As the number of dice increases, the range of potential outcomes increases.
As the number of dice increases, the likelihood of an extreme outcome decreases.
To show this, let's assume a WIN is 50%: 4,5,6 on a d6. If you roll 1d6, the likelihood of the max result (1) is 50% and the likelihood of the minimum result (0) is also 50%. It's very "swingy" within that range. With two dice, it's 25% minimum, 50% middle, and 25% maximum. As you add more dice, those results cluster more and more toward the middle. With enough dice the result smooths out to something that looks like a normal distribution (it isn't, but it looks a lot like one).
So, yes, the result has a higher potential range with more dice, but no, it actually becomes more predictable, not less.
You can find the odds yourself fairly easily. The formula looks complex, but it's actually something you can do on the back of an envelope for up to 10 or so dice (and trivially in a spreadsheet). I'll state it, walk briefly through the how it works, and show you how to generalize it:
For k dice with d sides and w winning faces, the probability of exactly n wins is:
[( n! / (k! * (n-k)! ) )] * [w/d]^[n] * [(d-w)/d]^[k-n]
First, each die is independent. Think of rolling them one at a time; the way each die lands does not affect the way the next die lands. So you have a giant binary decision tree: with two dice, the results could be:
- Loss then Loss (LL)
- Loss then Win (LW)
- Win then Loss (WL)
- Win then Win (WW)
Let's change the earlier assumption and say instead that a WIN is only when a die shows six. so to get the first result, you have a (5/6) event, then another (5/6) event: 25/36 two-dice rolls will end showing Loss-Loss.
Similarly, wins are (1/6), so WW is only 1/36.
LW is (5/6) then (1/6), which is 5/36. WL is (1/6) then (5/6) which is also 5/6.
Collectively, that's 36 out of 36 possible outcomes.
You can build a similar tree for any number of dice, but it's easier to generalize: the likelihood of each discrete outcome there is [probability of win]^[number of wins] * [probability of loss]^[number of losses]. (Anything raised to the power 0 is 1). So:
- WWWW would be (1/6)^4 * (5/6)^0 = (1/1296) * (1) = 1/1296
- WLWWWL would be (1/6)^4 * (5/6)^2 = (25/36) * (1) = 25/46656
- LLWWWW would be (1/6)^4 * (5/6)^2 = (25/36) * (1) = 25/46656
- LWWWWL would be (1/6)^4 * (5/6)^2 = (25/36) * (1) = 25/46656
(last three examples show that order doesn't matter, just the totals.)
Counting all the ways you could have 2 losses and four wins is hard. Fortunately, there are two ways to to figure that out. First is the "binomial coefficient": for k items, how many ways can you choose n of those items? Or more relevant here, for for k dice rolled, how many ways can n of those dice show wins?
You could calculate it using a formula: ( n! / (k! * (n-k)! ) )
If you're doing that, plug it into a spreadsheet instead.
Or you can use a little math trick to generate the coefficients: Pascal's Triangle. It's faster up to about 5 dice, and still good to about 10:
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Notice how on each line, the outside numbers are 1, then each other number is the sum of the two numbers diagonally up from it. So, going from the fourth line (1 3 3 1) to the fifth line ( 1 4 6 4 1):
- 4 is the sum of 1 and 3
- 6 is the sum of 3 and 3
- 4 is the sum of 3 and 1
...and you can extend that on down infinitely (next line would be 1,6,15,20,15,6,1).
The numbers on each row are the binomial coefficients for that number of dice, starting with 0. So, from the top:
- There is only one way to get a result with 0 dice.
- With 1 die, there is one way to get 1 win and one way to get 1 loss.
- With 2 dice, there is one way to get 2W, two ways to get 1W 1L, and one way to get 2L.
So then that brings us down to the final piece: to get the probability of exactly n wins on k dice with w winning faces, you calculate:
[binomial coefficient] * [probability of win]^[number of wins] *
[probability of loss]^[number of losses]
[binomial coefficient] * [w/6]^[n] * [(6-w)/6]^[k-n]
...which still seems complicated, but is actually pretty quick. For example, getting 2 wins on 3 dice, with 2 winning sides is:
 * [2/6]^ * [4/6] = 3*(2^2)*4 / (6^2 * 6)
The only difficulty there is 6^3 = 216 (which will be the denominator for everything with 3 dice), so with those inputs the probability of two wins is 48/216.