There are 17 people. We assume that a year has 365 days.

a) What is the probability that at least two of them have birthday at the same day of the year?

b) What is the probability that exactly two has birthday at 1st of January?

a) me guess is ${17 \choose 2} * (\frac{1}{365})^2 * (\frac{364}{365})^{15}$

b) no idea.

  • 2
    $\begingroup$ Read about the birthday surprise problem $\endgroup$ – Dilip Sarwate Feb 8 '14 at 21:37

a) We solve this problem by first finding the probability of NONE of the $17$ people having the same birthday.

Person $1$ has a birthday (let's call it $D_1$). Then Person $2$ must have a birthday other than $D_1$ (let's call it $D_2$), which leaves him $365-1 = 364$ options. Person $3$ must have a birthday other than $D_1$ and $D_2$ (let's call it $D_3$), which leaves him $365-2 = 363$ options. This goes on until you reach $D_{17}$.

Because there are $365^{17}$ arrangements of birthday possible, and you have $365 \cdot 364 \cdot 363 \cdots 349$ options, the answer is $\displaystyle \frac{365 \cdot 364 \cdot 363 \cdots 349}{365^{17}}$.

Note that we're not finished yet, however, because we found the probability that NONE of them have the same birthday when we really want the probability that AT LEAST TWO have the same birthday. Fortunately, the math is simple - we just subtract the value we found from $1$. So the final answer is $$ 1 - \frac{365 \cdot 364 \cdot 363 \cdots 349}{365^{17}}.$$

b) We can choose two people among the $17$ to have the birthday of January $1$st, and we don't care when the others have their birthdays as long as it's not January $1$st.

We have to choose the two people. Because order doesn't matter, this will be a combination - namely, $\dbinom{17}{2}$. The probability that both of them have the birthday of January $1$st is $\displaystyle \left(\frac{1}{365}\right)^2$.

We then have to assign a birthday to each of the others (N.B.: They can have the same birthdays), which means the probability will be $\displaystyle \left(\frac{364}{365}\right)^{15}$).

Multiplying all of these together, we get

$$\dbinom{17}{2} \left(\frac{1}{365}\right)^2 \left(\frac{364}{365}\right)^{15}.$$

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