easy math problem

[cmhardw]

Ok, so I don't know very many math problems to post here. But I always liked this one:

I have a deck of three cards. One card is black on both sides, one card is red on both sides, one card is black on one side and red on the other side.

I place a card on the table, and the side that you see is red. What is the probability that when I turn the card over, the other side is also red?

My guess is that many of you have probably already seen this before, so my disclaimer is that this is probably an easy problem for all you math experts out there

Chris

 

[not_kevin]

I'll hazard a guess of 2/3?

 

[sgowal]

I'll hazard another guess... 1/2

 

[nitrocan]

this is similar to the monty hall problem, so i won't spoil it more for you since i know the answer. (its not SO easy! )

 

[hawkmp4]

One half.
Highlight for quick explanation:

Spoiler

Well, since you can see that at least one side of the card is red, that eliminates the card with black on both sides. You have two possibilities left- it could be the card with one side black and one red, or both sides red. Since one side is red already, there's two possibilities, with equal weight, for the other side- black or red. We're looking for red, so P(red) = 1/2.

 

[nitrocan]

unfortunately that is not correct think more complex (you might want to search "conditional probability")

 

[cmhardw]

(its not SO easy! )

Yes, I didn't mean to imply that the problem was simple - it is kinda tricky. But I meant to imply that it is probably not as difficult as other math problems that have been posted on this board already.

Chris

 

[MistArts]

3/5...? I don't how I got it

 

[brunson]

Nitrocan, you're right and wrong.

The problem is a variation of Josef Bertrand's classic box problem and you're right, the probability isn't 1/2, it's 2/3 and it is absolutely based on conditional probablility.

But it isn't really anything like the Monte Hall problem. Monte Hall involves someone with external knowledge of the facts influencing the outcome of the game and what choice you should make knowing that Monte has information that you do not. The probability is actually very simple. What is the chance the you picked the wrong door? 2/3.

 

[rjohnson_8ball]

At first I thought it had to be 1/2, but now that I thought it through I know it is 2/3. This is assuming that each of the red sides has equal probability of being up in the first place.

Chris H, can you grab fixes (typos and stuff) I made for your web pages? See http://www.geocities.com/rjohnson_8ball/ Read my descriptions of fixes.

EDIT: By the way, I typed up this answer before I saw brunson's post. I also noticed this was not like the Monte Hall problem, which is why I first thought it was 1/2, until I thought it through. EDIT2: Corrected Monty to Monte.

 

[hawkmp4]

Hm, I don't see how its two-thirds...can anyone give an explanation?
I don't see how three comes into play at all. We're finding the probability of A given B, where A is that the second side is red and B is that the first side is red...correct?

 

[JBCM627]

Could we please use spoiler tags? It isn't hidden in the standard skin when you just change the font color.

Hawk mp4: right, ignore the black card, but consider that there are actually 3 red sides, not just 2 cards with red on them.

Spoiler

Continuing, from those 3 sides, 2 of them have red on the other side (2 sides from the all-red card), and 1 has black (from the red-black card). Hence 2/3 probability.

 

[hawkmp4]

Thank you JBCM, that makes sense. I thought of that as soon as I saw your post xD

 

[brunson]

@JCBM627, sorry for not using spoiler tags, I didn't think it was important because knowing the answer is 2/3 is nothing without knowing *why* it's 2/3.

Here's different spoiler

Spoiler

Conditional probability is defined by Bayes' Law which states that The probability of some event A occurring, written "P(A)", given the fact that B already occurred [which we say as, "the probability of A given B" or "P(A|B)"] is equal to the probability of B given A times the probability of A divided by the probability of B.

P(A|B) = P(B|A)*P(A)/P(B)

We can restate the problem such that we're asking, "Did we select the all red card?" because we've been shown red and want to know if the other side is red.

If we then define A as "we have the all red card" and define B as "we are shown a red face", then we are looking for the probability that we have the all red card given than we have been shown a red face, or P(A|B)

Now, given that you've been shown a red side of the card, the other side is red iff (read, "if and only if") the card is the all red card. And also given that the card you selected is the one with two red faces (A), then the likelihood of seeing a red face is 100%, i.e. P(B|A) = 1. That's the hard part over.

We know that the probability of picking the all red card is 1/3, that's simple, and since half the faces of the cards are red, then the probability of being shown a red card face is 1/2, so substituting we get:

P(A|B) = (1 * 1/3) / (1/2) = 2/3

BTW, JCBM627, at first I didn't think your reasoning was correct, but I thought about it on the train ride home and I like it a lot now. I like the intuitiveness of it.