Consider a lottery where $5$ balls are chosen randomly among $90$ balls numbered from $1$ to $90$. A man cheats adding to the $90$ balls, before the draw, three more balls numbered $1,2,3$. We say that he will get busted if among the five balls chosen, at least two have the same numbers. We say that he wins if at least three drawn balls have numbers $1,2,3$.
1) What is the probability he will get busted?
2) What is the probability he will win?
Considering ordered $5$-tuples and after a lot of counting, for $1)$ I obtain $$ \dfrac{5}{713} - \dfrac{1}{194649} = \dfrac{44}{6279}$$ and for $2)$ $$ \dfrac{3442}{5774587}.$$ I am especially dubious regarding point $2)$. Is there a clever and faster way to solve these?
Count by considering the originals and their counterfeits to be distinct in fact though not appearance.
Busted: Draw at least one of $\{1,2,3\}$ twice in 5 balls out of 93.
There are ${93 \choose 5}$ ways to choose 5 balls in 93, in total.
We need one of the pair of balls to turn up. There are ${3 \choose 1}$ choices of pairs and ${91\choose 3}$ equi-probable ways to select the rest.
Also, we need to exclude the case of 2 pairs turning up. There are ${3 \choose 2}$ choices of pairs and ${89 \choose 1}$ equi-probable ways to choose the rest.
$$\mathcal{Pr}(\text{busted})= \frac{{3\choose 1}{91 \choose 3}-{3\choose 2}{89\choose 1}}{93 \choose 5} = \frac{44}{6279} \approx 0.007007\dotsc $$
"Win": Draw all of $\{1,2,3\}$ exactly once in 5 balls out of 93.
There are $2^3$ ways to draw a 'win' and $87\choose 2$ ways to draw the rest (that don't include doubles).$$\mathcal{Pr}(\text{win})=\dfrac{2^3{87\choose 2}}{93\choose 5} = \dfrac{9976}{17323761} \approx 0.000576\dotsc$$
Edit: Well, that is unless you can 'win' while 'busted'; which does appear to be the case.
If so, we have to include the cases where 1 or 2 of the winning numbers turn up; which means the numbers can be 3,4,or 5 of the selected 5. $$\frac{{6\choose 3}{87\choose 2}+{6\choose 4}{87\choose 1}+ {6\choose 5}}{93\choose 5}=\frac{8459}{5774587}\approx 0.001465\dotsc$$
Edit 2: When the win condition is read as "each of 1,2,3 to appear at least once".
$$\frac{2^3{87\choose 2} + {3\choose 1}2^2{87\choose 1} + {3\choose 2}2^1}{93\choose 5}=\frac{3442}{5774587}\approx 0.0005961\dotsc$$