Three cards in a hat

Quote from igsi:



Of course it's 50%. It may seem like 66% is the probability to pick a red side out of three sides left, two reds and one white. However, you should not count one red side which is the other side of white/red card. That outcome is impossible because the side we see is red, not white and therefore that red on white/red is out. Thus, we have two possible outcomes, white and red, which makes the probability 50%.
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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
 
The way I figured it out was:

Card1 is Side1 of W and Side2 of W
Card2 is Side1 of W and Side2 of R
Card3 is Side1 of R and Side2 of R

So, the unique possibilities are:

Draw1: Card1 Side1 (W)
Draw2: Card1 Side2 (W)
Draw3: Card2 Side1 (W)
Draw4: Card2 Side2 (R)
Draw5: Card3 Side1 (R)
Draw6: Card3 Side2 (R)

Since we know it's showing Red, that leaves 3 possibilities that are equally likely:

Card2 Side2 (R)
Card3 Side1 (R)
Card3 Side2 (R)

And two out of three are Card3 so the answer is 2 out of 3 or .66666. . .
 
Bob still thinks it is 50%. He has 2 degrees, one in math and one in physics.
His thinking is that the double white card is totally eliminated in the equation because it does not meet the parameter of having a red side. That leaves the other card and the card on the floor. Giving it a 50/50 chance of having 2 red sides.

I am math deprived, so I abstain.
 
I'll try making it a little more intuitive for Bob, rs7 and igsi. An equivalent problem would be the following :

3 jars are placed into a hat. One jar contains 2 white balls, a second jar contains 1 white ball and 1 red ball, and the third jar contains 2 red balls. A ball is drawn out of one of the jars in the hat (it's a big hat and naturally we can't see the jars) and it is red. What is the probability that it was drawn from the jar with 2 red balls?

Forgetting for the moment the solution to this problem, hopefully you will agree that it is the equivalent of aphie's problem. If not, then let me know why. Assuming you accept this equivalence, your answer will be 50% while mine remains 66.67%. As Bob would say, the jar with the 2 white balls is totally eliminated in the equation leaving the other 2 jars, giving it a 50/50 chance of being the one containing 2 red balls.

Let us now multiply everything by a factor of 50. One jar now contains 100 white balls, a second jar contains 50 white balls and 50 red balls, and the third jar contains 100 red balls. This also is equivalent to aphie's problem and your answer should remain 50%. If not, speak out now.

Now consider the following problem :

3 jars are placed into a hat. One jar contains 100 white balls, a second jar contains 99 white balls and 1 red ball, and the third jar contains 100 red balls. A ball is drawn out of one of the jars in the hat and it is red. What is the probability that it was drawn from the jar with 100 red balls?

Still 50%?
 
Quote from TGregg:

The way I figured it out was:

Card1 is Side1 of W and Side2 of W
Card2 is Side1 of W and Side2 of R
Card3 is Side1 of R and Side2 of R

So, the unique possibilities are:

Draw1: Card1 Side1 (W)
Draw2: Card1 Side2 (W)
Draw3: Card2 Side1 (W)
Draw4: Card2 Side2 (R)
Draw5: Card3 Side1 (R)
Draw6: Card3 Side2 (R)

Since we know it's showing Red, that leaves 3 possibilities that are equally likely:

Card2 Side2 (R)
Card3 Side1 (R)
Card3 Side2 (R)

And two out of three are Card3 so the answer is 2 out of 3 or .66666. . .
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I must admit, this sounds very convincing. However, I believe that solutions of this type are methodologically flawed and, as a result, incorrect. The fault, as I see it, is that you are taking into account events which took place in the past and I think that just brings a confusion into analyzing a problem which is basically to assess the future outcome based on current conditions.

Thus, we have a red card lying on the floor. We have another card with a red side in the hat. Lets take that card and put it beside the first card. As you can see, we neither speculating about past events nor revealing any additional information about the cards. It is obvious that the other card with the red side exists, so we just putting together what we know.

Well, what do we have now? We have two red cards lying on the floor. What are the odds that either one is hiding red? Looks to me like 50/50. Now, I am putting one card back to the hat. Does that change the odds? No, it does not and we still have 50% probability. qed
 
Quote from Mr Subliminal:

I'll try making it a little more intuitive for Bob, rs7 and igsi.
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Mr Subliminal, that did not make it any more intuitive for me. If you are good at this, please reply with your comments to my previous post.
 
Again, the probability that the card with two red sides being on the floor is GREATER than that of the card with a white side and a red side. The problem is correct as it was presented. The odds are not 50/50. If both cards had two red sides, the odds would be 50/50, but you are neglecting the fact that the one card that is still in the hat that has not been eliminated only has one red side and not two.

Mr. Subliminal has provided a great example.
 
Quote from aphexcoil:

Again, the probability that the card with two red sides being on the floor is GREATER than that of the card with a white side and a red side. The problem is correct as it was presented. The odds are not 50/50.
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You said it twice. If you keep saying sugar, that does not make it any sweeter for me.

If both cards had two red sides, the odds would be 50/50,
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And I think I clearly demonstrated that this can be achieved without changing conditions, ie tweaking the odds.

but you are neglecting the fact that the one card that is still in the hat that has not been eliminated only has one red side and not two.
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This statement is a nonsense. First, we do not know how many red sides the card in the hat has. Second, what difference does it make?
 
Quote from igsi:
Mr Subliminal, that did not make it any more intuitive for me. If you are good at this, please reply with your comments to my previous post.
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igsi, you dismissed my "intuitive" explanation without saying why, yet I distinctly peppered the post with logical exits whereby you could express disagreement. I therefore return your courtesy and will not "reply with comments" to your previous post. Rather, I append a numbered proof showing why you're wrong. Disagree by all means, just give me the number and reason.

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)
 
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