@fabulousgrizzly I think minji meant that in a set of five rolls, it doesn't matter what the first four rolls are. It only matters what the parity (oddness or evenness) of the **sum** of the first four rolls is. See, if the sum of the first four rolls is even (e.g. \(1 + 2 + 3 +4 = 10),\) then we will want the fifth roll to be also **even** (e.g. \(10 + \text{even} = \text{even}\)), and if the sum of the first four rolls is **odd** (e.g. \(1 +2 + 3 + 3 = 9),\) then we will want the fifth roll to be **odd** as well (since \(9 + \text{odd} = \text{even}\)). So since we don't care what the first four numbers are, there are \(6\) choices for each of those, and since the last roll has to match the parity (oddness or evenness) of the running total, there are \(3\) choices for that number.

Actually, I have another way of solving this question!

Each dice has three **odd** numbers and three **even** numbers. The chance of getting an **even** is \(\frac{1}{2},\) and the chance of getting an odd is \(\frac{1}{2}.\) How about not worrying about what the numbers are, and only keeping track of whether the number on the dice is **even** or **odd**? Then, looking at a set of five consecutive rolls, we can describe this like this:

$$ EOEEO$$

Now, we want the sum of these five numbers to be even, so there are some cases:

There are zero odds (\(O\))

There are two odds (\(OO\))

There are four odds (\(OOOO\))

How many ways are there for each case?

If there are zero odds, there is \(1\) way (\(EEEEE\))

If there are two odds, this is like choosing \(2\) spots out of \(5\) to declare to be odds, which has \(\binom{5}{2} = \frac{5 \times 4 }{2} = 10\) ways

If there are four odds, this is like choosing \(4\) spots out of \(5\) to declare to be odds, or, more simply, like choosing \(1\) spot out of \(5\) to be even, so this is equal to \(\binom{5}{1} = 5\) ways

In all, there are \( 1 + 10 + 5 = 16\) ways to get an even sum after five consecutive rolls.

Now, the total number of ways is the ways to create a line of five letters like

$$ EOOEO$$

with no restriction. Each spot has \(2\) choices (\(E\) or \(O\),) so there are \(2 \times 2 \times 2 \times 2 \times 2 = 32\) ways total.

Thus the probability of getting an even sum after five consecutive rolls is equal to \(\frac{16}{32} = \frac{1}{2},\) which might also make sense if you just think about the fact that the parity of the sums should be balanced over even and odd!

Now we just need to stretch this to the case where we have \(100\) rolls. Well, if every five rolls has an even sum, then the total sum over \(100\) rolls should also be even! By the multiplicative rule, since we can do a set of five rolls \(20\) times, the probability overall is \(\boxed{\frac{1}{2}^{20}.}\)

I hope that made sense! This was a fun problem. I'm glad that you asked it!