Four standard six-sided dice are rolled and any side is equally likely to land face up. The highest value is dropped and the remaining three values are summed. What is the expected value of the result?
If all four dice have the same number, then the value from any one die can be dropped.
I have never seen the question asked in this way before, but I have seen a number of similar problems where the lowest of the four dice is dropped, so I would assume this problem can be solved in the same way. Here is my solution:
Start by finding the expected value of the highest die roll, which I will call $E(H)$:
There is only one way one can be the highest die roll, which is when all dice show 1.
There are 2^4 – 1^4 = 15 ways where 2 can be the highest
There are 3^4 – 2^4 = 65 ways where 3 can be the highest
There are 4^4 – 3^4 = 175 ways where 4 can be the highest
There are 5^4 – 4^4 = 369 ways where 5 can be the highest
There are 6^4 – 5^4 = 671 ways where 6 can be the highest.
E(H) = (1/1296 * 1) + (15/1296 * 2) + (65/1296 * 3) + (175/1296 * 4) + (369/1296 * 5) + (671/1296 * 6)
E(H) = 5.24
The expected value of the sum would be $$E(X_1 + X_2 + X_3 – H) = E(X_1) + E(X_2) + E(X_3) – E(H).$$
Since we know the expected value of a die is $3.5$, this comes out to $$3.5 + 3.5 + 3.5 +3.5 – 5.24 = 8.74.$$
I would appreciate it if someone could check my reasoning (and my math) to help me solve this question. Thank you.
Best Answer
I think you meant $\ldots = 8.76$, which is indeed correct.
You're missing an $X_4$ in your sum:
$$E(X_1 + X_2 + X_3 + X_4 - H) = E(X_1) + E(X_2) + E(X_3) + E(X_4) - E(H),$$
although you did account for it in your calculation.