Question: Three dice are rolled. If X represents the second largest value of the three rolls, derive the probability distribution of X.
Many thanks in advance for all your help!
What I have tried so far:
I tried at first to consider all unique rolls, of which there are 56, and my distribution was:
P(X=1)=6/56 P(X=2)=10/56 P(X=3)=12/56 P(X=4)=12/56 P(X=5)=10/56 P(X=6)=6/56
but then I realised that I had to take into account the different orders. So I then tried the following:
When all three numbers are the same, there are 3! possible ways these numbers can be ordered, so the number of possibilities increases by 36 (6 each for all of 1,2,3,4,5,6).
When two of the numbers are the same, there are 2! possible ways these numbers can be ordered for each case, but the numbers can be written in increasing/decreasing order, so I need to multiply by a factor of 2. 30*4=120 cases.
If the numbers are all different, the middle number is fixed, but the other two numbers can be swapped, so 20*2=40.
This gave me 196 cases to consider. At this point, I became stuck and didn’t see a systematic way to derive a distribution. I’m not even sure the number of cases as 196 is correct.
There are $6^3=216$ ways that three different dice can fall. To have the middle number be $1$ you have to throw $111,112,113,114,115, \text{ or } 116$ The first represents one combination, the others three, so the chance $1$ is the middle number is $1$ is $\frac {16}{216}$. By symmetry, this is the same as the chance $6$ is the middle number.
To have $2$ the middle number we need $122, 12x, 222, \text { or } 22x$ where $x$ represents a number greater than $2$ so is $4$ possibilities. $122$ is three throws, $12x$ is $24, 222$ is $1, 22x$ is $12$, so the chance $2$ is the middle number is $\frac {40}{216}$. By symmetry that is also the chance that $5$ is the middle number.
The chances $3$ and $4$ are the middle number are equal by symmetry and must represent all the rest of the probability. The chance it is $3$ is then $\frac 12-\frac {16+40}{216}=\frac {13}{54}$