I have a die with three possible outcomes. The three outcomes are win (+1), draw (0) and lose (-1). $P(w) + P(d) + P(l) = 1$.
(1) If I throw the die Y times, what is the probability I will win $X$ or more of those throws (assuming the dice is fair)?
(2) If I want to stand a $Z\%$ chance of winning $X$ or more throws out of $Y$, what bias must in introduce to the die (i.e. what must $P(w)$ be)?
(1) Coming at this form a tree perspective, my thoughts were;
Even if the dice is fair the outcome winning is not equally likely as losing. If I were to say that there are for this purpose two possible outcomes, I win ($p(w)$) or I do not win ($p(nw)$) where not winning is $p(l)+p(d)$.
In the case of $p(w) = p(nw)$ one can simply count the number of combinations that lead to a win out of all possible outcomes; i.e.
$$ P(\text{win $\geq x$ throws}) = \frac{\sum_{n=x}^y \frac{(y!)}{n!(y-n)!}} {2^y} $$
If the probabilities are not equally likely then one needs to adjust for them to get the probability of that string of outcomes.
$$ P(\text{win $\geq x$ throws}) = \sum_{n=x}^y \frac{(y!)}{n!(y-n)!} \times p(w)^n \times (1-p(w))^{(y-n)} $$
This as far as I can see does not simplify to a swiftly computable answer, especially for large $y$. I would have thought there is a more sophisticated way of doing it?
Side note: The complications for the combinations part involving factorials is that one has to use the sterling approximation in order to be able to compute it for large $y$.
Sterling Approximation
$$ \ln(n!) \approx n \ln(n) -n +\ln(\sqrt{2 \pi n}) $$
So we get;
$$ P(\text{win $\geq x$ throws}) = \sum_{n=x}^y exp( y\ln(y)+\ln(\sqrt{2\pi y})-n\ln(n)-\ln(\sqrt{2\pi n}) -(y-n)\ln(y-n)-\ln(\sqrt{2\pi (y-n)})) \times p(w)^n \times (1-p(w))^{(y-n)} $$
Simplifying
$$ P(\text{win $\geq x$ throws}) = \sum_{n=x}^y exp( y \ln(y)- \frac{1}{2} \ln(2) + \frac{1}{2} \ln(\pi y) - n \ln(n) - \frac{1}{2} \ln(\pi n) (n- y) \ln(y-n) - \frac{1}{2} \ln(\pi (y-n))) \times p(w)^n \times (1-p(w))^{(y-n)} $$
$$ P(\text{win $\geq x$ throws}) = \sum_{n=x}^y exp( y \ln(y)- \frac{1}{2} \ln(2) + \frac{1}{2} \ln(\pi y) - n \ln(n) - \frac{1}{2} \ln(\pi n) (n- y) \ln(y-n) - \frac{1}{2} \ln(\pi (y-n)) + n\ln(p(w)) + (y-n)\ln(1-p(w))) $$
(2) With the current form of (1) as far as I can see you are forced into numerical methods to solve what $p(w)$ must be for a given probability of winning $x$ or more throws out of $Y$, when I was looking for an analytical method.
Hint:
For large values of $Y(>30)$ you can approximate binomial distribution by the Gaussian distribution. See this. Consider each outcome of the dice roll be from Bernoulli distribution such that $P_{Win}=p(w)$ and $P_{NotWin}=p(d)+p(l)=1-p(w)$.