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
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Sorry, the above should have read:[hide]2 in 3[/hide]
Oh - you want the justification?[hide]There are 3 ways you could draw a white side: the white side of the Red/White card, and either of the White/White sides. Out of those 3, two can be flipped to show a white on the other side (the two white/white options). Thus 2 in 3 chances.[/hide]
Oh - you want the justification?[hide]There are 3 ways you could draw a white side: the white side of the Red/White card, and either of the White/White sides. Out of those 3, two can be flipped to show a white on the other side (the two white/white options). Thus 2 in 3 chances.[/hide]
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Hi
I am wondering how could this be solved programmatically ?
Feherke.
The card in the hand can not be the red/red. So we can take in consideration only two cards.
To answer the question "is the opposite face white", we have to know which face is the one we currently see. And we see one of three possible faces :
[ul]
[li]white/white's first face[/li]
[li]white/white's second face[/li]
[li]white/red's white face[/li]
[/ul]
From the above three possibilities the first two would mean the opposite face is also white.
So the chance is 2 to 3.
To answer the question "is the opposite face white", we have to know which face is the one we currently see. And we see one of three possible faces :
[ul]
[li]white/white's first face[/li]
[li]white/white's second face[/li]
[li]white/red's white face[/li]
[/ul]
From the above three possibilities the first two would mean the opposite face is also white.
So the chance is 2 to 3.
Feherke.
As I say, you've completely misread my answer, and tried to take one statement in isolation.
When I said "the two white/white options" I am talking about the (one) card with two white faces (as is made clear in the first part of my justification), either face of which we might see. Statistically, they are different faces: we can be seeing face A or face B
When I said "the two white/white options" I am talking about the (one) card with two white faces (as is made clear in the first part of my justification), either face of which we might see. Statistically, they are different faces: we can be seeing face A or face B
Hi
strongm, I understand your explanation only now, after I reread mine and I realized the similarities. Earlier I understood your explanation the same way IPGuru did. Maybe you indeed spared a few words there...
Feherke.
strongm, I understand your explanation only now, after I reread mine and I realized the similarities. Earlier I understood your explanation the same way IPGuru did. Maybe you indeed spared a few words there...
Feherke.
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CajunCenturion
Programmer
[hide]I agree with strongm - the odds are 2 in 3 that the other side is white. When the white card is shown, there are three possibilities.
1: you're looking at side 1 of the white/white card
2: you're looking at side 2 of the white/white card
3: you're looking at the white side of the white/red card
In two of those three cases, the other side of the card is white.[/hide]
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1: you're looking at side 1 of the white/white card
2: you're looking at side 2 of the white/white card
3: you're looking at the white side of the white/red card
In two of those three cases, the other side of the card is white.[/hide]
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the explanation is flawed
although the white/white card can be in one of 2 states they are implicitly linked
The card can be a/b or b/a but it is still the same card.
whichever side is uper most it is now a known state.
Therefore we only have 2 posibilities for which card is on the table
white/white
or white red
(red red has been iliminated by the setup conditions)
The only card that can be white on the reverse is the white white card
The answer is 1 in 2
the OP either has the answer wrong or has miss phrased it.
although the white/white card can be in one of 2 states they are implicitly linked
The card can be a/b or b/a but it is still the same card.
whichever side is uper most it is now a known state.
Therefore we only have 2 posibilities for which card is on the table
white/white
or white red
(red red has been iliminated by the setup conditions)
The only card that can be white on the reverse is the white white card
The answer is 1 in 2
the OP either has the answer wrong or has miss phrased it.
CajunCenturion
Programmer
[hide]Look at it this way.
There are six faces. Each card has a A-side and a B-side, and we can label them as follows:
Card 1: Wa and Wb (white on a side, white on b side)
Card 2: Wa and Rb (white on a side, red on b side)
Card 3: Ra and Rb (red on a side, red on b side)
What are the odds of pulling a card and showing a white face?
50/50 right?
There are six sides - three are white and three are red; therefore, the odds of showing a white side is 50/50.
Now, given that you're looking at one of those three white faces, how many are white on the other side? Two. Hence the odds of the other side being white is 2/3.[/hide]
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FAQ181-2886
As a circle of light increases so does the circumference of darkness around it. - Albert Einstein
There are six faces. Each card has a A-side and a B-side, and we can label them as follows:
Card 1: Wa and Wb (white on a side, white on b side)
Card 2: Wa and Rb (white on a side, red on b side)
Card 3: Ra and Rb (red on a side, red on b side)
What are the odds of pulling a card and showing a white face?
50/50 right?
There are six sides - three are white and three are red; therefore, the odds of showing a white side is 50/50.
Now, given that you're looking at one of those three white faces, how many are white on the other side? Two. Hence the odds of the other side being white is 2/3.[/hide]
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Strongm
your statistics are flawed
the initial status of the 1st card has already been defined
if it is white white then the upsied has already been identified.
alternativly if you wish to think of the whit/white card as 2 cards a/b & b/a then the red /white card also needs to be though of as 2 cards because ists white side can match both a & b
it is efectivly a/red & b/red
this then gives 4 cards 2 of which can have a white back
2 in 4 = 1 in 2
your statistics are flawed
the initial status of the 1st card has already been defined
if it is white white then the upsied has already been identified.
alternativly if you wish to think of the whit/white card as 2 cards a/b & b/a then the red /white card also needs to be though of as 2 cards because ists white side can match both a & b
it is efectivly a/red & b/red
this then gives 4 cards 2 of which can have a white back
2 in 4 = 1 in 2
Ok try an experiment to prove them
get 2 sheets of paper
on sheet 1 one side a & 1 side b
on sheet 2 mark one side red
these are the 2 cards
place sheet 1 on the table showing face a
now place next to it a sheet showing face b
it cannot be done because a & b are the same item
in the ecample it does not matter if it is face a or face be that is showing because a=b
or to use boolean loging the white card can show either a or not a
a or not a = 1 therefore it does not matter what face is showing.
or would you prefer it as a truth table
card 1 = white white (labled a & b) card 2 = red white
card 1 card 2 side 1 and side 2 = white
a white yes
b white yes
a red no
b red no
There are clearly 4 possible states only 2 give white/white
there are lies, damned lise & statistics
get 2 sheets of paper
on sheet 1 one side a & 1 side b
on sheet 2 mark one side red
these are the 2 cards
place sheet 1 on the table showing face a
now place next to it a sheet showing face b
it cannot be done because a & b are the same item
in the ecample it does not matter if it is face a or face be that is showing because a=b
or to use boolean loging the white card can show either a or not a
a or not a = 1 therefore it does not matter what face is showing.
or would you prefer it as a truth table
card 1 = white white (labled a & b) card 2 = red white
card 1 card 2 side 1 and side 2 = white
a white yes
b white yes
a red no
b red no
There are clearly 4 possible states only 2 give white/white
there are lies, damned lise & statistics
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Because you have all ready seen the white side, the red/red card is out of the picture.
This leaves 2 cards to work with: white/white and white/red.
You see the white side of 1 of these 2 cards.
The other side, obviously is either red (white/red card) or white (white/white card).
There is no 3rd possiblity.
The answer is 1 in 2.
This leaves 2 cards to work with: white/white and white/red.
You see the white side of 1 of these 2 cards.
The other side, obviously is either red (white/red card) or white (white/white card).
There is no 3rd possiblity.
The answer is 1 in 2.
Randy
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