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
- Thread starter BobRodes
- Start date
p5wizard, and all 50%ers.
Your premise only works if the cards have an equal chance of being on the table. ALL of you are saying that it does.
All of you are saying that the drawing of the card has already happened and how the card got there does not matter.
So, get yourself two cards/coins/whatever and draw them 10 times and place them on the table.
Out of ten attempts of drawing the w/w card how many times will it be placed and remain on the table. Answer = 10.
Out of 10 attempts of drawing the w/r card, how many times will it be placed and remainon the table. Answer ~ 5.
Conclusion - They don't have an Equal Chance of being on the table. The W/W card will be there twice as often. The initial conditions matter.
Probability does not ignore initial conditions. Example - A die is cast and it comes up as a 1. What is the probability of that happening. Your logic would state that it is already a 1 in this instance. Initial conditions do not matter. It has a 100% chance of being a 1. Really?
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Occam's Razor - All things being equal, the simplest solution is the right one.
Your premise only works if the cards have an equal chance of being on the table. ALL of you are saying that it does.
All of you are saying that the drawing of the card has already happened and how the card got there does not matter.
So, get yourself two cards/coins/whatever and draw them 10 times and place them on the table.
Out of ten attempts of drawing the w/w card how many times will it be placed and remain on the table. Answer = 10.
Out of 10 attempts of drawing the w/r card, how many times will it be placed and remainon the table. Answer ~ 5.
Conclusion - They don't have an Equal Chance of being on the table. The W/W card will be there twice as often. The initial conditions matter.
Probability does not ignore initial conditions. Example - A die is cast and it comes up as a 1. What is the probability of that happening. Your logic would state that it is already a 1 in this instance. Initial conditions do not matter. It has a 100% chance of being a 1. Really?
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Occam's Razor - All things being equal, the simplest solution is the right one.
Challenge to 50%ers.
I have provided several methods of manually duplicating the experiment without bias and asked for results. None have taken me up.
I challenge you to provide me with an experiment I can run multiple times using the conditions in this puzzle where my results will be 50%.
3 Possible outcomes:
1 - You will not be successful and be convinced of new paradigm.
2 - You will be successful but it will be provable that your experiment does not meet the conditions of this puzzle
3 - You will be successful and you will convince me of a new paradigm.
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Occam's Razor - All things being equal, the simplest solution is the right one.
I have provided several methods of manually duplicating the experiment without bias and asked for results. None have taken me up.
I challenge you to provide me with an experiment I can run multiple times using the conditions in this puzzle where my results will be 50%.
3 Possible outcomes:
1 - You will not be successful and be convinced of new paradigm.
2 - You will be successful but it will be provable that your experiment does not meet the conditions of this puzzle
3 - You will be successful and you will convince me of a new paradigm.
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Occam's Razor - All things being equal, the simplest solution is the right one.
kwbMitel - you may be disappointed. I've been asking the 50% fan club to demonstrate the flaw in the White A and White B illustration of the problem for some time. It has been studiously and repeatedly ignored. I dont see them picking up and running with your challenge either.
The empirical data that kwbMitel provided is good at illustrating that the 2:3 answer is correct, but really, the explanations provided by strongm, CajunCenturion, and myself give precise explanations of why 50% is wrong.
If the explanations given are wrong, where specifically are they wrong?
@lionelhill
>Proof by Google was a new one on me
Where did you see a "Proof by Google" in this entire thread?
If the explanations given are wrong, where specifically are they wrong?
@lionelhill
>Proof by Google was a new one on me
Where did you see a "Proof by Google" in this entire thread?
- Thread starter
- #85
This is certainly a record number of responses for a thread that I started. And by the way, it's called the "three card swindle" because people make money all the time off of people who insist that the odds are 50%. A good huckster will start off with even betting, and then improve the odds to maybe 5 to 4 and keep winning. Ipguru and Mmerlinn, if you're ever in California we'll have to play. 
BobRodes, this is nothing compared to Strongm and CajunCenturion trying to convince me that I was wrong on a probability question regarding Lottery Odds. All told I think that one exceeded 300 posts over 3 threads.
Strongm - I hear you. I was trying to show the same patience you've shown with me.
Guitarzan - Thanks for the boost.
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Occam's Razor - All things being equal, the simplest solution is the right one.
Strongm - I hear you. I was trying to show the same patience you've shown with me.
Guitarzan - Thanks for the boost.
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Occam's Razor - All things being equal, the simplest solution is the right one.
OMG, I've been swindled - and on the internet no less...
Hey, someone had to go the 50/50 route for this thread to kick off.
What worries me more: I even got a star for being swindled. Someone on the internet must really hate me! ;-)
Is from a UNIX fortune cookie. Here's the exact text:
HTH,
p5wizard
Hey, someone had to go the 50/50 route for this thread to kick off.
What worries me more: I even got a star for being swindled. Someone on the internet must really hate me! ;-)
me in this thread said:
Is from a UNIX fortune cookie. Here's the exact text:
/usr/games/fortune said:
HTH,
p5wizard
P5Wizard, Are you now a 2/3 convert? Does "I've been swindled" = 50% is wrong?
Oh, and you got your star from mmerlinn, who unilaterally decided he/she was right and also anyone who agreed with him/her.
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Occam's Razor - All things being equal, the simplest solution is the right one.
Oh, and you got your star from mmerlinn, who unilaterally decided he/she was right and also anyone who agreed with him/her.
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Occam's Razor - All things being equal, the simplest solution is the right one.
- Thread starter
- #91
![[ponder]](/data/assets/smilies/ponder.gif "[ponder] [ponder]")
Unilaterally - Definition as applies to my usage.
-Emphasizing or recognizing only one side of a subject.
-Having only one side.
In a sentence:
At the end of the meeting, with all parties not achieving consensus, the Chairperson unilaterally decided what was the correct answer.
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Occam's Razor - All things being equal, the simplest solution is the right one.
-Emphasizing or recognizing only one side of a subject.
-Having only one side.
In a sentence:
At the end of the meeting, with all parties not achieving consensus, the Chairperson unilaterally decided what was the correct answer.
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Occam's Razor - All things being equal, the simplest solution is the right one.
- Thread starter
- #93
Olaf Doschke
Programmer
And this whole thread could've been much short with this very simple explanation:
With no outset given the chance to pick the red/white card is 1/3. So the chance, the downside face of the card has the different color is 1/3 and therefore the chance the downside face is the same color is 2/3. That's true for either outset of a white or red face showing.
Did I make you blush
Bye, Olaf.
With no outset given the chance to pick the red/white card is 1/3. So the chance, the downside face of the card has the different color is 1/3 and therefore the chance the downside face is the same color is 2/3. That's true for either outset of a white or red face showing.
Did I make you blush
Bye, Olaf.
Olaf, The red/White card only has a 1/6 chance of being drawn and being white side up. The White/White Card has a 1/3 chance. The Red/Red card cannot be drawn with white side up.
Your example implies that the puzzle asks if the downside face is the same color regardless of the top side. The puzzle states that the top side is white.
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Occam's Razor - All things being equal, the simplest solution is the right one.
Your example implies that the puzzle asks if the downside face is the same color regardless of the top side. The puzzle states that the top side is white.
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Occam's Razor - All things being equal, the simplest solution is the right one.
- Thread starter
- #98
- Thread starter
- #99
Ok, to be fair, Olaf is explaining the probabilities accurately, using a more general example (the probability of the other side of a card being the same color as the color showing is 2/3) than the one in the puzzle. On the other hand, I doubt that the thread would have been shortened by this "simple explanation," given that it's not really any simpler than some of the previous ones.
Bob,
Olaf's example has absolutely no bearing on the original problem. None. The fact that his "simple" example comes out with 2/3 as the answer is coincidental. For example, add a fourth card, a blue/blue card, to your original question. The answer is still 2:3, but following Olaf's logic, the new "simple" answer becomes 3/4.
Olaf's example has absolutely no bearing on the original problem. None. The fact that his "simple" example comes out with 2/3 as the answer is coincidental. For example, add a fourth card, a blue/blue card, to your original question. The answer is still 2:3, but following Olaf's logic, the new "simple" answer becomes 3/4.
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