I'm trying to figure out what the probability of rolling a dice and getting exactly two of the same numbers after we throw the dice $n$ times is?
Exactly two of a specified number, or exactly two of any number?
Case 1 : The probability of getting exactly two sixes. $n\in \{2,3,....\}$
$$\mathsf P(T_6) = {n\choose 2} 5^{n-1}6^{-n} = \frac{5^{n-2}n(n-1)}{2\cdot 6^n}$$
Similarly, $\mathsf P(T_1)=\mathsf P(T_2) = \cdots = \mathsf P(T_6)$
Case 2 : The probability of getting any number exactly twice. $n\in \{2,3, ...\}$
We need a rather complicated inclusion and exclusion, depending on the size of $n$.
$\begin{align}
\mathsf P(\bigcup_{k=1}^6 T_k) & = \sum_{k=1}^6\mathsf P(T_k) - \sum_{k=1}^5\sum_{h=k+1}^6\mathsf P(T_k\cap T_h) + \sum_{k=1}^4\sum_{h=k+1}^5\sum_{j=h+1}^6 \mathsf P(T_k\cap T_h\cap T_j) - \cdots
\\[4ex] & =
6 {n\choose 2, n-2}\frac{5^{n-2}}{6^n} \operatorname{\bf 1}_{\{2,..,\infty\}}(n)
- 15 {n\choose 2,2,n-4}\frac{4^{n-4}}{6^n} \operatorname{\bf 1}_{\{4,..,\infty\}}(n)
+ 20{n\choose 2,2,2,n-6}\frac{3^{n-6}}{6^n} \operatorname{\bf 1}_{\{6,..,\infty\}}(n)
- 15 {n\choose 2,2,2,2,n-8}\frac{2^{n-8}}{6^n} \operatorname{\bf 1}_{\{8,..,\infty\}}(n)
+ 6{n\choose 2,2,2,2,2,n-10}\frac 1{6^n} \operatorname{\bf 1}_{\{10,..,\infty\}}(n)
- {n\choose 2,2,2,2,2,2}\frac 1{6^n} \operatorname{\bf 1}_{\{12\}}(n)
\\[4ex] & =
\frac{6\cdot 5^{n-2}n(n-1)}{2\cdot 6^n} \operatorname{\bf 1}_{\{2,..,\infty\}}(n)
- \frac{15\cdot 4^{n-4}n(n-1)(n-2)(n-3)}{4\cdot 6^n} \operatorname{\bf 1}_{\{4,..,\infty\}}(n)
+ \frac{20\cdot 3^{n-6}n(n-1)(n-2)\cdots(n-5)}{8\cdot 6^n} \operatorname{\bf 1}_{\{6,..,\infty\}}(n)
- \frac{15\cdot 2^{n-8}n(n-1)\cdots(n-7)}{16\cdot 6^n} \operatorname{\bf 1}_{\{8,..,\infty\}}(n)
+ \frac {6\cdot n(n-1)\cdots(n-9)}{32\cdot 6^n} \operatorname{\bf 1}_{\{10,..,\infty\}}(n)
- \frac {12!}{64\cdot 6^{12}} \operatorname{\bf 1}_{\{12\}}(n)
\end{align}$
Assuming that the results of each dart throw are independent*, then the probability of hitting double-double-treble 20, in that order, is indeed simply the product of the individual probabilities that you’ve derived from your simulation. I would argue that the resulting very small value does in fact match your experience: you’ve computed the probability of hitting that particular combination while aiming for the bullseye. The relevant real-life experience to which you should compare this isn’t the overall frequency with which you can land this particular combination, but instead the number of times that you’ve gotten it accidentally while trying to shoot bulls. I’ll go out on a limb here and say that’s never happened. The first two darts both have to land in a fairly small area that’s half a board width away from your aiming point.
Note, too, that your simulated probability of a treble 20 is almost three times that of a double 20. This certainly makes sense for Gaussian scatter from the center of the board, but is backwards from what one might expect when actually trying to hit those regions. Cetera paribus the probability of a treble 20 should be smaller since it covers a significantly smaller area than the double 20.
* Assuming independence for such small regions of the dart board doesn’t seem like a good approximation to me. Each dart that lands significantly reduces in interesting ways the available area for the next one. There’s not a lot of leeway for three darts in any of the treble zones, for instance.
Best Answer
If by "the exact same spot" you mean a point, the chances are zero even with a countable infinity of throws, as there are uncountably many points. Any countable set of points has no area. If you mean within a small error, take the ratio of the area to the area of the dartboard (presumably you are assuming a uniform distribution over the dartboard and nowhere else) and that is the chance on the next throw. Of course, those who can hit the dartboard every time don't have a uniform distribution of throws over it, so this is unrealistic.