Fortunately for you, you have only *one* continuous covariate. Thus, you can just make four (i.e., 2 SEX x 2 AGE) plots, each with the relation between BID and $p(Y=1)$. Alternatively, you could make one plot with four different lines on it (you could use different line styles, weights, or colors to distinguish them). You can get these predicted lines by solving the regression equation at each of the four combinations for a range of BID values.

A more complicated situation is where you have more than one continuous covariate. In a case like this, often there is a particular covariate that is 'primary' in some sense. That covariate can be used for the X axis. Then you solve for several pre-specified values of the other covariates, typically the mean and +/- 1SD. Other options include various types of 3D plots, coplots, or interactive plots.

My answer to a different question here has information on a range of plots for exploring data in more than 2 dimensions. Your case is essentially analogous, except that you are interested in presenting the model's predicted values, rather than the raw values.

**Update:**

I have written some simple example code in R to make these plots. Let me note a few things: Because the 'action' takes place early, I only ran BID through 700 (but feel free to extend it to 2000). In this example, I am using the function you specify and taking the first category (i.e., female and young) as the reference category (which is the default in R). As @whuber notes in his comment, LR models are linear in log odds, thus you can use the first block of predicted values and plot as you might with OLS regression if you choose. The logit is the link function, which allows you to connect the model to probabilities; the second block converts log odds into probabilities via the inverse of the logit function, that is, by exponentiating (turning into odds) and then dividing the odds by 1+odds. (I discuss the nature of link functions and this type of model here, if you want more info.)

```
BID = seq(from=0, to=700, by=10)
logOdds.F.young = -3.92 + .014*BID
logOdds.M.young = -3.92 + .014*BID + .25*1
logOdds.F.old = -3.92 + .014*BID + .15*1
logOdds.M.old = -3.92 + .014*BID + .25*1 + .15*1
pY.F.young = exp(logOdds.F.young)/(1+ exp(logOdds.F.young))
pY.M.young = exp(logOdds.M.young)/(1+ exp(logOdds.M.young))
pY.F.old = exp(logOdds.F.old) /(1+ exp(logOdds.F.old))
pY.M.old = exp(logOdds.M.old) /(1+ exp(logOdds.M.old))
windows()
par(mfrow=c(2,2))
plot(x=BID, y=pY.F.young, type="l", col="blue", lwd=2,
ylab="Pr(Y=1)", main="predicted probabilities for young women")
plot(x=BID, y=pY.M.young, type="l", col="blue", lwd=2,
ylab="Pr(Y=1)", main="predicted probabilities for young men")
plot(x=BID, y=pY.F.old, type="l", col="blue", lwd=2,
ylab="Pr(Y=1)", main="predicted probabilities for old women")
plot(x=BID, y=pY.M.old, type="l", col="blue", lwd=2,
ylab="Pr(Y=1)", main="predicted probabilities for old men")
```

Which produces the following plot:

These functions are sufficiently similar that the four-parallel plot approach I outlined initially is not very distinctive. The following code implements my 'alternative' approach:

```
windows()
plot(x=BID, y=pY.F.young, type="l", col="red", lwd=1,
ylab="Pr(Y=1)", main="predicted probabilities")
lines(x=BID, y=pY.M.young, col="blue", lwd=1)
lines(x=BID, y=pY.F.old, col="red", lwd=2, lty="dotted")
lines(x=BID, y=pY.M.old, col="blue", lwd=2, lty="dotted")
legend("bottomright", legend=c("young women", "young men",
"old women", "old men"), lty=c("solid", "solid", "dotted",
"dotted"), lwd=c(1,1,2,2), col=c("red", "blue", "red", "blue"))
```

producing in turn, this plot:

## Best Answer

If you're fitting a binomial GLM with a logit link (i.e. a logistic regression model), then your regression equation is the log odds that the response value is a '1' (or a 'success'), conditioned on the predictor values.

Exponentiating the log odds gives you the odds ratio for a one-unit increase in your variable. So for example, with "gender", if Female = 0 and Male = 1 and a logistic regression coefficient of 0.014, then you can assert that the odds of your outcome for men are exp(0.014) = 1.01 times that of the odds of your outcome in women.