Three cards in a hat

Quote from Mr Subliminal:

Let the cards be WW, RW and RR.

(1) You maintain that with a red card lying on the floor, the probability that it is RR is 50%, and the probability it is RW is 50%.

(2) By symmetry, it would follow that if the white card is lying on the floor, the probability that it is WW is also 50%, and the probability it is RW is 50%.

(3) Now you must also agree with me that even before this experiment is conducted, the probability that the card lying on the floor will be red (or white) is 50%.

(4) So what in effect you're saying is that there is a 50% chance the card on the floor will be red and if this is the case (ie. the card is red) then there is a 50% chance it's either RR or RW. Similarly, there is a 50% chance the card on the floor will be white and if this is the case (ie. the card is white) then there is a 50% chance it's either WW or RW.

(5) From (4) it follows directly that the probability of drawing RR is 50%*50% = 1/4, the probability of drawing WW is 50%*50% = 1/4, and the probability of drawing RW is 50%*50% + 50%*50% = 1/2.

(6) But we know that the probability of drawing RR = 1/3, the probability of drawing WW = 1/3 and the probability of drawing RW = 1/3.

(7) From (5) and (6) it follows that (1) and (2) must be false.

(Proof ad absurdum)

(QED)

Thanks, I can see that this is true. I still fail to see a mistake in my reasoning, though. :(
 
Quote from Mr Subliminal:

So, as a trader, what do you do?

I am not a trader, so I do nothing. As a trader-wannabe, I give myself a little credit for admitting that I was wrong. I am also going to give it another thought trying to see where I was wrong which hopefully will help me better understand myself and the world around me.
 
If the three cards are placed into the hat, and a gambler reaches in and removes one. Places it face down, and a red side shows , he will bet it is the all red card. Giving even odds, he will consistently win-----because BEFORE he pulled out a card, his chances were two out of three that any card removed would be a solid color. He merely bet that a solid color card had been removed. This was written up in Scientific American magazine in 1950 while I was working on my MS degree in Chem. My college prof (when I showed him the article)started betting with the other profs and LHAO at them. :-)
 
Quote from genebort:

If the three cards are placed into the hat, and a gambler reaches in and removes one. Places it face down, and a red side shows , he will bet it is the all red card. Giving even odds, he will consistently win-----because BEFORE he pulled out a card, his chances were two out of three that any card removed would be a solid color. He merely bet that a solid color card had been removed. This was written up in Scientific American magazine in 1950 while I was working on my MS degree in Chem. My college prof (when I showed him the article)started betting with the other profs and LHAO at them. :-)

I got this problem from my psychology textbook and it was used as an example of how human decision making relies more on wholes than on parts.
 
Quote from aphexcoil:



If you can't see it now, you never will.

Putting aside intended insult, this statement is foolish and logically flawed. Moreover, you are evidently wrong. I did not see it then, but I see it now. However, that did not happen due to your input. I say "input" because I think that you post, which consisted of two parts, Ad Nauseam argument and nonsense, does not deserve to be called explanation. Meaningless input is all I've seen from you.
 
Quote from igsi:



Putting aside intended insult, this statement is foolish and logically flawed. Moreover, you are evidently wrong. I did not see it then, but I see it now. However, that did not happen due to your input. I say "input" because I think that you post, which consisted of two parts, Ad Nauseam argument and nonsense, does not deserve to be called explanation. Meaningless input is all I've seen from you.

I didn't mean to offend you. I am sorry.
 
Quote from igsi:

Thanks, I can see that this is true. I still fail to see a mistake in my reasoning, though. :(

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! :)
 
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