mathematics - Probability of nobody being born in a specific month

Assume 11 persons are born in 11 different months. What is the probability $P(n)$ that none of them is born in the $n$-th month?

This would be pretty easy if all months had the same length, then $P(n)$ would be $\tfrac{1}{12}$ for all months. It gets interesting when one considers that different months have different length.

Assumptions:

• The probability to be born on a specific day is the same for all days.


• In the time span all these peole are born February has 28¼ days on average. (This would not be the case, if these people could have been born in 1900.)

Bonus: Find a way of avoiding a brute-force summation over all possibilities.

I'm asking this question because people seem to be stuck at one part of another question. I thought, instead of adding a hint there, it would be more fun to outsource it into a separate question. As both questions are now linked, people can use the result of this question, which should work almost the same as providing a hint.

Answer

Suppose the month lengths are $a_1,a_2,\dots,a_{12}$, and the probability of nobody being born in month $i$ as in the question statement is $P_i$ for $1\le i\le12$.

Let the probability of an arbitrary $10$ people being born randomly around the year being born in all months except $i$ and $j$ with $i\ne j$ given that they are in $10$ different months be $Q_{\{i,j\}}$ (same problem with $10$ months).

Notice that for all $i\ne j$, $P_i:P_j=\frac{a_j}{a_i+a_j}:\frac{a_i}{a_i+a_j}=a_j:a_i$, since $P_i=Q_{\{i,j\}}\cdot\frac{a_j}{a_i+a_j}$ and $P_j=Q_{\{i,j\}}\cdot\frac{a_i}{a_i+a_j}$.

Let $X=a_1+a_2+\dots+a_{12}$.

Then $P_1:P_2:\dots:P_{12}=\frac{X}{a_1}:\frac{X}{a_2}:\dots:\frac{X}{a_{12}}$.

Now, since $P_1+P_2+\dots+P_{12}=1$, for $1\le i\le12$:

$$P_i=\frac{\frac{X}{a_i}}{\frac{X}{a_1}+\frac{X}{a_2}+\ldots+\frac{X}{a_{12}}}=\frac{\frac{1}{a_i}}{\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_{12}}}$$ $$=\frac{1}{a_i\left(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_{12}}\right)}$$

Now in our case, the year, these probabilities turn out to be about (according to Python 3.6):

1. 0.08176161304326855


2. 0.089720708118277


3. 0.08176161304326855


4. 0.08448700014471083


5. 0.08176161304326855


6. 0.08448700014471083


7. 0.08176161304326855


8. 0.08176161304326855


9. 0.08448700014471083



10. 0.08176161304326855


11. 0.08448700014471083


12. 0.08176161304326855

Code:

x=[31,28.25,31,30,31,30,31,31,30,31,30,31]
y=sum([1/i for i in x])
for i in x:print(1/i/y)