Conditional probability with two coin tosses with two possible coins Let's say I have in front of me two coins. One of them is unbiased, and the second is biased with P(H) = 0.9
I do not know which coin is which. Let's say I pick up 1 coin and have to flip it twice. The first flip shows up as Heads. Does this increase my probability of the second flip also showing up as Heads? If so, why? I would like the logical explanation as well as supporting math.
Also, can this be reconciled as similar to the following? Or does the unbiased nature of one of the coins also bring something to the table
Two fair coins are flipped once at the same time. The probability of getting two heads (1/4) is less than the probability of getting a heads on the one of the flips given the the other flip also resulted in heads (1/3)
Hope this concept can be cleared to me.
Reference: https://www.youtube.com/watch?v=pCLQ-BzGk-o&index=8&list=PLBAGcD3siRDjiQ5VZQ8t0C7jkHQ8fhuq8 8:28
 A: To get some intuition, let's make things more extreme. With the same two coins, you pick one of them and flip it ten times. All of the results are heads. Right now, how confident would you be that you are holding the biased coin?
Well, it would be very unlikely to observe ten heads in a row with a fair coin, but the probability of it happening with the biased coin is about $34\%$. So it stands to reason that the probability of you having the biased coin has increased significantly. But then this must increase the probability that the next flip will be heads; the more likely you are to have the biased coin, the more likely you are to flip heads. 
Back the original problem. For a math explanation, use Bayes' theorem. Let $H_1$ be the event that the first flip is heads.
$$
P(\text{ biased }|H_1)=\frac{P(H_1|\text{ biased })P(\text{ biased })}{P(H_1|\text{ biased })P(\text{ biased })+P(H_1|\text{ fair })P(\text{ fair })}=\frac{0.9\cdot 0.5}{0.9\cdot 0.5+0.5\cdot 0.5}=0.64
$$
Then, let $H_2$ be the probability the second flip is heads.
$$
\begin{align}
P(H_2\,|H_1)
&=P(H_2\,|H_1,\text{ biased })P(\text{ biased }|H_1)+P(H_2|H_1,\text{ fair })P(\text{ fair }|H_1)
\\&=0.9\cdot 0.64+0.5\cdot 0.36=0.756
\end{align}
$$
On the other hand, the unconditional probability of $H_2$ is
$$
\begin{align}
P(H_2)
&=P(H_2\,|\text{ biased })P(\text{ biased })+P(H_2|\text{ fair })P(\text{ fair })
\\&=0.9\cdot 0.5+0.5\cdot0.5=0.7
\end{align}
$$
so the probability of second heads has increased.
A: The reason the probability of heads increases is that, once you get a heads on the first toss, you might have picked the biassed coin.  Imagine that you flip the chosen coin ten times, and it comes up heads every time.  Surely, by now you strongly suspect that you've picked the biassed coin, and you would estimate the probability of heads on the next toss as very, very close to .9.  Well, the increase in probability doesn't happen all at once, but increases little by little.  Here, I'm considering probability to be a measure of belief.
If I haven't made a calculation error, the probability of heads goes up from $.7$ to $.757$ from the first toss to the second.
The second example seems to be very different.  I think you are asking what is the probability that both of them are heads, given that at least one of them is heads, which ought to be more than the a priori probability that both are heads, since we have ruled out the possibility that both are tails.  At least, this interpretation gives the $1/3$ probability you mention.  Notice that this is very different from the situation where you flip two coins and randomly choose one of them to look at.  If it turns out to be heads that tells you nothing about the other coin.  It's more like a situation where you toss the coins, and I inspect them and tell you that at least one of them is heads.  If you are to bet that both are heads, what are fair odds?  (If it turns out that I lied and both are tails, you win.)
If we perform this second experiment with the original two coins, we still find that the probability increases.  That is, if we toss the fair coin and the biassed coin simultaneously, the probability of two heads is $.45.$  If we are told that one came up heads, the probability that both came up heads rises to $.45/.95\approx.4736.$
A: Probability that the first flip is heads:
Choose the biased coin and get heads: $0.5\cdot0.9=0.45$
Choose the unbiased coin and get heads: $0.5\cdot0.5=0.25$
Thus, the probability that the first flip is heads: $0.45+0.25=\frac7{10}$
Given that the first flip was heads:
The probability that we chose the biased coin: $\frac{0.45}{0.45+0.25}=\frac9{14}$
The probability that next flip is heads: $\frac9{14}\cdot0.9+\frac5{14}\cdot0.5=\frac{53}{70}$
Note that $\frac{53}{70}\gt\frac7{10}$. The fact that the second flip has a greater probability makes sense because the probability that we chose the biased coin increased from $\frac12$ to $\frac9{14}$ when the first flip was a head.
A: Refer to the probability tree diagram below ($H,T$ - fair, $h,t$ - biased):

We have:
$$P(1H)+P(2h)=0.25+0.45=0.7 \\
\frac{P(1HH)+P(2hh)}{P(1HH)+P(1TH)+P(2hh)+P(2th)}=\frac{0.125+0.405}{0.125+0.405+0.405+0.045}=\frac{53}{70}\approx 0.757.$$
Thus, the probability increased slightly.
