15 dice, chances of getting three either 1 or 4 I have 15 dice, by rolling three either 1 or 4 can win me the game. It doesn't have to be three 1's or three 4's, it could also be 1-1-4 or 1-4-4, etc. As long as I have three 1 or 4, I will win the game.
What is the probability of me losing the game?
I know the total outcome of rolling 15 dice is 6^15. I tried to simplify the question by just rolling 2 dice, total outcome is 6^2 = 36. Drew a table to list all 36 possible outcomes and counted the cases with at least a 1 or 4, and the probability of such case is 20/36 = 55%.
But I just don't know the proper way to deal with this question without drawing the table out.
Thank you!
P.S.
To clarify my question, here is an example.
Let say you have 15 dices, and you roll them all at once. Afterwards, you look at each dice and see if it is either 1 or 4. If it is, then you add a point to the scoreboard. If it's not, then your score remains the same. Your score starts at 0. The goal is to get 3 points, and you will win the game.
So, what is the probability of losing the game? Meaning after looking at all 15 dices, your score is 2 or less.
I think Bruce has already answered the question, but I added this for extra clarification just in case.
 A: Comment continued with a simulation of a million such games in R. The simulated answer
$0.0791 \pm 0.0005$ matches the exact answer in my Comment to three places.
set.seed(804)
m = 10^6; s = numeric(m)
for (i in 1:m) {
  x = sample(1:6, 15, rep=T)
  s[i] = sum(x==1)+sum(x==3) }
mean(s <= 2)
[1] 0.079097      # aprx P(Lose) = 0.07936
2*sd(s <= 2)/1000
[1] 0.0005397805  # aprx 95% margin of simulation error

The histogram bars show simulated values of $\mathsf{Binom}(15, 1/3)$
and the red dots show exact binomial probabilities.
hist(s, prob=T, br = (-.5:15.5), col="skyblue2")
k=0:15; pdf=dbinom(k,15,1/3)
 points(k, pdf, col="red", pch=19)


The figure below adds the density function of $\mathsf{Norm}(\mu = np, \sigma=\sqrt{np(1-p)}),$ where $n = 15, p = 1/3,$ which approximates your
binomial distribution.

We can get a reasonable approximation to $P(X \le 2) = P(X \le 2.5) \approx 0.0856$ by
using this normal distribution. Normally, you shouldn't expect more than two places of accuracy from a normal approximation, unless $n$ is large or $p$ is near $1/2$ or both. (The normal approximation is not as
close to the truth as is the simulation above.)
mu = 15/3;  mu
[1] 5
sg = sqrt(2*mu/3); sg
[1] 1.825742
dnorm(2.5, mu, sg)
[1] 0.08556962

You can standardize and use printed normal tables to get about this same
answer from the normal approximation. (It will be a little different because
some rounding is necessary when using printed tables.)
