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I'll use BoT to denote a boy born on a Tuesday, BnT for a boy born not on Tuesday, and G for a girl. The possible combinations for the two kids are

 BoT, BoT    1/14 * 1/14 = 1/196
 BoT, BnT    1/14 * 6/14 = 6/196
 BnT, BoT    6/14 * 1/14 = 6/196
 BoT, G      1/14 * 7/14 = 7/196
 G  , BoT    7/14 * 1/14 = 7/196

plus various combinations that do not involve any BoT, but those are not important.
The respective probabilities for each these 5 combinations is given. Note that I changed $1/2$ into $7/14$ in order to give them all the same denominator.

We want to know the probability of two boys, given that there is (at least) one BoT. We know that:

$P(2\ boys\ \&\ BoT) = \frac{1+6+6}{196} = \frac{13}{196}$

$P(BoT) = \frac{1+6+6+7+7}{196} = \frac{27}{196}$

Therefore $P(2\ boys\ |\ BoT) = \frac{P(2\ boys\ \&\ BoT)}{P(BoT)} = \frac{13}{27}$, or almost 50%.


Without that day of the week, and just knowing there was at least one boy, the answer would be $\frac{1}{3}$.

On the other hand, if you knew that the oldest child was a boy, then the answer would be $\frac{1}{2}$.

By saying the day of the week the boy you happen to know was born on, it is almost as good as specifying a particular child of the two, almost like saying that the oldest child is definitely a boy, and thereby raises the probability from $\frac{1}{3}$ to almost but not quite $\frac{1}{2}$.