Inspired by the Birthday Problem

Let

${n}$ = Number of people in my friends list = 336 people;

${X}$ = Event no one shares the same birthday in $n$.

${Y}$ = Event everyone has distinct birthdays in $n$

$$P(X) + P(Y) = 1$$

$$P(Y) = \frac{365!}{(365-n)!(365)^n}$$

I understand the birthday problem, but the following question was driving me crazy:

My question is, what is the probability that a person- James, has a distinct birthday from the others.

• The wanted probability is the complementary probability of the event someone that is not James was born the same day as James', i.e. the probability that every person different from James was born in a day different from James'. – Jack D'Aurizio Mar 23 '17 at 20:13
• So $$\left(\frac{364}{365}\right)^{335}\approx 39.89\%$$ – Jack D'Aurizio Mar 23 '17 at 20:15
• A lot simpler than I expected... Jeez, thanks! – Navy Seal Mar 23 '17 at 20:16
• @JackD'Aurizio, did you raise to the power because any of the 335 remaining people can be born in any of the remaining days of the year? – Navy Seal Mar 23 '17 at 20:19
• Exactly. $\phantom{}$ – Jack D'Aurizio Mar 23 '17 at 20:32

$X=\text{James shares birthday with atleast one more person}$
$Y=\text{All other }n-1\text{ people have different birthday than James}$
Again, $P(X)+P(Y)=1$.
To find $P(Y)$ note that birthday of James can be any of $365$ days. Fixing birthday of James, birthdays for the other $n-1$ people can be chosen from any of the remaining $364$ days, in $364^{n-1}$ ways, giving total $365\times364^{n-1}$ ways for birthdays.
Without any restriction, there are $365^n$ ways the birthdays can be selected.
So $P(Y)=\frac{365\times364^{n-1}}{365^n} = \left(\frac{364}{365}\right)^{n-1}$, and hence $P(X)=1-\left(\frac{364}{365}\right)^{n-1}$.