Q. (the probability that the total after rolling 4 fair dice is 21) > (the probability that the total after rolling 4 fair dice is 22)
A. Explanation: All ordered outcomes are equally likely here. So for example with two dice, obtaining a total of 9 is more likely than obtaining a total of 10 since there are two ways to get a 5 and a 4, and only one way to get two 5’s. To get a 21, the outcome must be a permutation of (6, 6, 6, 3) (4 possibilities), (6, 5, 5, 5) (4 possibilities), or (6, 6, 5, 4) (4!/2 = 12 possibilities). To get a 22, the outcome must be a permutation of (6, 6, 6, 4) (4 possibilities), or (6, 6, 5, 5) (4!/22 = 6 possibilities). So getting a 21 is more likely; in fact, it is exactly twice as likely as getting a 22.
What I have understood so far is that after rolling 4 fair dice you can get 22 in the following ways -
- 6,6,6,4 - $4C_1$ - 4
- 6,6,5,5 - $4C_2$ - 6
So the total ways in which you can get 22 is 10, where you can get 21 in the following ways -
- 6,6,6,3 - $4C_1$ - 4
- 6,6,5,4 - $4C_2$ - 6
- 6,5,5,5 - $4C_1$ - 4 and hence the total ways in which you can choose is 14.
Am I doing something wrong here? Can someone please help me understand the answer (A.) a bit better?
Thanks in advance.
There are $12$, not $6$ ways to get a permutation of $6,6,5,4$. You can choose the position of the $5$ in $4$ ways, then choose the position of the $4$ in $3$ ways.
This gives us $4+12+4 = 20$ ways to get a total of $21$, and $4+6=10$ ways to get a total of $22$, and so we conclude that rolling a $21$ is twice as likely.