Proving the probability of the sum of 500 dice being an even number

Question

Prove that the sum of rolling 500 dice being an even number is 1/2. The hint is to define A as the event that "the sum of the first 499 dice is even" and define B as the event that "the 500th dice is even".

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My Attempt: Define X as the event "the sum of 500 dice is even"

Then: $$P(X) = P(B|A) + P(B^c|A^c)$$ $$=\frac{P(B \cap A)}{P(A)} +\frac{P(B^c \cap A^c)}{P(A^c)}$$ $$=\frac{P(B)P(A)}{P(A)} +\frac{P(B^c)P(A^c)}{P(A^c)}$$ $$P(B) + P(B^c) = \frac{1}{2}+\frac{1}{2}=1$$ Where the 2nd to 3rd step is because events B and A are independent of each other. Clearly this is wrong since $P(X)$ should be $\frac{1}{2}$ and not $1$.

I'm assuming there is a mistake in my first step. My thinking is that if the sum of the 499 dice is even, then the 500th has to also be even for the sum of 500 dice to be even, and if the sum of 499 dice is odd, then the 500th dice has to be odd.

Any help is greatly appreciated :)

Answer 1

Define the event $E_k = \{\text{the sum of} \ k \ \text{dice is even}\}$.

But the sum of $k$ dice is even if and only if the sum of first $k−1$ dice is even and the $k$-th die is even or that the sum of first $k−1$ dice is odd and $k$-th die is odd, then

$$P(E_k) = \frac{1}{2}P(E_{k-1})+\frac12P(E_{k-1}^c) = $$

$$\frac{1}{2}(P(E_{k-1})+P(E_{k-1}^c)) = \frac{1}{2}(P(E_{k-1})+1-P(E_{k-1})) = \frac12$$

So in particular $P(E_{500})= \frac12$.

Answer 2

When you throw a dice you'll add a number $N\in\{1,2,3,4,5,6\}$ to your current number S. Half of these outcomes in N will make your number S change parity of odd$\rightarrow$even or even$\rightarrow$odd ($N\in\{1,3,5\}$) while adding the other half will make it have the same parity ($N\in\{2,4,6\}$). This means that adding the dice of any single roll to an already odd or even number will make it equally likely for the next outcome to be odd or even.

Now since the first outcome is $P(\text{even}) = P(N\in\{2,4,6\}) =1/2$ and $P(\text{odd})=P(N\in\{1,3,5\} = 1/2$ we know that the probability the sum of the final number is even will be $1/2$.