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
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@CC
I see your point but don't have another argument in favor of my point other than what I've all ready presented.
Not being an expert in probabilities, I can only go with what feels right.
Although your argument makes perfect sense, I still think my does as well.
Just wish I knew of another way to present it!
Randy
I see your point but don't have another argument in favor of my point other than what I've all ready presented.
Not being an expert in probabilities, I can only go with what feels right.
Although your argument makes perfect sense, I still think my does as well.
Just wish I knew of another way to present it!
Randy
CajunCenturion
Programmer
And does it not accurately and completely model the question as posed?randy700 said:
The modifications you suggest are for a different problem, one that is independent of the initial conditions of this problem. You can't ignore the initial conditions. They do matter. The initial conditions and initial draw determine the valid states from which you make the flip.
I understand that you want to redefine the valid states for the flip, but it's simply not appropriate for THIS problem. The initial conditions matter.
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Randy700 try taking all theory out of the equation, create 3 cards and write red on 3 sides of 2 cards. Now do the experiment.
Here is what will result.
Every time you pick the W / W card you will keep it.
Half the time you pick the W / R card you will keep it
None of the times you pick the R / R card will you keep it.
You will keep the W / W card twice as many times as you will keep the W / R card.
2:1 ratio = 2/3
Don't believe me, try it!
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Occam's Razor - All things being equal, the simplest solution is the right one.
Here is what will result.
Every time you pick the W / W card you will keep it.
Half the time you pick the W / R card you will keep it
None of the times you pick the R / R card will you keep it.
You will keep the W / W card twice as many times as you will keep the W / R card.
2:1 ratio = 2/3
Don't believe me, try it!
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Occam's Razor - All things being equal, the simplest solution is the right one.
kbwmitel that is not the question (although it may have been what was intended)
the question is how often it will be white when when a white back is selected.
any time the red/red is selected it is removed from the test & ignored
likewise when the red white is selected showing the red back it is ignored
the white card can be selected in either orientation which leads to 2-3 BEFORE the card is selected but once the card has been chosen & revealed a white side (the conditions of the experiment present at the time the question is now being asked) the probabiliy for the white white card to have 2 states colapses.
there are now only 2 options
it is the white/white card in which case we pass
or it is the white red card & we fail 50:50
The question is not how many times will we pick the white/white card in preference to the white side of the red/white card.
it is once we have chosen a white card what colour will the opposite side be.
In the 1st q the white card can be either white A or white B giving 2 options in the second an actual question the card can only be white white.
the question is how often it will be white when when a white back is selected.
any time the red/red is selected it is removed from the test & ignored
likewise when the red white is selected showing the red back it is ignored
the white card can be selected in either orientation which leads to 2-3 BEFORE the card is selected but once the card has been chosen & revealed a white side (the conditions of the experiment present at the time the question is now being asked) the probabiliy for the white white card to have 2 states colapses.
there are now only 2 options
it is the white/white card in which case we pass
or it is the white red card & we fail 50:50
The question is not how many times will we pick the white/white card in preference to the white side of the red/white card.
it is once we have chosen a white card what colour will the opposite side be.
In the 1st q the white card can be either white A or white B giving 2 options in the second an actual question the card can only be white white.
CajunCenturion
Programmer
==> it is once we have chosen a white card what colour will the opposite side be.
Yes, and the reason that the answer is white 2/3 of the time is because you're going to asking that question twice as often when holding the white/white card than you would be when holding the red/white card. Therefore, twice as often, you're going to see a white background.
You're going to pick the white/white card the same number of times that you're going to pick the red/white card. Agreed?
Let's say you pick each card 100 times. In all 100 picks of the white/white card, you're going to then ask the question about the opposite side. But you're only going to ask that question 50 times from the 100 picks of the red/white card because half the time, it will come up red, and you're not going to ask the opposite question in those cases.
So you're going to ask the opposite side question 150 times: 100 times from picking the white/white card and 50 times from picking the white side of the red/white card. You'll see the opposite side be white 100 times and be red 50 times. That's a 2/3 of the opposite being white and and 1/3 chance of the opposite being red. And that's after having chosen a white card.
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Yes, and the reason that the answer is white 2/3 of the time is because you're going to asking that question twice as often when holding the white/white card than you would be when holding the red/white card. Therefore, twice as often, you're going to see a white background.
You're going to pick the white/white card the same number of times that you're going to pick the red/white card. Agreed?
Let's say you pick each card 100 times. In all 100 picks of the white/white card, you're going to then ask the question about the opposite side. But you're only going to ask that question 50 times from the 100 picks of the red/white card because half the time, it will come up red, and you're not going to ask the opposite question in those cases.
So you're going to ask the opposite side question 150 times: 100 times from picking the white/white card and 50 times from picking the white side of the red/white card. You'll see the opposite side be white 100 times and be red 50 times. That's a 2/3 of the opposite being white and and 1/3 chance of the opposite being red. And that's after having chosen a white card.
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Ipguru - Empirical Data trumps theory.
Try the experiment and post your results.
I'll even save you time. Only use 2 cards. One that is White-White and one that is White-Red. The Red-Red card is superfluous to the experiment as it is eliminated every time.
I won't predict the result and I trust you to be honest in your results. Try it and see, I have.
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Occam's Razor - All things being equal, the simplest solution is the right one.
Try the experiment and post your results.
I'll even save you time. Only use 2 cards. One that is White-White and one that is White-Red. The Red-Red card is superfluous to the experiment as it is eliminated every time.
I won't predict the result and I trust you to be honest in your results. Try it and see, I have.
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Occam's Razor - All things being equal, the simplest solution is the right one.
Strongm - I don't agree. This thread is virtually identical to the Monty Hall puzzle with the same contentious points. I can see how they get the 50% value and I know why it is wrong but explaining it is difficult when the math doesn't speak for itself. That being said, I will try one more "Different" explaination.
For the 50% value to be correct the cards with white faces must have an equal chance to be on the table. With only one instance in view, it is easy to lose sight of the fact that they do not have an equal chance.
An example of an equal chance puzzle would be to place both cards with white faces on the table at the same time. Now picking 1 would have a 50% chance of being white on the back.
This puzzle requires that a single card be picked and placed on the table with a white face up. Half of the times the W/R card is picked it will have a red face and would not fit the puzzle. To get back to my 50% example, this is the same as having Both cards on the table the first time but only the dual white face card the next, then both then only dual white face. Each time the dual white face is by itself it is certain to be white on the back and every alternate try it is 50%. This results in 2/3 Probability.
For the 50% adherants don't forget the question is about probability. Also Empirical Data trumps Theory. Set up an experiment as in the puzzle, it won't take long for the trend to become apparent.
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Occam's Razor - All things being equal, the simplest solution is the right one.
For the 50% value to be correct the cards with white faces must have an equal chance to be on the table. With only one instance in view, it is easy to lose sight of the fact that they do not have an equal chance.
An example of an equal chance puzzle would be to place both cards with white faces on the table at the same time. Now picking 1 would have a 50% chance of being white on the back.
This puzzle requires that a single card be picked and placed on the table with a white face up. Half of the times the W/R card is picked it will have a red face and would not fit the puzzle. To get back to my 50% example, this is the same as having Both cards on the table the first time but only the dual white face card the next, then both then only dual white face. Each time the dual white face is by itself it is certain to be white on the back and every alternate try it is 50%. This results in 2/3 Probability.
For the 50% adherants don't forget the question is about probability. Also Empirical Data trumps Theory. Set up an experiment as in the puzzle, it won't take long for the trend to become apparent.
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Occam's Razor - All things being equal, the simplest solution is the right one.
>Virtually identical to the Monty Hall Puzzle
MH - You have 3 Doors 1 of which has the car
3C - You have 3 cards 1 of which has a white face and a white back
MH - Choose a door
3C - Choose a card
MH - Eliminate 1 door (That does not contain the prize)
3C - Eliminate 1 card (Red - Red)
MH - Keep your pick or switch?
3C - Have the dual White card or not?
MH - Chances increase by double if you switch
3C - Chances are double that card on table is White White
Virtually Identical? maybe not, but pretty damn similar.
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Occam's Razor - All things being equal, the simplest solution is the right one.
MH - You have 3 Doors 1 of which has the car
3C - You have 3 cards 1 of which has a white face and a white back
MH - Choose a door
3C - Choose a card
MH - Eliminate 1 door (That does not contain the prize)
3C - Eliminate 1 card (Red - Red)
MH - Keep your pick or switch?
3C - Have the dual White card or not?
MH - Chances increase by double if you switch
3C - Chances are double that card on table is White White
Virtually Identical? maybe not, but pretty damn similar.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
- Thread starter
- #53
<The Monty Hall problem is in a whole different league ... this is an easy one.
Seems to me the difference between this problem and the Monty Hall problem is that this one is more specifically defined. Suppose the Monty Hall problem is defined as (1) 3 doors, one with a car behind it and the other two with goats behind them, (2) Contestant picks one door, (3) Monty reveals a door with a goat behind it (4) question is what are the odds of finding a car if you switch. It seems to me pretty much analogous. What's the deeper complication?
Seems to me the difference between this problem and the Monty Hall problem is that this one is more specifically defined. Suppose the Monty Hall problem is defined as (1) 3 doors, one with a car behind it and the other two with goats behind them, (2) Contestant picks one door, (3) Monty reveals a door with a goat behind it (4) question is what are the odds of finding a car if you switch. It seems to me pretty much analogous. What's the deeper complication?
- Thread starter
- #54
- Thread starter
- #55
Googling "three-card swindle" will give the answer. Even after being told, I had to chew on this for a while before I was able to understand why 2/3 is indeed the correct answer. I found it helpful personally to visualize a die. Let's say 1-3 are colored white, and 4-6 red. Ignore any rolls of 5 or 6, analogous to not drawing a red/red card. First roll is a 1, analogous to showing a white card face. The die face analogues to the remaining card faces are 2, 3, and 4, two of which are white.
The reason it's counterintuitive is because the fact that the sides are on three cards is irrelevant. Only the six sides matter when calculating the probability.
The reason it's counterintuitive is because the fact that the sides are on three cards is irrelevant. Only the six sides matter when calculating the probability.
- Thread starter
- #56
- Thread starter
- #57
<but once the card has been chosen & revealed a white side (the conditions of the experiment present at the time the question is now being asked) the probabiliy for the white white card to have 2 states colapses.
That's only true if you know which of the white sides has been revealed, isn't it?
That's only true if you know which of the white sides has been revealed, isn't it?
>2) it dosnt mater because for the sake of the comparison the card in question has only one state - white up / white down.
>There are now only 2 posible options (direction of white card does not matter as both match)
It is this leap of faith that you kep making that is wrong. It is vitally important mathematically that the two sides are in fact different sides. It matters not that a human is unable to see that difference. Please, please go back to my illustration involving putting a big A and a big B on each side of the white/white card. This changes absolutely nothing about the conditions of the problem stated
It then becomes indisputably clear that we have 3 possible starting states that meet the condition stated in the problem.
Face up Face Down
White Red
White with an A White with a B
White with a B White with an A
You cannot possibly argue that this is not the case. It is now visible, in front of your eyes.
The question asked in the problem is, given this starting point, what are the odds of getting a white face when we turn the card over?
>There are now only 2 posible options (direction of white card does not matter as both match)
It is this leap of faith that you kep making that is wrong. It is vitally important mathematically that the two sides are in fact different sides. It matters not that a human is unable to see that difference. Please, please go back to my illustration involving putting a big A and a big B on each side of the white/white card. This changes absolutely nothing about the conditions of the problem stated
It then becomes indisputably clear that we have 3 possible starting states that meet the condition stated in the problem.
Face up Face Down
White Red
White with an A White with a B
White with a B White with an A
You cannot possibly argue that this is not the case. It is now visible, in front of your eyes.
The question asked in the problem is, given this starting point, what are the odds of getting a white face when we turn the card over?
CajunCenturion
Programmer
==> we are all correct but providing answers for different parts of the question.
No, we are not all correct. This question has only one right answer, and either you have it or you don't.
==> after selection, when the card is on the table
There are now only 2 posible options (direction of white card does not matter as both match)
There have always been only two possibilities for the backside (red/white), but they don't have an equal probability of occurring because 2/3 of the white cards that make it to the table have white backsides. Only 1/3 of the white cards that make it to the table have red backsides. You cannot ignore that fact.
Ignoring that fact doesn't mean you're answer a different part of the question; it means your answer is wrong.
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No, we are not all correct. This question has only one right answer, and either you have it or you don't.
==> after selection, when the card is on the table
There are now only 2 posible options (direction of white card does not matter as both match)
There have always been only two possibilities for the backside (red/white), but they don't have an equal probability of occurring because 2/3 of the white cards that make it to the table have white backsides. Only 1/3 of the white cards that make it to the table have red backsides. You cannot ignore that fact.
Ignoring that fact doesn't mean you're answer a different part of the question; it means your answer is wrong.
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