Question 582299
Base case n=1 is obviously true.


Suppose that the statement is true for n dice (n>1). If we add another die that is distinguishable from the other n dice, then for each sequence of n dice rolls *[tex \LARGE a_1, a_2, ..., a_n], we can let


*[tex \LARGE S = \sum_{i=1}^{n}a_i]


*[tex \LARGE S + a_{n+1}] is even if and only if S is even and the (n+1)th roll is even, or if S is odd and the (n+1)th roll is odd. If we look at this from a probabilistic view, we have


P(S even) = 1/2 because there are as many even outcomes as odd outcomes
P((n+1)th roll even) = 1/2 because 2,4,6 are even


Therefore P(S even and (n+1)th roll even) = 1/4. Similarly, P(S odd and (n+1)th roll odd) = 1/4, so P(S + a_(n+1) even) = 1/2, which is half of all the possible outcomes. Hence we are done.