I think I have invented a formula that allows a computer to very easily calculate the probability of at least n1-consecutive die rolls, on an n2-sided die, rolling it n3-times.
For example, for a 3-sided die being rolled 4 times, the probability of at least 3 consecutive rolls being the same is:
$$\frac{5}{27} \sim 0.185185$$
A 10-sided die being rolled 20 times with at least 4 consecutive rolls:
$$\frac{153252438815221561}{10000000000000000000} \sim 0.0153252$$
A 20-sided die being rolled 100 times with at least 5 consecutive rolls:
$$\frac{36138486362801675395834082841530471263391618236217471764311872542282160082804618163213
4714483039586709049484138205953646876021}{63382530011411470074835160268800000000000000000
0000000000000000000000000000000000000000000000000000000000000000000000000000000000} \sim 0.000570165$$
A 150-sided die being rolled 250 times with at least 10 consecutive rolls:
$$\frac{43754862099840059340989164536890668843600275210242353790609200399332108157129005621344
12966072844123998821529817954285993344643635690087672932957210052124849484632371945364241
27895214917314522967829996314996884843354909465711479333655125328467972639354192054002381
80358736161798175079981214320161396998878382245814510025222948918658240716181935621089269
06271521762936897812401688121481273594338138312959838934408957524646299446591373165468391
26633170992252043228387167654509762247790434963321680468677569650750302475087706401}{7026
24848833633473725832814569816725466578833488064526319504046334823913293570611014402352480
20759777065059629450925139424048788112889589987529495486017499085597652471999291372698929
85667366792663663798677273390781336908763915319543616880317198383190582106072596957346831
91403746604919433593750000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
0000} \sim \text{6.2273433927754916$\grave{ }$*${}^{\wedge}$-18}$$
What I want to know, is there a method out there that already exists which is finding what I am already able to find? I am really scared that I have wasted my time 'inventing' something that someone has already done before as my literature review has come up empty. I am also a bit weary to share my method at the moment because I would ideally want to write a paper on this if this has not been done before.
Edit: You can generate tables with this easily too. For a 6-sided die for up to 15-rolls and -consecutive:
$$
\left(
\begin{array}{ccccccccccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & \frac{1}{6} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & \frac{11}{36} & \frac{1}{36} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & \frac{91}{216} & \frac{11}{216} & \frac{1}{216} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & \frac{671}{1296} & \frac{2}{27} & \frac{11}{1296} & \frac{1}{1296} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & \frac{4651}{7776} & \frac{751}{7776} & \frac{1}{81} & \frac{11}{7776} & \frac{1}{7776} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & \frac{31031}{46656} & \frac{5531}{46656} & \frac{7}{432} & \frac{1}{486} & \frac{11}{46656} & \frac{1}{46656} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & \frac{201811}{279936} & \frac{2177}{15552} & \frac{5611}{279936} & \frac{7}{2592} & \frac{1}{2916} & \frac{11}{279936} & \frac{1}{279936} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & \frac{1288991}{1679616} & \frac{270241}{1679616} & \frac{40091}{1679616} & \frac{13}{3888} & \frac{7}{15552} & \frac{1}{17496} & \frac{11}{1679616} & \frac{1}{1679616} & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & \frac{8124571}{10077696} & \frac{1827071}{10077696} & \frac{15497}{559872} & \frac{40171}{10077696} & \frac{13}{23328} & \frac{7}{93312} & \frac{1}{104976} & \frac{11}{10077696} & \frac{1}{10077696} & 0 & 0 & 0 & 0 & 0 \\
1 & \frac{50700551}{60466176} & \frac{126731}{629856} & \frac{979}{31104} & \frac{279851}{60466176} & \frac{31}{46656} & \frac{13}{139968} & \frac{7}{559872} & \frac{1}{629856} & \frac{11}{60466176} & \frac{1}{60466176} & 0 & 0 & 0 & 0 \\
1 & \frac{313968931}{362797056} & \frac{80043931}{362797056} & \frac{12790681}{362797056} & \frac{106217}{20155392} & \frac{279931}{362797056} & \frac{31}{279936} & \frac{13}{839808} & \frac{7}{3359232} & \frac{1}{3779136} & \frac{11}{362797056} &
\frac{1}{362797056} & 0 & 0 & 0 \\
1 & \frac{1932641711}{2176782336} & \frac{521516711}{2176782336} & \frac{84941711}{2176782336} & \frac{6619}{1119744} & \frac{1912811}{2176782336} & \frac{1}{7776} & \frac{31}{1679616} & \frac{13}{5038848} & \frac{7}{20155392} & \frac{1}{22674816} &
\frac{11}{2176782336} & \frac{1}{2176782336} & 0 & 0 \\
1 & \frac{11839990891}{13060694016} & \frac{561766711}{2176782336} & \frac{11638417}{272097792} & \frac{24761}{3779136} & \frac{715337}{725594112} & \frac{1912891}{13060694016} & \frac{1}{46656} & \frac{31}{10077696} & \frac{13}{30233088} &
\frac{7}{120932352} & \frac{1}{136048896} & \frac{11}{13060694016} & \frac{1}{13060694016} & 0 \\
1 & \frac{72260648471}{78364164096} & \frac{21637367221}{78364164096} & \frac{50620543}{1088391168} & \frac{563631721}{78364164096} & \frac{44059}{40310784} & \frac{12876971}{78364164096} & \frac{41}{1679616} & \frac{1}{279936} & \frac{31}{60466176} &
\frac{13}{181398528} & \frac{7}{725594112} & \frac{1}{816293376} & \frac{11}{78364164096} & \frac{1}{78364164096} \\
\end{array}
\right)
$$
Best Answer
As an actual research result, this isn't particularly interesting - there are many easy techniques that would be able to generate the result (recurrence relations, dynamic programming, etc). It ends up being a straightforward exercise.
The problem itself is not particularly difficult and therefore anyone who needs the result would generally be able to just derive it on the spot (as a calculation). On the other hand, it's not fundamental enough that people want the solution (to how to compute it) easily accessible all the time. Hence the lack of publication of this.
[Edit] If what you have is a semi-closed (closed) form, this would be a (really) good article for a combinatorics journal. If it's purely an algorithm that would generate it, that's not enough. If you write it well enough and sell the pedagogy/beauty of the derivation, this might still work for a recreational mathematics publication.