Coin pair betting paradox (NOT!) If we throw two fair coins then there are 4 equally probable possibilities: HH, TT, HT, TH.
Suppose we can't see the result, but we can check one of those two coins. (doesn't matter which one)
Suppose the checked one is H. Then we know that both of them cannot be TT. So there are now only 3 possibilities: HH, HT, TH.

******And here is my mistake******
Therefore, the probability that the other coin in the pair is T (when we know that one of them is H) is 2/3.

Actually the probability is 1/2, because by checking one coin at random we remove only half of the cases in which they are different, so we are left with equal probabilities that both are same, and that they are different.
So there is no paradox.

Similarly, if we get T when we check one of those two coins, then the possibilities are: TT, HT, TH.
So, in this case the probability that the other coin in the pair is H is 2/3.
Let say that we have N such random pairs, and for each pair we check one coin and than place a bet on the state of the other. We would naturally always choose the opposite value for the other coin because, as demonstrated above, the probability that the other one will be opposite of the one we checked is always 2/3.
However, if those N pairs are truly random, then for large N there will be about N/2 pairs with the coins of same value (TT or HH), and about N/2 pairs with the coins of opposite values (TH or HT). And since we are effectively always betting that the coins have the opposite values, then this implies that we would win about half of the time.
How can that be? We are constantly betting on the results which have 2/3 probability, but we win only half of the times. What is wrong with my reasoning?
 A: When you specify which of the two coins is heads (here the coin is specified as 'the one you looked at'), the other coin has a .5 chance of being tails.  
For your situation to work, you need someone to tell you 'at least one of the coins is heads', whenever this is indeed the case.  Then there is a $\frac23$ probability that one of the coins is tails.
A: 
Let say that we have N such random pairs, and for each pair we check one coin and than place a bet on the state of the other. We would naturally always choose the opposite value for the other coin because, as demonstrated above, the probability that the other one will be opposite of the one we checked is always 2/3.

$2/3$ is the probability that the coins are different when you know "at least one coin is a head".
However that is not the same thing as knowing "the coin looked at is a head", which is what you are really being told.
The coin is being tossed, and the person looks at either the left or the right coin, so there are actually eight equally probable outcomes.  Underlining the coin looked at, these are: $$\Large \{\mathsf {\underbrace{\overbrace{\underline{H}H, H\underline{H}, \underline{H}T, T\underline{H}}^{\text{the coin looked at is a head}}, H\underline{T}, \underline{T}H}_{\text{at least one coin is a head}}, \underline{T}T, T\underline{T}}\}$$
So the coins will be different on only $1/2$ of the times the person tells you they looked at a head.

Now, if we change the set up so that instead of looking at just one coin the other person looks at both coins and tells you "at least one coin is $\underline{\quad}$".  
Suppose further that when the coins are different the other person guarantees to say "heads" or "tails" without bias.   So when you hear "at least one coin is a heads" you know you will know the probability that the coins are different is only $1/2$ -- because on the $1/4$ chance they are the same you will always hear the words, but on the $1/2$ probability that they are different you will only hear those words $1/2$ the time.   So your expected chance of being right by guessing opposite of what you're told is: $(1/2)(1/2)+(1/2)(1/2)=1/2$
Suppose instead that when the coins are different the other person is biased to always say "at least one coin is a head".   Then there will be a $2/3$ probability that the coins are different when you hear those words, and you will hear those words $3/4$ of the time.   However, there will also be a $0$ probability that they are different when you hear "at least one coin is a tail", and you will hear that $1/4$ of the time.   So your expected chance to be right by guessing opposite of what you're told is: $(3/4)(2/3)+(1/4)(0) = 1/2$
You could improve your odds by noticing such a bias and changing your strategy accordingly -- if the other person didn't change their bias in response.

Thus the apparent paradox is avoided.
A: I think regardless if you view the first coin or not, the 2nd coin has a 50% chance of being a tail.  Why would that ever change?
Also, how would your experiment be any different than just tossing one coin first, seeing that it is heads, and then tossing the 2nd coin?  Would you say if the first coin in the individual coin toss experiment is heads that the 2nd toss will then have a 75% chance of being tails?  Those two coin tosses, whether done at the same time or sequentially are independent of each other.
There are only 4 heads you can see, namely HH, HT, and TH.  In the first case, if either of those heads are seen, then other coin is also a H.  In the HT and TH cases, if an H is seen, the other coin flip is a T.  So if you see an H for one coin, there is a 2 out of 4 chance the other coin will be an H and a 2 out of 4 chance the other coin will be a T. 
Here are the possible cases where we see a H on the one coin checked:
HH (we see the left H)
HH (we see the right H)
TH
HT  
Upon seeing a H for one coin, there are 2 cases where the other coin is an H and 2 cases where the other coin is a T.
Your initial statement that if a H is seen then there are only 3 possibilities (HH, HT, and TH) is correct but they are not equiprobable because you could see the H in the HH case twice as often as the others, thus making the effective outcomes HH, HH, HT, and TH.  The HH case is "doubly weighted" since no matter what H we see, the other one is also an H and there are 2 ways to do that.
I don't see any paradox.
