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 derfloh 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
>What's the deeper complication?
The difference is that in the MH problem each event (choice of original door and Monty's revelation) is important to the final result. More importantly is the subtle point that Monty always reveals a goat (i.e. he knows where the prize is), and it is this point that most people don't realise switches the answer between it not making any difference whether you change your mind, and that changing your mind doubles your chances of winning.
In 3C problem, it is completely irrelevant on how we get to the starting point of a face-up white card. We then have a very simple state machine, that my big A and big B example help visualise. No real need to do any maths at all.
The difference is that in the MH problem each event (choice of original door and Monty's revelation) is important to the final result. More importantly is the subtle point that Monty always reveals a goat (i.e. he knows where the prize is), and it is this point that most people don't realise switches the answer between it not making any difference whether you change your mind, and that changing your mind doubles your chances of winning.
In 3C problem, it is completely irrelevant on how we get to the starting point of a face-up white card. We then have a very simple state machine, that my big A and big B example help visualise. No real need to do any maths at all.
Read my above post regarding why we had the disagreement
Chanses of selecting white white as aposed to white/red(white top) is indeed 2 to 1 as you state
but once laid on the table the up side has been defined & if fixed. the white side is up ( in your experiment it now says A or B but cannot be indeterminate)
the queston was not worded as what were the odds of selecting the white white card but what are the odds of the card beeing white white after the selection paramaterd have been defined.
The question was ambigous resulting in 2 possible answers
As this is at least partila agrrement (based on the intended interpretation of the question) I think we should call it a day
Chanses of selecting white white as aposed to white/red(white top) is indeed 2 to 1 as you state
but once laid on the table the up side has been defined & if fixed. the white side is up ( in your experiment it now says A or B but cannot be indeterminate)
the queston was not worded as what were the odds of selecting the white white card but what are the odds of the card beeing white white after the selection paramaterd have been defined.
The question was ambigous resulting in 2 possible answers
As this is at least partila agrrement (based on the intended interpretation of the question) I think we should call it a day
CajunCenturion
Programmer
The question reads:
--------------
Good Luck
To get the most from your Tek-Tips experience, please read
FAQ181-2886
As a circle of light increases so does the circumference of darkness around it. - Albert Einstein
The question asks for the backside of the card that you just pulled out. You've already acknowledged that the card you just pulled out has a 2:1 chance of being white/white vs white/red. It's still the same card. It still has a 2:1 chance of a white backside.
--------------
Good Luck
To get the most from your Tek-Tips experience, please read
FAQ181-2886
As a circle of light increases so does the circumference of darkness around it. - Albert Einstein
>The question was ambigous resulting in 2 possible answers
No, it isn't. Doesn't matter how many times you want to argue this. The question is very specific and has one answer only.
And that question is not:
>what are the odds of the card beeing white white
It very carefully asks a subtly different question.
No, it isn't. Doesn't matter how many times you want to argue this. The question is very specific and has one answer only.
And that question is not:
>what are the odds of the card beeing white white
It very carefully asks a subtly different question.
IPGuru - I've asked twice for you to try the experiment, now a third time. All the theorizing in the world is meaningless against hard data. Please do the following, I promise it won't take you more than 10 minutes. In fact it took me longer to write this post so please, pretty please.
To include your reasoning as part of the experiment. Once you see a white side on the table the result is "Defined and Fixed". So, each time you achieve a valid result (white up) start the experiment as defined and fixed. Out of 30 tries, how many times will you be right? 1/2 or 2/3?
Find 3 coins of same denomination with different dates. I used quarters.
[tt]
Date 1 Either = W/W = True
Date 2 Heads = W/R = False
Date 2 Tails = R/W = ignore
Date 3 Either = R/R = ignore
[/tt]
Shake coins in your cupped hands, choose 1 without looking at them and observe. Record results.
My results[tt]
Quarter 2004 = 11 times
Quarter 2007 Heads = 5 times
Quarter 2007 Tails = 7 times
Quarter 2009 = 7 times[/tt]
16 times out of 30 a white face occured
14 trials out of 30 ignored due to red face
11 times of 16 the reverse side was white
5 times of 16 the reverse side was red
10/15 = Expected result = .6666666
11/16 = .6825
5/16 = .3175
Try it yourself, Pretty Please.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
To include your reasoning as part of the experiment. Once you see a white side on the table the result is "Defined and Fixed". So, each time you achieve a valid result (white up) start the experiment as defined and fixed. Out of 30 tries, how many times will you be right? 1/2 or 2/3?
Find 3 coins of same denomination with different dates. I used quarters.
[tt]
Date 1 Either = W/W = True
Date 2 Heads = W/R = False
Date 2 Tails = R/W = ignore
Date 3 Either = R/R = ignore
[/tt]
Shake coins in your cupped hands, choose 1 without looking at them and observe. Record results.
My results[tt]
Quarter 2004 = 11 times
Quarter 2007 Heads = 5 times
Quarter 2007 Tails = 7 times
Quarter 2009 = 7 times[/tt]
16 times out of 30 a white face occured
14 trials out of 30 ignored due to red face
11 times of 16 the reverse side was white
5 times of 16 the reverse side was red
10/15 = Expected result = .6666666
11/16 = .6825
5/16 = .3175
Try it yourself, Pretty Please.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
lionelhill
Technical User
I'm giving Cajun a star for patience and because he seems to be about the only person who is still starless.
I'm really enjoying a sub-plot on methods of mathematical proof here. Proof by Google was a new one on me.
I'm really enjoying a sub-plot on methods of mathematical proof here. Proof by Google was a new one on me.
When you pull the card out, you have eliminated the red/red card from the equation.
Therefore you have only two possibilities, white/white or white/red.
The question says, "What are the chances that the other side of the card is ALSO white?" That implies that there are only TWO possibilities in the universe of possibilities as the red/red card does NOT have a white face.
So, stars to those who correctly said 50%.
mmerlinn
"We've found by experience that people who are careless and sloppy writers are usually also careless and sloppy at thinking and coding. Answering questions for careless and sloppy thinkers is not rewarding." - Eric Raymond
Poor people do not hire employees. If you soak the rich, who are you going to work for?
Therefore you have only two possibilities, white/white or white/red.
The question says, "What are the chances that the other side of the card is ALSO white?" That implies that there are only TWO possibilities in the universe of possibilities as the red/red card does NOT have a white face.
So, stars to those who correctly said 50%.
mmerlinn
"We've found by experience that people who are careless and sloppy writers are usually also careless and sloppy at thinking and coding. Answering questions for careless and sloppy thinkers is not rewarding." - Eric Raymond
Poor people do not hire employees. If you soak the rich, who are you going to work for?
Also, it does not matter that there are 3 white faces out of the original 6 faces. Once you have chosen the card, there are only 2 possibilities left, white/white or white/red.
mmerlinn
"We've found by experience that people who are careless and sloppy writers are usually also careless and sloppy at thinking and coding. Answering questions for careless and sloppy thinkers is not rewarding." - Eric Raymond
Poor people do not hire employees. If you soak the rich, who are you going to work for?
mmerlinn
"We've found by experience that people who are careless and sloppy writers are usually also careless and sloppy at thinking and coding. Answering questions for careless and sloppy thinkers is not rewarding." - Eric Raymond
Poor people do not hire employees. If you soak the rich, who are you going to work for?
mmerlinn, you are ignoring the likehood of which card is present with a white face up. The mathmatical arguments are not working for those that are inclined to believe the 50% model. I have provided a simple and effective means to run the scenario without bias. Get your hands dirty and take 10 minutes to run thru as many as you can. Let us know the results.
To prove my willingness to not only ask but to do it as well I am starting another scenario. I will post results in about 10 minutes
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
To prove my willingness to not only ask but to do it as well I am starting another scenario. I will post results in about 10 minutes
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
-
1
- #71
>So, stars to those who correctly said 50%.
Sadly, incorrecty awarded stars. But there we go. You are not the first person in this thread to award a star to the incorrect answer.
>When you pull the card out, you have eliminated the red/red card from the equation
The red/red card is a red herring. It has no bearing whatsoever on the problem. Frankly it is there specifically to help fool people.
>Also, it does not matter that there are 3 white faces out of the original 6 faces
This is true, since the original 6 faces are completely irrelevant. The ony important thing is the starting condition
Let's state the problem as:
You have two cards in a hat. One is white on both sides, second 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?
You make the mathematically incorrect statement that: "therefore you have only two possibilities, white/white or white/red"
As I have repeatedly shown with no maths whatsoever (and has been repeatedly ignored by the 50% fans), there are 3 starting possibilities, two of which result in a (different) white face when flipped.
Sadly, incorrecty awarded stars. But there we go. You are not the first person in this thread to award a star to the incorrect answer.
>When you pull the card out, you have eliminated the red/red card from the equation
The red/red card is a red herring. It has no bearing whatsoever on the problem. Frankly it is there specifically to help fool people.
>Also, it does not matter that there are 3 white faces out of the original 6 faces
This is true, since the original 6 faces are completely irrelevant. The ony important thing is the starting condition
Let's state the problem as:
You have two cards in a hat. One is white on both sides, second 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?
You make the mathematically incorrect statement that: "therefore you have only two possibilities, white/white or white/red"
As I have repeatedly shown with no maths whatsoever (and has been repeatedly ignored by the 50% fans), there are 3 starting possibilities, two of which result in a (different) white face when flipped.
Results:
w/w 9 time True
w/r 6 times False
r/w 5 times ignored
r/r 10 times ignored
15 valid test of 30
9 of 15 true
6 of 15 false
Result = 60%
Combined totals with previous test:
20 of 31 = true
11 of 31 = false
Result = 64.5%
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
w/w 9 time True
w/r 6 times False
r/w 5 times ignored
r/r 10 times ignored
15 valid test of 30
9 of 15 true
6 of 15 false
Result = 60%
Combined totals with previous test:
20 of 31 = true
11 of 31 = false
Result = 64.5%
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
The red/red card is a red herring. It has no bearing whatsoever on the problem. Frankly it is there specifically to help fool people.
EXACTLY...
When the test is performed, there is no longer a red/red card. You only have TWO cards with which to work - white/white and white/red.
Therefore...
1. If you are looking at the white side of the white/red card, the opposite side is red.
2. If you are looking at either white side of the white/white card, the opposite side is also white.
There are no other possibilites....... 50%
Randy
EXACTLY...
When the test is performed, there is no longer a red/red card. You only have TWO cards with which to work - white/white and white/red.
Therefore...
1. If you are looking at the white side of the white/red card, the opposite side is red.
2. If you are looking at either white side of the white/white card, the opposite side is also white.
There are no other possibilites....... 50%
Randy
CajunCenturion
Programmer
You continue to disregard the fact that you're going to be looking at the white/white card twice as often as you'll be looking at the white/red card.
--------------
Good Luck
To get the most from your Tek-Tips experience, please read
FAQ181-2886
As a circle of light increases so does the circumference of darkness around it. - Albert Einstein
mmerlinn:
When you pull the card out, you have eliminated the red/red card from the equation.
Yes.
Therefore you have only two possibilities, white/white or white/red
NO!!! You have FOUR possibilities at this point; white/white with side A facing up, white/white with side B facing up, white/red with white side facing up, and white/red with red side facing up.
The next given is that the card is placed down, and white is facing up. So, that eliminates "white/red with red side facing up". Leaving you with only THREE possibilities. And two of those are favorable (white/white with side A facing up, and white/white with side B facing up).
randy700
>1. If you are looking at the white side of the white/red card, the opposite side is red.
YES!
2. If you are looking at either white side of the white/white card, the opposite side is also white.
The operating word here being "EITHER"!!!!! YES, the white/white card has TWO sides... both white! They both count!
When you pull the card out, you have eliminated the red/red card from the equation.
Yes.
Therefore you have only two possibilities, white/white or white/red
NO!!! You have FOUR possibilities at this point; white/white with side A facing up, white/white with side B facing up, white/red with white side facing up, and white/red with red side facing up.
The next given is that the card is placed down, and white is facing up. So, that eliminates "white/red with red side facing up". Leaving you with only THREE possibilities. And two of those are favorable (white/white with side A facing up, and white/white with side B facing up).
randy700
>1. If you are looking at the white side of the white/red card, the opposite side is red.
YES!
2. If you are looking at either white side of the white/white card, the opposite side is also white.
The operating word here being "EITHER"!!!!! YES, the white/white card has TWO sides... both white! They both count!
>You only have TWO cards with which to work
But four sides.
Nope
Therefore...
1. If you are looking at the white side of the white/red card, the opposite side is red.
2. If you are looking at white side A of the white/white card, the opposite side is white side B.
3. If you are looking at white side B of the white/white card, the opposite side is white side A.
Hence 2/3
Go back and look carefully at my example that prints a big A and a big B on the cards to help visualise this.
But four sides.
Nope
Therefore...
1. If you are looking at the white side of the white/red card, the opposite side is red.
2. If you are looking at white side A of the white/white card, the opposite side is white side B.
3. If you are looking at white side B of the white/white card, the opposite side is white side A.
Hence 2/3
Go back and look carefully at my example that prints a big A and a big B on the cards to help visualise this.
So, 1 hour of rolling dice later (300 times I might add)
[tt]Rolled Times
1 52
2 39
3 51
4 47
5 56
6 55[/tt]
Take your pick which 2 faces = opposite white faces and then pick which of the remaining sides = white
Hint: there are 12 combinations
Now calculate how many times the opposite side was white.
Maximum value = 73.29% using combination 1&6 = WW and 2 is white
Minimum value = 63.33% using combination 2&5 = WW and 6 is White
Average of all 12 combinations = 66.72%
Can't argue with hard data folks. I did this the old fashioned way. Literally rolled a die 300 times.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
[tt]Rolled Times
1 52
2 39
3 51
4 47
5 56
6 55[/tt]
Take your pick which 2 faces = opposite white faces and then pick which of the remaining sides = white
Hint: there are 12 combinations
Now calculate how many times the opposite side was white.
Maximum value = 73.29% using combination 1&6 = WW and 2 is white
Minimum value = 63.33% using combination 2&5 = WW and 6 is White
Average of all 12 combinations = 66.72%
Can't argue with hard data folks. I did this the old fashioned way. Literally rolled a die 300 times.
*******************************************************
Occam's Razor - All things being equal, the simplest solution is the right one.
> What are the chances that the other side of the card is also white?
It either is white or it isn't. So 50/50 chance.
The chance of any one event happening is always 50/50. Either it happens or it doesn't. OP never said to do that pick 300 times ;-)
HTH,
p5wizard
It either is white or it isn't. So 50/50 chance.
The chance of any one event happening is always 50/50. Either it happens or it doesn't. OP never said to do that pick 300 times ;-)
HTH,
p5wizard
>It either is white or it isn't. So 50/50 chance
No, it is either White A, White B, or Red. So 2/3 chance of being white.
Let's take a slightly different approach with my big printed A and big printed B.
Conditions of the original problem remain, but the question becomes "What is the probability of turning the card over and finding White B?
No, it is either White A, White B, or Red. So 2/3 chance of being white.
Let's take a slightly different approach with my big printed A and big printed B.
Conditions of the original problem remain, but the question becomes "What is the probability of turning the card over and finding White B?
- Status
Similar threads
- Locked
- Question