# Birthday paradox solving for small n values?

This is for the probability that in a group of n people at least two have the same birthday, n = 3.

Hi, So for those who are familiar with the birthday paradox could you check my work:).

P(E) = [Event]/[SampleSpace]

Sample space = 365^3

Event = (365*364*363

P(E) = 0.991796

1-P(E) = 8.204 x 10^-3

This is where i am a bit lost, i dont know how to prove they are equal, any help is appreciated :).

Notice that $P(E)$ is the probability that in a group of 3 people, at least 2 people share the same birthday. What you are calculating is $P(\overline E)$, the probability that each person has a different birthday.

So you've calculated $1 - P(\overline E) = 0.008204$. Now you have to use counting principles to find $P(E)$.

$$P(E) = [\mathrm{Event}]/[\mathrm{SampleSpace}]$$

Now let's find the number of events in which at least 2 people share the same birthday. Consider person 1, who has $365$ possible birthdays. Now if person 1 only shares a birthday with person 2, then person 2 can only have $1$ possible birthday—the same day as person 1! Since person 1 only shares a birthday with person 2, then person 3 must be born on a different day, any of the $364$ other days.

The same logic applies if person 1 only shares a birthday with person 3. There are also only $365$ events in which all three people have the same birthday, one event for each day of the year.

Finally, if person 1 does not share a birthday with either person 2 nor person 3, then person 2 and person 3 must share a birthday on a different day than person 1's birthday. So person 2 has $364$ different possible days to have a birthday, and person 3 has only $1$ possible day, the same day as person 2.

When we add up all these events, we get:

$365 \times 1 \times 364 + 365 \times 364 \times 1 + 365 + 365 \times 364 \times 1 = 1093 \times 365$

And, $\frac{1093 \times 365}{365^3} = 0.008204$, so we are done.

• You found out $1 - P(\overline E) = 0.008204$. Now $1 - P(\overline E) = E$, so why do we need to find $E$ again in a roundabout way ? – true blue anil Oct 21 '15 at 6:58
• The link says, "Calculate $P(E)$ for $n = 3$ using counting principles, and confirm that it is the same as $1 - P(\overline E)$." – eyqs Oct 21 '15 at 7:32
• Ok, sorry, I didn't see the link. – true blue anil Oct 21 '15 at 7:49

You should define what $E$ is.

Any way, P(at least 2 have a common birthday) = $1 - 0.991796 = 0.008204$

which in scientific notation, becomes $8.204\times 10^{-3}$

• You have already reached (365-n+1) = (365-3+1) = 363 in writing 365*364*363 – true blue anil Oct 21 '15 at 4:48