Conditional Probability with Normal Distributions? Let's say that I have $3$ independent random normal variables, $A$, $B$ and $C$.
They all have a standard deviation of $17.526$, while $A$ has a mean of $143$, $B$ of $139$, and $C$ of $129$.
My question is, how would I go about calculating $P(A>C\mid A>B)$? I know how to calculate $P(A>C)$ and $P(A>B)$, but it's the conditional part I'm struggling with. I'm not sure how I would manage to apply Bayes' theorem to this, so I'm stuck.
If it helps speed-wise, I have $P(A>C)$ to be $0.713912$, and $P(A>B)$ to be $0.564105$ with the difference distribution in each case having a stdev of $24.785$.
Thanks.
 A: Given $3$ independent normal random variables $A, B$, and $C$, we have that
$$P\{A > C \mid A > B\} = \frac{P\left(\{A > C\}\cap \{A > B\}\right)}{P\{A > B\}}
\tag{1}$$
where the denominator of the fraction on the right side of $(1)$ is easy to compute by observing that 
$B-A \sim N(\mu_B-\mu_A, \sigma_B^2+\sigma_A^2)$, and so
$$P\{A > B\} = P\{B-A < 0\} 
= \Phi\left(\frac{\mu_A-\mu_B}{\sqrt{\sigma_B^2+\sigma_A^2}}\right).\tag{2}$$
The OP has made this observation and done this computation already.  However, the numerator of the fraction on the right side of $(1)$ is not easily computed,
and numerical integration might be necessary. One approach using conditional
distributions is as as follows. Given that $A=a$
$$\begin{align}
P\left(\{A > C\}\cap \{A > B\}\mid \{A = a\}\right)
&= P\left(\{a > C\}\cap \{a > B\}\right)\\
&= P\{a > C\}P\{a > B\} & \scriptstyle{B~ \text{and}~C~\text{are independent}}\\
&= \Phi\left(\frac{a-\mu_C}{\sigma_C}\right)\Phi\left(\frac{a-\mu_B}{\sigma_B}\right).
\end{align}$$ 
We can now use the law of total probability to write that
$$\begin{align}
P\left(\{A > C\}\cap \{A > B\}\right) 
&= \int_{-\infty}^{\infty} P\left(\{A > C\}\cap \{A > B\}\mid \{A = a\}\right)f_A(a)\,\mathrm da\\
&= \int_{-\infty}^{\infty} \Phi\left(\frac{a-\mu_C}{\sigma_C}\right)\Phi\left(\frac{a-\mu_B}{\sigma_B}\right)
\frac{1}{\sigma_A}\phi\left(\frac{a-\mu_A}{\sigma_A}\right)\,\mathrm da.
\end{align}$$
To the best of my knowledge, there is no "closed-form" expression for
this probability and the value of the above integral has to be computed
via numerical integration. Using the value provided by the simulation by @MasatoNakazawa with the value $P(A>B) = 0.564\ldots$ provided by the OP,
it would appear that for the values of the parameters in the OP's problem,
$$P\left(\{A > C\}\cap \{A > B\}\right)  \approx 0.468\ldots$$
Crude bounds on the above probability can be obtained for those not inclined
to simulate or numerically integrate. We have
$$P\left(\{A > C\}\cap \{A > B\}\right) 
\leq \min \left\{P\{A > C\}, P\{A > B\}\right\} = 0.564105$$
Also, the complementary probability $P\left(\{A < C\}\cup \{A < B\}\right)$
is bounded above as $$P\left(\{A < C\}\cup \{A < B\}\right)
\leq P\{A < C\} + P\{A < B\} = 0.721983$$
and so we have that 
$$0.278017 \leq P\left(\{A > C\}\cap \{A > B\}\right) \leq 0.564105.$$
A: You can use the definition of conditional probability:
$$P(X|Y)=\frac{P(X,Y)}{P(Y)}$$
In your case, we have
$$P(A>C|A>B) = \frac{P(A>C,A>B)}{P(A>B)} = \frac{P(A>\max(B,C) )}{P(A>B)}$$
You could find the distribution of $\max(B,C)$ and then find the probability of the numerator.
A: I just want to supplement @Comp_Warrior's great answer (an empirical estimate of $P(A > max(B, C))/P(A>B)$):
s <- 17.526
N <- 10000

set.seed(1)
A <- rnorm(N, 143, s) 
B <- rnorm(N, 139, s) 
C <- rnorm(N, 129, s) 

sum(A > pmax(B, C))/sum(A>B) ## ~0.83

