SOLUTION: Prove by induction that if n distinct dice are rolled, then the number of outcomes where the sum of the faces is an even integer equals the number of outcomes where the sum of the

Question 582299: Prove by induction that if n distinct dice are rolled, then the number of outcomes where the
sum of the faces is an even integer equals the number of outcomes where the sum of the faces
is an odd integer?
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
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

, we can let

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.

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