Probability of different people being born on the same day of the week What is the chance that at least two people were born on the same day of the week if there are 3 people in the room?
I know how to get the answer which is 19/49 when considering all 3 people not being born on the same day. However, when I try to calculate the answer directly I seem to get it wrong.
Considering exactly 2 people being born on the same day I get 1*1/7*6/7. And then, exactly 3 people is 1*1/7*1/7. Thus, the total is 6/49 + 1/49 = 7/49. This must be something fairly simple, but I was just wondering where I'm going wrong. 
Thanks
 A: Here's how to arrive at the answer $P(\text{at least two people on same day of week}) = 1 - P(\text{all people on different days of the week}) = 1 - \frac{7 * 6 * 5}{7 * 7 * 7} = 1 - \frac{30}{49} = \frac{19}{49}$.
The main thing there is $P(\text{all people on different days of the week}) = \frac{\text{no of ways to assign three different birthdays to three guys}}{\text{total no of ways to assign three birthdays to three guys}}$.
In your calculation, you are missing some cases. Exactly two people born of the same day is $7 * \frac{1}{7} *\frac{1}{7} * \frac{6}{7} * 3$. And like you said, all three people on one day is $7 * \frac{1}{7} * \frac{1}{7}$.
Your calculation is wrong here: 

Considering exactly 2 people being born on the same day I get
  1*1/7*6/7

You must multiply by 3 because you can choose any of the three guys to not have the identical birthday.
A: The only mistake you have committed is that you have not included the combinations while calculation of 2 were born on the same day
If you name the persons A, B ,C
Since the two of them may be
A B,
B C,
C A
Therefore you must also multiply it by a factor of 3
