how many people for a 50% of two or more have the same sign?

Question

there are 12 signs of zodiac. how many people must be present for there to be at least a 50% chance that two or more of them were born under the same sign.

im not sure what formula i should be using to start this

Answer 1

$$ 1 - \frac{12}{12}*\frac{11}{12}*\frac{10}{12}*\frac{9}{12} = 0.5722 >0.5 $$ $$ 1 - \frac{12}{12}*\frac{11}{12}*\frac{10}{12}*\frac{9}{12}*\frac{8}{12} = 0.382 <0.5 $$

Hence, 5.

Answer 2

Assuming the usual assumptions (the people are random for general population; each zodiac sign is equally likely etc.)

The probability of at least two having the same sign is $1 - $ the probability of all having different signs.

If there are $n$ people there are $12^n$ possible ways for the $n$ people to have signs. There are $12$ signs the first person can have and $11$ signs the second person can have. So there are $12*.....*(12-n+1) = \frac {12!}{(12-n)!}$ ways for all the signs to be different. (Note: this assumes $n \le 12$. If there are more than $12$ people then at least two must share signs.)

So the probability of all the birthadays being differents is $\frac {12!}{(12-n)^2*12^n}$ . And the probability that at least two share is $1 - \frac {12!}{(12-n)^2*12^n}$

So we need $1 - \frac {12!}{(12-n)!*12^n} \ge \frac 12$ so

$\frac {12!}{(12-n)!*12^n} \le \frac 12$
For $n = 1$ LHS = $\frac {12}{12} = 1$. (which makes sense. If
there is only one person it is $100\%$ certain that no two have the
same sign.)

For $n = 2$ LHS=$\frac {12*11}{12^2} = \frac {11}{12}$ (which makes
sense. If there are only one person it is $\frac 1{12}$ they have the
same sign.)

For $n = 3$ LHS= $\frac {12*11*10}{12^3} = \frac{11*10}{12^2} = \frac {11*5}{12*6} = \frac {55}{72}$.
For $n = 4$ LHS = $\frac {12*11*10*9}{12^4} = \frac {11*5*3}{12*6*4}=\frac {55}{72}\frac 34 =\frac {55}{96}$.

For $n = 5$ LHS = $\frac {12*11*10*9*8}{12^5} = \frac {55}{96}\frac 8{12} = \frac {55}{96}\frac 23 = \frac {55}{144} < \frac 12$.

So $n = 5$.