I was at the store the other day and saw a game called Rubik's Race. The game consist of: two 5x5 sliding puzzle boards each populated with 4 tiles of each of the 6 colors of a Rubik's cube, 9 six sided dice each side has a color (I am uncertain of the distribution of the colors), and a three by three Boggle style dice shaker. To play the dice are loaded into the shaker and shook. When the dice settle the the players race to organize the center nine tiles of their sliding puzzle to match the layout in the shaker. The rules indicate that if the dice land in such a way that more than four of a single color is present they should be shook up again.
Supposing that each die has all six colors of the Rubik's cube how do you find the probability of getting an unplayable layout?
It has been a long time since I have done any meaningful probability and I was thinking that it was something like: $$1 - 6 \sum_{i=1}^4 {9 \choose i} \frac 16^i \frac 56^{9-i}$$ but that is not right at all.
It is easier to calculate the chance to get too many squares of a color, as you can't have 5 of each of two colors. To get exactly $i$ squares of a given color, the chance is ${9 \choose i}(\frac 16)^{i}(\frac 56)^{9-i}$. There are six different colors, so the chance of an unplayable layout is $6\sum_{i=5}^9{9 \choose i}(\frac 16)^{i}(\frac 56)^{9-i}=\sum_{i=5}^9{9 \choose i}\frac {5^{9-i}}{6^8}$
Wolfram Alpha gives this as $\frac{22549}{419904}\approx 5.37\%$