Looking for a layman's explanation of how to manually calculate log odds? I will start that I am not as math oriented as I would like to and could use a layman's / non-staticians explanation walk through of how to calculate the log odds.
I am reading Hosmer, Lemeshow, and Sturdivant's Appliged Logistical Regression which is helpful, but I could use a primer / tutorial to help solidify the concepts for me. 
Given something like the following equation (values randomly chosen):
$$
\log\bigg(\frac \pi {(1-\pi)}\bigg) = 0.3211 + 0.27 X_1 + 0.732 X_2
$$
And the following information:


*

*Indicator Variable Group A $X_2 = 1$, Group B $X_2 = 0$.  

*Sample Size 1000

*Likelihood Value for Model = 0.0598


How would I compute the following from the above information manually without using R or another application.


*

*Log Odds for $X_1 = 3$ for each group.

*Odds for $X_1 = 3$ for each group.

*Probability $X_1 = 3$ for each group.


If I understand correctly the Log Odds is the $\ln(p/(1-p))$ and log-odds and odds are different; but I am NOT clear on how to apply the above information to calculate the above information and am looking for a step-by-step walk through that covers most of the steps required of how to apply this information and perform the calculation.
Note: While I am utilizing this to assist me in a class it is not part of an assignment or test and the equation is made up by me purely for example purposes as I feel I am in need of a starting point example as I've spent some time reading the book (particularly chapter 3) but it is not clicking like I need it to. 
 A: You are right that the "$\ln(p/(1−p))$" is the log odds.  That means all you have to do to calculate the log odds is plug in the values you want for your $X$'s and do the arithmetic.  Here is the calculation for #1:
\begin{align}
\text{log odds}(X_1 = 3; A) &= 0.3211 + 0.27\times X_1 + 0.732\times X_2  \\
                            &= 0.3211 + 0.27\times 3\ \ \ + 0.732\times 1  \\
                            &= 0.3211 + 0.81\quad\quad\ \  + 0.732  \\
                            &= 1.8631  \\
\ \\
\text{log odds}(X_1 = 3; B) &= 0.3211 + 0.27\times X_1 + 0.732\times X_2  \\
                            &= 0.3211 + 0.27\times 3\ \ \ + 0.732\times 0  \\
                            &= 0.3211 + 0.81   \\
                            &= 1.1311  \\
\end{align}
To get the odds from the log odds, you exponentiate the log odds.  That means you raise the special number e ($\approx 2.718281828$) to the power of the log odds.  Here is the calculation for #2:
\begin{align}
\text{odds}(X_1 = 3; A) &= e^{1.8631}  \\
                        &= 6.443681    \\
\ \\
\text{odds}(X_1 = 3; B) &= e^{1.1311}  \\
                        &= 3.099064
\end{align}
(Full disclosure: I used R to do the arithmetic on that one.)  
To get the probability, you divide the odds by the quantity odds plus one.  Here is the calculation for #3:
\begin{align}
\text{probability}(X_1 = 3; A) &= \frac{6.443681}{6.443681 + 1}  \\
\ \\
                               &= 0.8656579    \\
\ \\
\text{probability}(X_1 = 3; B) &= \frac{3.099064}{3.099064 + 1}  \\
\ \\
                               &= 0.7560419
\end{align}
(I used R for this one, too.)  
