Test for exhaustion is definitely out of the question

When rolling a die once, the probability of getting an even number (1, 3, 5) is the same as the probability of getting an odd number (2, 4, 6).

Assume that when rolling $n$ dice, the probability of getting an even sum and the probability of getting an odd sum are both $\frac{1}{2}$ . Now, roll $n+1$ dice.

• An odd result occurs when the sum of the first $n$ dice and the final die have opposite parity. This happens with probability $\frac{1}{2}\times\frac{1}{2} + \frac{1}{2}\times\frac{1}{2} = \frac{1}{2}$ (you can think of this as the probability the first $n$ are odd times the probability the last one is even, plus the probability the first $n$ are even times the probability the last one is odd).

• Similarly, an even result occurs when the sum of the first $n$ dice and the final die have the same parity. This also happens with probability $\frac{1}{2}\times\frac{1}{2} + \frac{1}{2}\times\frac{1}{2} = \frac{1}{2}$ .

The statement is True .

Considering the implied question in the problem title.

$\text{FindFit}\left[\text{Table}\left[\text{First}\left[\text{Timing}\left[\log _2(\text{Plus}\text{@@}\text{Table}[(\text{Plus}\text{@@}t \bmod 2),\{t,\text{Tuples}[\{0,1\},n]\}])\right]\right],\{n,25\}\right],e^{a x+b}+c,\{a,b,c\},x\right]$

${a -> 0.716967962881, b -> -14.4048401877, c -> -0.0089040378436}$

$e^{a x+b}+c\text{/.}\, \{a\to 0.716968,b\to -14.4048,c\to -0.00890404,x\to 1000\}$

$1.3161\times 10^{305}$

The last is the estimated time on my computer using Wolfram Mathematica in seconds for 1000 dice, which is vastly greater than the age of the universe.