that this differnce does not matter.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
Olaf Doschke
Programmer
In mathematical problems like equation systems it's totally normal to simply problems by equivalency transformations. It's even part of the solution. If you don't see the three different formulations are stating the same problem in terms of this problem equivalency, I can only give up to explain this to you.
Bye, Olaf.
Bye, Olaf.
traingamer
Programmer
Given: You have 498 red/red cards, one red/white card, and one white/white card. 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?
[hide]It is still 2/3[/hide]
Greg
People demand freedom of speech as a compensation for the freedom of thought which they seldom use. Kierkegaard
What are the chances that the other side of the card is also white?
[hide]It is still 2/3[/hide]
Greg
People demand freedom of speech as a compensation for the freedom of thought which they seldom use. Kierkegaard
- Thread starter
- #117
<Olaf's example has absolutely no bearing on the original problem. None.
Well, I believe I disagree, the strongm imprimatur notwithstanding. Olaf is talking about the same set of cards. (Not 4, not 7, but the same set I proposed.) He's simply saying what the probability that if you turn up a white face, the next face will be white, or conversely if you turn up a red face the next face will be red. This is what he means by "with no outset" (granted, not an entirely accurate use of the word): he's saying that if the puzzle does NOT state the color of the card, his probabilities apply. My example is simply an instance of that more general question.
So, I guess it's a matter of how you define "bearing." No, it's not the same question. But I would opine that it has more bearing on the puzzle than, say, the dietary habits of birds does. As such, since the dietary habits of birds has zero bearing, his explanation must have nonzero bearing. Suppose you refute that logic.![[lol]](/data/assets/smilies/lol.gif "[lol] [lol]")
Well, I believe I disagree, the strongm imprimatur notwithstanding. Olaf is talking about the same set of cards. (Not 4, not 7, but the same set I proposed.) He's simply saying what the probability that if you turn up a white face, the next face will be white, or conversely if you turn up a red face the next face will be red. This is what he means by "with no outset" (granted, not an entirely accurate use of the word): he's saying that if the puzzle does NOT state the color of the card, his probabilities apply. My example is simply an instance of that more general question.
So, I guess it's a matter of how you define "bearing." No, it's not the same question. But I would opine that it has more bearing on the puzzle than, say, the dietary habits of birds does. As such, since the dietary habits of birds has zero bearing, his explanation must have nonzero bearing. Suppose you refute that logic.
>Adding a blue/blue card is unfair
No, it isn't. Adding a blue/blue card, a tauep/taupe card or a billion other x/x cards makes NO DIFFERENCE to the original problem at all, but it COMPLETELY changes your solution (traingamer illustrates this well, as did guitarzan earlier) because your solution (no matter what claims of symetry that you make) does NOT answer the problem as stated . It answers a completely different problem which, in one specific case, happens to match the correct answer for that original problem. In all other cases it does not and cannot match the correct solution becasue it is answering a different question.
> And that's what I used to simplify the problem
You didn't simplify. You changed the question.
No, it isn't. Adding a blue/blue card, a tauep/taupe card or a billion other x/x cards makes NO DIFFERENCE to the original problem at all, but it COMPLETELY changes your solution (traingamer illustrates this well, as did guitarzan earlier) because your solution (no matter what claims of symetry that you make) does NOT answer the problem as stated . It answers a completely different problem which, in one specific case, happens to match the correct answer for that original problem. In all other cases it does not and cannot match the correct solution becasue it is answering a different question.
> And that's what I used to simplify the problem
You didn't simplify. You changed the question.
- Thread starter
- #119
<Your re-statement of the premise of the puzzle is equivelent to:
Given the 3 cards how many have the same face front and back.
I disagree. IMO, what he is saying is "If you turn up a card, what is the probability that the other side of the card will be the same color"? (Your question has no bearing on that question. None.)
That said:
<As others have already said it was a coincidence only.
It strikes me as a more general formulation of the problem, of which my example is a specific instance.
<It is not a special case.
Agreed. I would say the reverse is true.
<It is not a solution to the problem posed.
Agreed.
<And, in my opinion, it isn't simpler than a couple of the correct solutions already put forward in the thread.
I share that opinion.
THAT said, it's not unreasonable to suggest one can gain understanding of problems extrapolating general principles from them and then applying the principles back to the problem. Perhaps that's the way Olaf approaches problem solving.
Given the 3 cards how many have the same face front and back.
I disagree. IMO, what he is saying is "If you turn up a card, what is the probability that the other side of the card will be the same color"? (Your question has no bearing on that question. None.
)That said:
<As others have already said it was a coincidence only.
It strikes me as a more general formulation of the problem, of which my example is a specific instance.
<It is not a special case.
Agreed. I would say the reverse is true.
<It is not a solution to the problem posed.
Agreed.
<And, in my opinion, it isn't simpler than a couple of the correct solutions already put forward in the thread.
I share that opinion.
THAT said, it's not unreasonable to suggest one can gain understanding of problems extrapolating general principles from them and then applying the principles back to the problem. Perhaps that's the way Olaf approaches problem solving.
- Thread starter
- #120
<Yes I do insist the outset of a white face, the puzzle clearly states it. Your solution ignores the puzzle and answers a completely different question. The fact that the answers are the same is meaningless.
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.
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.
- Status
Similar threads
- Locked
- Question