A famous three card trick. 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?
It sounds fifty fifty, there are only two cards relevant.
But you could be looking at one of three sides.
The red side of the red/white one
One Side of the red/red
The other side of the red/red
Sounds like another tricky 66.66% thing
It has to be 50-50, and I don't care if Einstein personally brings a different answer and tatoos it on my buttocks.
So what are the chances he'll choose the left buttock first?
(https://www.happyatheistforum.com/forum/proxy.php?request=http%3A%2F%2Fi46.photobucket.com%2Falbums%2Ff131%2Faasaliman%2F6744c15a.jpg&hash=43bbb0fa82549a3536eed149dbd0d70e576d89bb)
Ha! You can't fool me this time! It is 4/6 aka 66.66%. Bazinga!
WHY, WHY has no one taken the time to congratulate me on my brilliance in solving this puzzle?!? Is it possible that you scoundrels don't realize this may well be the only "logic" question I will ever answer correctly on this board?!? This is a once in a lifetime occurance. Appreciate it, or I swear to Asmo I will level this place.
*long hard squinty glare at each of you*
;D
By Jove, I think she's got it!
Congratulations Ali. So smart.
I love that you wrote all the permutations down. There were only 6 of them so it was well worth the time.
We need a graduation cap for you to throw into the air!
I grovel, O Ali, but I still don't get it.
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Now here's a good one:
There are three crows sitting on a fence. You shoot one crow. How many crows remain?
Quote from: OldGit on May 11, 2012, 09:37:39 AM
I grovel, O Ali, but I still don't get it.
***********************
Now here's a good one:
There are three crows sitting on a fence. You shoot one crow. How many crows remain?
None if they have any sense.
^ No, sorry. ;)
Quote from: Ali on May 10, 2012, 04:24:42 PM
(https://www.happyatheistforum.com/forum/proxy.php?request=http%3A%2F%2Fi46.photobucket.com%2Falbums%2Ff131%2Faasaliman%2F6744c15a.jpg&hash=43bbb0fa82549a3536eed149dbd0d70e576d89bb)
Ha! You can't fool me this time! It is 4/6 aka 66.66%. Bazinga!
Ali,
I, too, grovel before your superior intelligence. I have been staring at your solution and am having the worst time understanding it. I am completely clueless, for example, as to what criteria you used to write "1" or "0" next to each permutation. It is very possible that I am just an utter cretin and am missing what is obvious to everyone, and I apologize for asking you to spell out what seems transparent to others. But, is there any other way you can explain it? When I tried the original problem, I too came up with an answer of "2/3" (since, obviously, 4/6 reduces to 2/3) but only because I reasoned clumsily that of the 3 red sides, 2 belonged to the card with a red side on both, so there was a 2/3 possibility. But I'm guessing that isn't a correct avenue to the solution. And, please, if others besides Ali want to jump in and school this patent imbecile, please feel free to do so.
Quote from: OldGit on May 11, 2012, 11:52:58 AM
^ No, sorry. ;)
The dead/injured one, I suppose?
Quote from: The Black Jester on May 11, 2012, 06:04:43 PM
Ali,
I, too, grovel before your superior intelligence. I have been staring at your solution and am having the worst time understanding it. I am completely clueless, for example, as to what criteria you used to write "1" or "0" next to each permutation. It is very possible that I am just an utter cretin and am missing what is obvious to everyone, and I apologize for asking you to spell out what seems transparent to others. But, is there any other way you can explain it? When I tried the original problem, I too came up with an answer of "2/3" (since, obviously, 4/6 reduces to 2/3) but only because I reasoned clumsily that of the 3 red sides, 2 belonged to the card with a red side on both, so there was a 2/3 possibility. But I'm guessing that isn't a correct avenue to the solution. And, please, if others besides Ali want to jump in and school this patent imbecile, please feel free to do so.
Groveling before my superior intellect! That I like!
1 or 0 was kind of shorthand for yes or no. The question for each variation is "does this scenario show a card with a red face next to another card with a red face?" 1 = yes, 0 = no. So for the first possible layout, the card that has red/red is on the left, and the card that has red/white is in the middle. The two cards both have red and are next to each other, so that's a yes. For the second scenario, red/red is on the left, white/white is in the middle, and red/white is on the right. The two cards that have red on them are not next to each other (white/white is in between them) so that's a no. There are 6 possible scenarios, and out of that, 4 of them have the two cards that feature red next to each other (marked by a 1). Makes better sense?
Quote from: The AsmoThe dead/injured one, I suppose?
Right, oh Asmo! One crow remains: the dead one. The other two flew away. Please, nobody tell me it's 0.66 crows or I'll go
MAAAAAD.
Quote from: Ali on May 11, 2012, 08:14:57 PM
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
Quote from: Stevil on May 11, 2012, 08:58:03 PM
Quote from: Ali on May 11, 2012, 08:14:57 PM
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.
QuoteThere 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.
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.
Nope. Don't get it. *Sigh*
Quote from: Stevil on May 11, 2012, 09:17:41 PM
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.
Quote from: Stevil on May 11, 2012, 09:17:41 PM
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%.
Quote from: Ali on May 11, 2012, 09:01:29 PM
Quote from: Stevil on May 11, 2012, 08:58:03 PM
Quote from: Ali on May 11, 2012, 08:14:57 PM
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.
QuoteThere 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.
Quote from: En_Route on May 11, 2012, 10:07:09 PM
Quote from: Stevil on May 11, 2012, 09:17:41 PM
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.