[Guest]

I have two cards, both with two sides. One card has a black side and a white side, the other one has two white sides. I put them into a bag and shuffle and flip them randomly until they are solidly mixed up. You reach into the bag, pull out one card, and look at one side. It is white. What is the probability that the other side is white too?

[brainz]

I have two cards, both with two sides. One card has a black side and a white side, the other one has two white sides. I put them into a bag and shuffle and flip them randomly until they are solidly mixed up. You reach into the bag, pull out one card, and look at one side. It is white. What is the probability that the other side is white too?

wouldnt it be 50%?

[akaslickster]

75% chance

Edited August 11, 2008 by akaslickster

[Guest]

I have two cards, both with two sides. One card has a black side and a white side, the other one has two white sides. I put them into a bag and shuffle and flip them randomly until they are solidly mixed up. You reach into the bag, pull out one card, and look at one side. It is white. What is the probability that the other side is white too?

This is one of those given questions

This is one of those given questions

It's late, so I won't explain how, probably someone else will, but its 2/3

Edited August 11, 2008 by Chaplam13

[Gambit]

2/3

[Guest]

3 white sides, so all possible events 3. 1 card has 2 white sides, so number of events favourable to the outcome 2. Probability 2/3.

[Guest]

I have two cards, both with two sides. One card has a black side and a white side, the other one has two white sides. I put them into a bag and shuffle and flip them randomly until they are solidly mixed up. You reach into the bag, pull out one card, and look at one side. It is white. What is the probability that the other side is white too?

Hi, I'm new to this (just registered) but I think it is 50%

[Gambit]

I have two cards, both with two sides. One card has a black side and a white side, the other one has two white sides. I put them into a bag and shuffle and flip them randomly until they are solidly mixed up. You reach into the bag, pull out one card, and look at one side. It is white. What is the probability that the other side is white too?

2/3

the probabilty should be 1/2.

there's a 50/50 chance of the back of that card been white...

[Guest]

50%

It's 50%. The trick is that one card has already been selected, so, either the back of the card is white, which means that the card picked was white on both sides, or the back of the card is black, which means the other. It would be 2/3 only if you were to go back in the bag again, and pick out a card, and measure whether the first side you looked at was white.

[Guest]

This is one of those given questions

This is one of those given questions

It's late, so I won't explain how, probably someone else will, but its 2/3

We know we have picked a card with one white side. The total number of white sides in the bag is three, so the chances of that card being WW is 2/3, and WB is 1/3

Thus the probability of the other side being white, is the same as the probability that the card is WW which is 2/3.

There's proper maths available, but I think that sums things up.

[Guest]

The chance is 50%.

You have a card and see white. There are only two chances now, either you have a double white, or a white-black. Seeing as you can see white, the other side has a 50% chance of being black, a 50% chance of being white,

[Prof. Templeton]

The probability of seeing a White face is 3/4, and the probability of drawing the White card is 1/2. Therfore

1/2

---- = 2/3

3/4

[Guest]

With one white exposed ....

the probability for white is 2/3 since the remainder is 1 black & 2 whites

Edited August 11, 2008 by Ben Law

[Guest]

This makes no sense..

People are talking about the sides as if they are independent of the cards. I think it would be 50% because its irrelevant of the amount of white sides, but the amount of white/white cards, which is one, out of a total of 2 cards. So the probablity of picking the white/white card is 1/2, and therefore the probability of both sides on a single card being white is 50%.

[Prof. Templeton]

This makes no sense..

People are talking about the sides as if they are independent of the cards. I think it would be 50% because its irrelevant of the amount of white sides, but the amount of white/white cards, which is one, out of a total of 2 cards. So the probablity of picking the white/white card is 1/2, and therefore the probability of both sides on a single card being white is 50%.

This is a good example of "Bertrand's Box Paradox", where a person's intuition may tell them that they are picking cards so the probablity is 50/50, but you are picking card faces. There are 3 white faces. Two of the white faces are on 1 card. The chance of getting 1 of those 2 is 2/3, therefore the chance of flipping the card over to find another white face is 2/3. If you have drawn a white faced card, it is twice as likely that the face belongs to the all white card as opposed to the mixed card.

[Guest]

This is a good example of "Bertrand's Box Paradox", where a person's intuition may tell them that they are picking cards so the probablity is 50/50, but you are picking card faces. There are 3 white faces. Two of the white faces are on 1 card. The chance of getting 1 of those 2 is 2/3, therefore the chance of flipping the card over to find another white face is 2/3. If you have drawn a white faced card, it is twice as likely that the face belongs to the all white card as opposed to the mixed card.

I can see what you're saying. But I won't look at it like that because while truly I have a 75% chance of picking a white face at first, it is no more likely that the other side will be black than white. I disagree with what you are saying because I see the cards and not the faces as the basis upon which the probability is decided. Evidently if the puzzle were to read "I have 4 plastic coins, 1 black and 3 white. I take out a white coin, what is the probability I will pick another white?" the answer would be 2/3. However because the choice is limited to 2 options with each having a 50% chance of being picked, the colour of the reverse of a card is related to the probability of each card being picked, not the number of each colour faces.

[Guest]

33.3%

[Guest]

I can see what you're saying. But I won't look at it like that because while truly I have a 75% chance of picking a white face at first, it is no more likely that the other side will be black than white. I disagree with what you are saying because I see the cards and not the faces as the basis upon which the probability is decided. Evidently if the puzzle were to read "I have 4 plastic coins, 1 black and 3 white. I take out a white coin, what is the probability I will pick another white?" the answer would be 2/3. However because the choice is limited to 2 options with each having a 50% chance of being picked, the colour of the reverse of a card is related to the probability of each card being picked, not the number of each colour faces.

Think about it this way - at the beginning, before you pick any card, there are four possible outcomes:

1. Choose white face of mixed card

2. Choose black face of mixed card

3. Choose one white face of white card

4. Choose other white face of white card

These four possibilities are all equally likely. Once you choose a card and see a white face, you have eliminated option 2, leaving 1, 3 and 4 as possibilities. These are all equally likely, so the probability of you holding the all-white card is 2/3.

What's difficult for people to grasp here is that the face you see gives you information about what that card is likely to be. The other difficulty is understanding that options 3 and 4 are two distinct possibilities which each have a probability of 1/4.

[Guest]

Yes, but in your example you only have 2 out of the three left to draw because you are not drawing a card from the bag...you are flipping the one you have and you cannot flip the card and find the white side of the black and white card. Otherwise you would see black. Therefore you only have two options ouot of 4 left, 50%.

That isnt clear is it?

In the beginnning there are 4 choices. After you draw there are only 2 choices left. Because you are flipping the card in your hand and not drawing a new one your choices are...

1) The black side of the mixed card 1/2

2) The other side of the all white card 1/2

I think that I am right having read all the arguments and being of sound mind.It would be different if you were drawing the last card but that is not the question.

Ray

Edited August 11, 2008 by gatorray

[Guest]

Yes, but in your example you only have 2 out of the three left to draw because you are not drawing a card from the bag...you are flipping the one you have and you cannot flip the card and find the white side of the black and white card. Otherwise you would see black. Therefore you only have two options ouot of 4 left, 50%.

That isnt clear is it?

In the beginnning there are 4 choices. After you draw there are only 2 choices left. Because you are flipping the card in your hand and not drawing a new one your choices are...

1) The black side of the mixed card 1/2

2) The other side of the all white card 1/2

I think that I am right having read all the arguments and being of sound mind.It would be different if you were drawing the last card but that is not the question.

Ray

I don't understand what you mean by "drawing the last card" or how that might make 2/3 the right answer.

Once you have drawn the card, that affects the possible scenarios available to you regarding the contents of the card. You seem to have ignored Chuck Rampart's four starting conditions.

1. Choose white face of mixed card

2. Choose black face of mixed card

3. Choose one white face of white card

4. Choose other white face of white card

When you see a white face, that only eliminates 2 from the possible scenarios. The fact that 3 and 4 are different conditions is what the answer hinges on. If you can understand that, then the rest is easy. If you accept these four possible starting conditions as equally likely, then you will see that your #2 consists of two possible conditions. Double-white side A is different from double-white side B. You could do an experiment to test this fairly easily.

Take two notecards (or equivalent,) mark one side of one card with an X. Then put both cards in a hat or other receptacle and draw one out at random and look at only one side. If you see an X, ignore it and put the card back in and try again. If the face is blank, mark whether it was the X card or not and keep track. You shouldn't have to do very many tests to begin to see the pattern. If you keep track of all the draws, X and non-X, you should see a blank drawn three times more than an X, and of the blanks, they are twice as likely to be drawn from the non-X than the X card (which oddly enough should be about 2/3 of the time. )

[Guest]

ok, once you choose a card and see a white side there are 2 options

1) you are holding the mixed card

2) you are holding the all white card

if 1 is true, the chance of the other side being white = 0%

if 2 is true, the chance of the other side being white = 100%

There are only 2 options either you have drawn the mixed card or you havent.

IN your example conditions 3 and 4 are the same, you are counting the same card twice when you have already seen one side.

[bonanova]

This puzzle has been posted.

Please read this spoiler.

Since this thread has achieved a reasonable life, I'll keep it open.

[Guest]

This puzzle has been posted.

Please read this spoiler.

Since this thread has achieved a reasonable life, I'll keep it open.

That thread, and the interesting differences between that problem and this problem was the reason I posted this one in the first place. Yes, if you look at that problem as you do in that post, the math is the same, but the logical quandary is different in each.

In that post the differences arise from the way you pick children, in this one the logical misstep usually arises from treating the white/white card as an independent single event instead of two possible draws.

[Guest]

I like this coz it is pure compared to the effects of random effects on birth - takes out fuzzy obstacles such as chemistry - but that's just me not being able to let go in the g/b pzzl

nothing to add to the general consensus/answers