Yet another probability question

[yzb25]

Allow me to toss a question I've wondered about for a while into the plague of recent probability questions on this forum.

There's a magical die. Every single time you get a side, one of the dots gets rubbed off on that side. So, for example, if you're rolling the die for the first time and you roll a 5, that 5 will turn into a 4. And then your die will have 2 4s, 0 5s, and 1 of every other number. Then you roll a 1, the 1 magically turns into a 6, cuz that's just the way of the world. Then you have 2 6s, 2 4s, 0 5s, 0 1s, 1 2s and 1 3s.

So we have a pretty little cycle here. If you roll any side 6 times, you end up back where you started, as if nuthin' ever happened! For example, roll a 5 once and it turns into a 4. Then it turns into a 3, then a 2, then a 1, then a 6, then a 5 again!

Anyway, the question is this: If I roll the die 99 times, what is the probability every single side of the die will have the same number on it?

A friend proposed this to me a year ago. I've spent many hours on it, but I've honestly only made minimal progress - I'm simply too dumb to finish it. I've gotten an answer for 15, 21 and 27, and developed a basic understanding of the summations you'd have to do to get 99. 99 is crazy tho.

Anyway, no fancy computers (raw decimal form is overrated), and no looking shit up, or a vagina will grow from your penis and a penis will grow from your vagina. If you can't get the answer into closed form, just leave it with a shittonne of summa signs. It's nice to see how far you can get all by yourself.

P.S. The die is unbias :P

[SuperJack]

You use the word die.
How amusing.

[Yukitaka Oni]

You use the word die.
How amusing.

v)x.x)v

[oops_ur_dead]

All you need to do is enumerate all the solutions of 6a + 5b + 4c + 3d + 2e + f = 99 where b, c, d, e, and f are > 1. Then multiply the number of solutions by 6, and divide it by 6^99.

I don't think there's a way to do that without using a computer program.

[yzb25]

All you need to do is enumerate all the solutions of 6a + 5b + 4c + 3d + 2e + f = 99 where b, c, d, e, and f are > 1. Then multiply the number of solutions by 6, and divide it by 6^99.

I don't think there's a way to do that without using a computer program.

yeah, you need a 012345 (0 3s, 1 2s, 2 1s, 3 6s, exc. for example), and then, of the remaining throws, the remaining throws need to come in multiples of sixes or "one of each"s. (say you're on 93 and every side is a 3, if you get every single side, then every side will just be a 4. You'll still get a win) And not necessarily in this order.

Of course, you can program a computer/graphic calculator to sum up all the permutations, but that ruins the fun :3