My Question pertains to an uncalculated part of Example 3.8.14 found on page 133 of Blitzstein-Hwang's free online probability textbook which can be found here: http://probabilitybook.net
I am self-teaching myself probability using Blitzstein's Harvard lecture series Stat110 and must stress that this is not assigned homework.
Their chosen example serves to highlight why conditional independence does not imply independence - which I now understand.
The final conclusion of $$P(Y=1|X=1)>P(Y=1)$$ is intuitively very clear to me. Especially if we change the probabilities to an extreme (say $\frac{1}{100}$ and $\frac{99}{100}$) instead of the original $\frac{1}{2}$ and $\frac{3}{4}$.
My confusion is $\pmb{HOW}$ I would go about explicitly calculating the values of $P(Y=1|X=1)$ and $P(Y=1)$.
MY ATTEMPT:
By applying the Law of Total Probability I obtain that $P(Y=1)=P(Y=1|A)P(A)+P(Y=1|B)P(B)=(\frac{1}{2})\cdot(\frac12)+(\frac34)\cdot(\frac12) = \frac58 $
Where $A$ is the event of playing the evenly matched twin and $B$ is the event of playing the weaker twin (the twin for which I have a 3/4 chance of beating). Now one of the assumptions I am making is that I am no more likely to play one of the twins than the other. i.e. that $P(A)=P(B)=\frac12$
However by appling the Law of Total Probability to calculate $P(Y=1|X=1)$ I also got $\frac58$. I am new to probability (as is likely quite clear) and know there is a part of my understanding here which is wrong. Any insight into explicitly calculating these probabilities would be greatly appreciated.
I hope you are enjoying the book and lecture videos! This is an interesting example of conditional independence vs. unconditional independence. Your calculation that $$P(Y = 1) = \frac{5}{8} = 0.625$$ is correct. To find $P(Y = 1 | X = 1)$, we can use the definition of conditional probability: $$P(Y = 1 | X = 1) = \frac{P(X=1,Y=1)}{P(X=1)}.$$ We already know that the denominator is $5/8$. For the numerator, let's use LOTP to condition on which twin we are playing against. This is helpful since once we know which twin we are playing against, $X$ and $Y$ are (conditionally) independent. Then $$P(X=1,Y=1) = P(X=1,Y=1|A)P(A) + P(X=1,Y=1|B)P(B),$$ which is $$\left(\frac{1}{2}\right)^2 \cdot \frac{1}{2} + \left(\frac{3}{4}\right)^2 \cdot \frac{1}{2} = \frac{13}{32}.$$ So $$P(Y = 1 | X = 1) = \frac{13/32}{5/8} = \frac{13}{20} = 0.65.$$ Note that $P(Y=1 | X = 1)$ is indeed greater than $P(Y=1)$, as expected, but only slightly greater since winning the first game only provides weak evidence that we are playing against the weaker twin.