Three cards in a hat

Quote from igsi:



Now I see what was my mistake. I was not taking into account that the card itself and what side is up are both choosen randomly. So, my reasoning would be correct if the problem was stated differently.

"Three cards are placed into a hat. One card has a white face on both sides. A second card has a white side and a red side and the third card has two red sides.

A card is drawn out of the hat and we are told that one side of that cards is red. What is the probability that that card is the card with two red sides?"

The answer to this problem is 50%. Bottom line, the history is important! :)

I still agree. This was your original answer and my original resonse was posted as below. I believe it ALL is exactly as you explained, and I agreed. It is about the probabilities AFTER an action has occured, which changes the entire "problem".

Quote from rs7:



Makes sense to me! Especially since the problem was stated as it was. If the problem was worded differently, such as PRIOR to any event, then yes, I could see how the 66% answer would make sense. But the wording seemed to clearly indicate that one event had already occurred. Which would change the probability (as far as I can understand this).
On the other hand, it is never wise to be on the wrong side of an assessment by MrSub. He is rarely if ever wrong. So that has to be taken into serious consideration.

Peace,
:)rs7
 
Quote from igsi:
A card is drawn out of the hat and we are told that one side of that cards is red. What is the probability that that card is the card with two red sides?"
I don't see the difference between the above and the original problem. Unless "one side of that card is red" means only one side is red, in which case the probability of it being the card with two red sides is zero.
 
Quote from Mr Subliminal:

I don't see the difference between the above and the original problem.

The difference is that in the original problem the side of the card is chosen randomly while in my problem the card is chosen randomly. That changes problem's sample space.
 
Quote from Mr Subliminal:

OK, you draw your card randomly.

(1) Then what do you do?

(2) How do you formulate (1) into a probability question?

(1) The subject asks if one of the sides is red and receives "yes".
(2) The same is in original problem:
"What is the probability that that card is the card with two red sides?"
 
Well, you could just do this:

A guy walks into a room with a card laying on the floor. The card looks red to the person, but the other side cannot be seen. The person is told that before he walked into the room, three cards were put into a machine in the ceiling. A random card was dropped from the machine and landed on the floor. The three cards that were originally in the machine were a red/red, red/white and a white/white card.

The person is than asked, "What is the probability that the card on the floor is the red/red card."

That would be 2/3'rds, or 66.666 (ad infinitum) percent.
 
Quote from aphexcoil:

Well, you could just do this:

A guy walks into a room with a card laying on the floor. The card looks red to the person, but the other side cannot be seen. The person is told that before he walked into the room, three cards were put into a machine in the ceiling. A random card was dropped from the machine and landed on the floor. The three cards that were originally in the machine were a red/red, red/white and a white/white card.

The person is than asked, "What is the probability that the card on the floor is the red/red card."

That would be 2/3'rds, or 66.666 (ad infinitum) percent.

So, What's new? Assuming the person is told the truth, that's the same as you posted starting this thread.
 
Quote from igsi:



So, What's new? Assuming the person is told the truth, that's the same as you posted starting this thread.

Right. So the solution is something we all agree on.
 
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