Odds $11$ to $3$ against a month selected randomly containing portions of $6$ different weeks

Question

Here's a problem from my probability textbook:

Show that the odds are eleven to three against a month selected at random containing portions of six different weeks.

Here are some initial thoughts. Showing that the odds against are $11$ to $3$ suggests that the denominator is divisible by $7$, which suggests we should be traversing $7$ years and so it'll be $84$ months before the cycle repeats. But I'm not sure what to do next. Any help would be welcome.

Answer 1

The successful cases to consider are:

• 31-day months that start on Friday or Saturday
• 30-day months that start on Saturday

Each of the 12 months can start on any of the 7 days, yielding success probability $$p=\frac{7 \cdot 2 + 4 \cdot 1}{12 \cdot 7} = \frac{18}{84} = \frac{3}{14}$$

So the odds against are $$\frac{1-p}{p}=\frac{11/14}{3/14}=\frac{11}{3}$$