Probability no male- female pairs share same birthday

Question

There are 8 people in a room. There are 4 males(M) and 4 females(F). What is the probability that there are no M-F pairs that have the same birthday ? It is OK for males to share a birthday and for females to share a birthday. Assume there are $10$ total birthdays.

I give a solution below. Not sure if is correct and is there a more general way to approach it ? I break it into 5 cases-summing these cases gives the total ways M-F do not share. If divide the sum by $10^8$ would obtain desired probability.

Case 1: all men have different birthdays $N_1 = 10 \cdot 9 \cdot 8 \cdot 7 \cdot (10-4)^4$

Case 2: one pair men exact + two single men $N_2 = {\sideset{_{10}}{_1} C} \cdot {\sideset{_4}{_2} C} \cdot 9 \cdot 8 \cdot (10-3)^4$

• the first term chooses the single BD for the pair of men.
• The second term selects the 2 men in the pair.
• The $9\cdot 8$ are the number of ways the two single men can choose their birthdays.
• The final term is the number of ways the $4$ woman can select the remaining $10-3 = 7$ birthdays which do not equal the men which have used $3$ birthdays.

Case 3: two pair men exact $N_3 = {\sideset{_{10}}{_2} C} \cdot {\sideset{_4}{_2} C} \cdot {\sideset{_2}{_2} C} \cdot (10-2)^4$

Case 4: one triple and one single man $N_4 = {\sideset{_{10}}{_1} C} \cdot {\sideset{_4}{_3} C} \cdot {\sideset{_1}{_1} C} \cdot {\sideset{_9}{_1} C} \cdot (10-2)^4$

Case 5: all men have same birthday $N_5 = {\sideset{_{10}}{_1} C} \cdot (10-1)^4$

The sum of Case $1$ to $5$ is the total ways for no M-F pairs. The last term in each case is the number of permutations of the 4 woman with $(10-k)^4$ choices where $k$ is the number of unique birthdays used up for the men. I do not believe the order of the people matters: I calculate assuming all the men come first. Please comment on my approach.

I have not found an understandable solution on this website.

Answer 1

Let $A$ be the event no M-F pair share the same birthday. Let $B_1$ be the event all females share ONE birthday.

Let $N(A \cap B_1)$ be the number of possible configurations realizing the event $A \cap B_1$.

I think $$N(A \cap B_1) = {10 \choose 1} {9 \choose 1} + \left [ {10 \choose 2}* {4 \choose 2} \right ] * {8 \choose 1} + \left [ {10 \choose 3} * 3! * 3 \right ] *{7 \choose 1} + \left [ {10 \choose 4} * 4! \right ] * {6 \choose 1} $$

The first term is {all men share the same birthday} $\cap B_1$

the second term is {all men share two distinct birthdays } $\cap B_1$

the third term is {all men share three distinct birthdays } $\cap B_1$

the fourth term is {all men share 4 distinct birthdays} $\cap B_1$.

I think we can calculate $N(A \cap B_i)$ for $i = 2,3,4$ and then the result would be:

$$\frac{N(A \cap B_1) + N(A \cap B_2) + N(A \cap B_3) + N(A \cap B_4)}{10^8}$$

Let me know any errors.

Answer 2

Your calculation has a minor error in that case 4 should end with *(10-3)^4 rather than *(10-2)^4.

If you correct that and add the numbers up then you would get $19550250$. Dividing by $10^8$ would then give the probability of $0.1955025$

Generalising this is a little messy because your cases 2 and 3 each count possibilities for the mean having two birthdays between them. There is a way round this by using Stirling numbers of the second kind and you could say something like

If there are $d$ days in a year, and $m$ men and $w$ women with their birthdays independently and uniformly distributed across these days, then the probability that there are no cases of a man and a woman sharing a birthday is $$\frac{d! }{d^m}\sum\limits_{n=1}^{\min(m,d)} \frac{S_2(m,n) }{(d-n)!}\left(1-\frac{n}{d}\right)^w $$
where $S_2(x,y)$ is the corresponding Stirling number of the second kind.

If you applied this to your example with $d=10, m=4,n=4$, it would give $$362.88\left(\frac{1\times 0.9^{4}}{362880} + \frac{7\times 0.8^{4}}{40320} + \frac{6\times 0.7^{4}}{5040} + \frac{1\times 0.6^{4}}{720}\right)=0.1955025$$