Probability of Tails on Other Side of Heads Coin Flip

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The discussion revolves around a probability problem involving three coins: a two-headed coin, a two-tailed coin, and a normal coin. After tossing a coin and getting heads, the question is about the probability that the other side is tails. The original answer given was 2/3, while the poster believes the correct probability is 1/3. The poster analyzes the outcomes and concludes that, given the result of heads, the likelihood of the other side being tails is indeed 1/3, leading to a conflict with their professor. The poster expresses concern about addressing this disagreement with their professor.
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First of all I want to say hello. This is my first time posting on these forums, but they have (well actually YOU have :D) helped me solve certain questions of mine.


For the last week or so, I have been obssessed with a problem that we solved during class.
And I have trouble finding common ground with my professor.

The problem is as following:

We are given three coins: one of them has a head on each side, one of them has a tail on each side, and one "normal" coin with a head on one side and a tail on the other. Now we randomly choose a coin, we toss it and the result is heads.


What is the probabilty that the other side is tails?

The answer that was given to us is 2/3, I managed to come up with 1/3. Cross-checking my results by using common everyday logic, points out to me, that I am actually right. I really need your opinion on this one.


Let's call the two-headed coin C1, the two-tailed coin C2 and the normal coin C3. I start by accepting that fact that, since the result is heads, the coin we tossed is either coin C1 or coin C3.

Now let's say that C1's heads are respectively H1 and H2. And C2's head is H3 and it's tail is just plain T.

The result of tossing heads implies one of three possibilites:

1) the result is H1 and the other side is H2

2) the result is H2 and the other side is H1

3) the result is H3 and the other side is T



So what am I doing wrong?
 
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Look at all the possibilities

1) You picked the H-H coin and the result was H
2) You picked the H-H coin and the result was H (the other side)

3) You picked the H-T coin and the result was H
4) You picked the H-T coin and the result was T

5) You picked the T-T coin and the result was T
6) You picked the T-T coin and the result was T (the other side)

Obviously, all these things have probability 1/6 to happen.

You are now given information that the result is H. So we go to the above diagram and we eliminate all results that are T. This gives us

1) You picked the H-H coin and the result was H
2) You picked the H-H coin and the result was H (the other side)

3) You picked the H-T coin and the result was H

We have agreed these 3 things to have equal probability. So 1&2 imply the other side to be H. Thus the other side is H in 2/3 of the cases. The other side is T is 1/3 of the cases.
 


Thank you for your contribution, as I see it now, my real problem is confronting my professor that will not be too fond of me correcting her.
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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