How to find total possibilities of 4 events?

Question

Say you had 4 coins being tossed simultaneously, and a total score (X) is given by 3 points for each head and 1 point for each tail, added together. How would you find the total possibilities and the the probably of getting 8 for example?

This is easy enough if it is 2 coins, but going through each event individually is tedious with 4 coins.

Answer 1

First, split it into disjoint events of equal probability:

 C 1 | C 2 | C 3 | C 4 | Score
-----|-----|-----|-------------
  1  |  1  |  1  |  1  |   4
  1  |  1  |  1  |  3  |   6
  1  |  1  |  3  |  1  |   6
  1  |  1  |  3  |  3  |   8
  1  |  3  |  1  |  1  |   6
  1  |  3  |  1  |  3  |   8
  1  |  3  |  3  |  1  |   8
  1  |  3  |  3  |  3  |  10
  3  |  1  |  1  |  1  |   6
  3  |  1  |  1  |  3  |   8
  3  |  1  |  3  |  1  |   8
  3  |  1  |  3  |  3  |  10
  3  |  3  |  1  |  1  |   8
  3  |  3  |  1  |  3  |  10
  3  |  3  |  3  |  1  |  10
  3  |  3  |  3  |  3  |  12

Then, count the number of combinations of the specific score that you're interested in:

• The number of combinations that sum up to $ 4$ is $1$
• The number of combinations that sum up to $ 6$ is $4$
• The number of combinations that sum up to $ 8$ is $6$
• The number of combinations that sum up to $10$ is $4$
• The number of combinations that sum up to $12$ is $1$

Finally, divide this number by the total number of combinations, which happens to be $16$:

• $P(\text{score}= 4)=\frac{1}{16}$
• $P(\text{score}= 6)=\frac{4}{16}$
• $P(\text{score}= 8)=\frac{6}{16}$
• $P(\text{score}=10)=\frac{4}{16}$
• $P(\text{score}=12)=\frac{1}{16}$