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# The ever elusive coin problem

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Consider you have two heads-up coins on the table in front of you. Someone then tells you that on the other side of one of the coins is heads, and on the other side of the other is tails. Thus, you have a H/H coin and a H/T coin. Suppose you pick up, at random, one of the heads-up coins. It is already given that the side facing you is heads. What is the probability the other side is also heads? I say 1/2. My reasoning is thus: the problem is analogous to that of hidden items under cups. You know what's on top, but not what's on bottom. Under one cup is one thing, under another cup is another. The probability of getting either of those things is 1/2. Thus, the probability of getting another heads is 1/2. Another way to think of it is that your not dealing with four sides, but with two defined pairs of sides, namely H/H and H/T. The problem ultimately boils down to which of those two pairs you pick up. Thus, once again the probability that the other side is heads is equal to the probability that you pick up the H/H coin: 1/2. However, the book I got this problem out of says the answer is 2/3. Which makes sense to a certain extent: you have one heads already, so three sides remain: H, H, and T. Thus, 2 H's out of 3 sides means 2/3. However, I argue that you can't pick up individual sides -- you can only pick up entire coins. There are two coins, H/H and H/T. This leads to my previous reasoning above for the probabiliy being 1/2. So I told my friend this, and alas he disagrees with me. I can't seem to come to a conclusion on this. I've actually went so far as to do the experiment -- once I got my favored 1/2 probabiliy, the other time I got 2/3. So, which reasoning is correct: mine, or the book's? EDIT: I just realized this may be more appropriate in the Math forum, so feel free to move it.

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I think you're correct. The coin you picked is either the H/H or H/T, and it's only one of those two possibilities.

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How can there be three possibilities left been H/H/T ? You never can pick up what's on the front of the coins.. only what's on the back.. so the only two possibilities ever are H or T.
So it definitly is 1/2.

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Sounds like someone heard about the monty hall problem and tried to apply it to coins. But it's definately 1/2

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I agree with your solution of 1/2. The problem (at least as you've put it) actually only deals with the unknown sides of the coins, the top should be completely irrelevant.

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The probability that it will be a heads is 1/2. It's obvious and verifiable. The back of the coin is completely tangent to the problem.

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After some more thought, I know where the confusion came from. The author checked the possible outcomes from picking up one of these two coins.

Heads side 1
Heads side 2
Tails

And from that he concluded, that of the three possible outcomes, two of them yeilded heads, therefore the probability of it being heads is 2/3. However, he failed to realise that the fact that the side facing up is guaranteed to be heads consolidated some of his outcomes. Consider the case were we have a H/H coin and a H/T coin, and we don't know what sides are facing up. The probability of the side underneath one of the coints being a heads is 3/4, and the probability of it being a tails is 1/4. As soon as he places a restriction that both sides facing up must be heads, he eliminates the heads side of the tails coin from play. He did that. There is also the added restriction that it is now impossible to obtain a result where two heads are facing down. That means based on the result of one coin, we know what the result of the other coin will be. It consolidates the two heads results, like so:

Heads (1/2)
-Side 1 (1/4)
-Side 2 (1/4)
Tails (1/2)

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The description of the problem is ambiguous.

* Are you presented with one of the coins, showing heads, and asked what the probability of the back being tails is?
* Or are your presented with both coins, both showing heads, pick one, and asked what the probability if it being tails is?

The author's result is for the first case. Yours is for the second.

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Considering it's a probability it's asking for more than one time; the either heads or tails(50%) might work if this was just once but if done more than once this is what should be correct:

Let's first lable the coin sides.

h1/t1
h2/h3

and list all possible combinations with heads on top.

h1/t1
h2/h3
h3/h2

When you flip the coin over you have a 1/3 chance to get tails.

Remember the coin picked up can be one of the three, you must consider all three heads when concluding what the chance is.

But I think this pretty much sums up the arguments
"The description of the problem is ambiguous.

* Are you presented with one of the coins, showing heads, and asked what the probability of the back being tails is?
* Or are your presented with both coins, both showing heads, pick one, and asked what the probability if it being tails is?

The author's result is for the first case. Yours is for the second."

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Quote:
 The description of the problem is ambiguous.* Are you presented with one of the coins, showing heads, and asked what the probability of the back being tails is?* Or are your presented with both coins, both showing heads, pick one, and asked what the probability if it being tails is?The author's result is for the first case. Yours is for the second.

The description isn't ambiguous, it indicates the second case. It would be best to post the original problem as it was stated in the book however, to be sure it isn't a misinterpretation of the question.

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