More improbability

[Stevil]

The question for each variation is "does this scenario show a card with a red face next to another card with a red face?"

But, that wasn't the question

[Ali]

The question for each variation is "does this scenario show a card with a red face next to another card with a red face?"

But, that wasn't the question

Really?  Hahahahahaha  I thought that's what En_Route was asking.  Having re-read the question, I have no idea what he's even asking. 

There  are inded three cards; one has two white faces, one has two red faces, one has a red face and a white face. What are the chances of a card with a red face showing also having a red face on its other side?

Drats.  I got all excited about knowing the answer to the wrong question.

[Stevil]

The permutations are as follows
Perm1 - Card1(R1 showing, R2 hidden), Card2(W1 showing, W2 hidden), Card3(R1 showing, W2 hidden) = 2
Perm2 - Card1(R2 showing, R1 hidden), Card2(W1 showing, W2 hidden), Card3(R1 showing, W2 hidden) = 2
Perm3 - Card1(R2 showing, R1 hidden), Card2(W2 showing, W1 hidden), Card3(R1 showing, W2 hidden) = 2
Perm4 - Card1(R2 showing, R1 hidden), Card2(W2 showing, W1 hidden), Card3(W2 showing, R1 hidden) = 1
Perm5 - Card1(R2 showing, R1 hidden), Card2(W1 showing, W2 hidden), Card3(W2 showing, R1 hidden) = 1
Perm6 - Card1(R1 showing, R2 hidden), Card2(W1 showing, W2 hidden), Card3(W2 showing, R1 hidden)  = 1
Perm7 - Card1(R1 showing, R2 hidden), Card2(W2 showing, W1 hidden), Card3(W2 showing, R1 hidden) = 1
Perm8 - Card1(R1 showing, R2 hidden), Card2(W2 showing, W1 hidden), Card3(R1 showing, W2 hidden) = 2

You ought to have 8 perms because each card has two faces so we have 2x2x2 = 8

We have 4 instances where two reds are showing and 4 instances where only 1 red is showing, so that is 12 red cards showing.
Of those 12 red cards, 8 of them have a red back and 4 of them have a white back
So if you have a red card then it is 8/12*100 = 66% likely to have a red back.

[Ali]

Nope.  Don't get it.  *Sigh*

[The Black Jester]

The permutations are as follows
Perm1 - Card1(R1 showing, R2 hidden), Card2(W1 showing, W2 hidden), Card3(R1 showing, W2 hidden) = 2
Perm2 - Card1(R2 showing, R1 hidden), Card2(W1 showing, W2 hidden), Card3(R1 showing, W2 hidden) = 2
Perm3 - Card1(R2 showing, R1 hidden), Card2(W2 showing, W1 hidden), Card3(R1 showing, W2 hidden) = 2
Perm4 - Card1(R2 showing, R1 hidden), Card2(W2 showing, W1 hidden), Card3(W2 showing, R1 hidden) = 1
Perm5 - Card1(R2 showing, R1 hidden), Card2(W1 showing, W2 hidden), Card3(W2 showing, R1 hidden) = 1
Perm6 - Card1(R1 showing, R2 hidden), Card2(W1 showing, W2 hidden), Card3(W2 showing, R1 hidden)  = 1
Perm7 - Card1(R1 showing, R2 hidden), Card2(W2 showing, W1 hidden), Card3(W2 showing, R1 hidden) = 1
Perm8 - Card1(R1 showing, R2 hidden), Card2(W2 showing, W1 hidden), Card3(R1 showing, W2 hidden) = 2

You ought to have 8 perms because each card has two faces so we have 2x2x2 = 8

We have 4 instances where two reds are showing and 4 instances where only 1 red is showing, so that is 12 red cards showing.
Of those 12 red cards, 8 of them have a red back and 4 of them have a white back
So if you have a red card then it is 8/12*100 = 66% likely to have a red back.

Yep, this makes sense to me now...thanks Stevil.  I had caught on to the fact that Ali's permutations were calculating whether or not two cards with red faces were standing next to one another, but for the life of me couldn't figure out how that developed into a successful solution to the problem.  Which is why I asked my question.  Your explanation makes perfect sense.  Thank you.

[En_Route]

The permutations are as follows
Perm1 - Card1(R1 showing, R2 hidden), Card2(W1 showing, W2 hidden), Card3(R1 showing, W2 hidden) = 2
Perm2 - Card1(R2 showing, R1 hidden), Card2(W1 showing, W2 hidden), Card3(R1 showing, W2 hidden) = 2
Perm3 - Card1(R2 showing, R1 hidden), Card2(W2 showing, W1 hidden), Card3(R1 showing, W2 hidden) = 2
Perm4 - Card1(R2 showing, R1 hidden), Card2(W2 showing, W1 hidden), Card3(W2 showing, R1 hidden) = 1
Perm5 - Card1(R2 showing, R1 hidden), Card2(W1 showing, W2 hidden), Card3(W2 showing, R1 hidden) = 1
Perm6 - Card1(R1 showing, R2 hidden), Card2(W1 showing, W2 hidden), Card3(W2 showing, R1 hidden)  = 1
Perm7 - Card1(R1 showing, R2 hidden), Card2(W2 showing, W1 hidden), Card3(W2 showing, R1 hidden) = 1
Perm8 - Card1(R1 showing, R2 hidden), Card2(W2 showing, W1 hidden), Card3(R1 showing, W2 hidden) = 2

You ought to have 8 perms because each card has two faces so we have 2x2x2 = 8

We have 4 instances where two reds are showing and 4 instances where only 1 red is showing, so that is 12 red cards showing.
Of those 12 red cards, 8 of them have a red back and 4 of them have a white back
So if you have a red card then it is 8/12*100 = 66% likely to have a red back.

You can simplify this. A card showing a red face can only be either the card with two red faces or the card with one red face and one white face. There are three equally possible scenarios when a red face will show upwards.The card with two red faces-  1.red face one  2.red face two; the card with one red face and one white face-  3. red face.  2/3 = 66%.

[En_Route]

The question for each variation is "does this scenario show a card with a red face next to another card with a red face?"

But, that wasn't the question

Really?  Hahahahahaha  I thought that's what En_Route was asking.  Having re-read the question, I have no idea what he's even asking. 

There  are inded three cards; one has two white faces, one has two red faces, one has a red face and a white face. What are the chances of a card with a red face showing also having a red face on its other side?

Drats.  I got all excited about knowing the answer to the wrong question.

Look on the bright side. it was the right answer to the right question.

[The Black Jester]

The permutations are as follows
Perm1 - Card1(R1 showing, R2 hidden), Card2(W1 showing, W2 hidden), Card3(R1 showing, W2 hidden) = 2
Perm2 - Card1(R2 showing, R1 hidden), Card2(W1 showing, W2 hidden), Card3(R1 showing, W2 hidden) = 2
Perm3 - Card1(R2 showing, R1 hidden), Card2(W2 showing, W1 hidden), Card3(R1 showing, W2 hidden) = 2
Perm4 - Card1(R2 showing, R1 hidden), Card2(W2 showing, W1 hidden), Card3(W2 showing, R1 hidden) = 1
Perm5 - Card1(R2 showing, R1 hidden), Card2(W1 showing, W2 hidden), Card3(W2 showing, R1 hidden) = 1
Perm6 - Card1(R1 showing, R2 hidden), Card2(W1 showing, W2 hidden), Card3(W2 showing, R1 hidden)  = 1
Perm7 - Card1(R1 showing, R2 hidden), Card2(W2 showing, W1 hidden), Card3(W2 showing, R1 hidden) = 1
Perm8 - Card1(R1 showing, R2 hidden), Card2(W2 showing, W1 hidden), Card3(R1 showing, W2 hidden) = 2

You ought to have 8 perms because each card has two faces so we have 2x2x2 = 8

We have 4 instances where two reds are showing and 4 instances where only 1 red is showing, so that is 12 red cards showing.
Of those 12 red cards, 8 of them have a red back and 4 of them have a white back
So if you have a red card then it is 8/12*100 = 66% likely to have a red back.

You can simplify this. A card showing a red face can only be either the card with two red faces or the card with one red face and one white face. There are three equally possible scenarios when a red face will show upwards.The card with two red faces-  1.red face one  2.red face two; the card with one red face and one white face-  3. red face.  2/3 = 66%.

This is what I was thinking originally, but then everyone began getting more and more complex, so I thought I must be wrong.