Chance of $7$ of a kind with $10$ dice

Solution 1:

If you want exactly $7$ of a kind, you use binomial distribution to first find the probability that you roll $7$ of the same "good" number in $10$ trials. The binomial distribution shows that this probability is $$ \binom{10}{7}\left(\frac{1}{6}\right)^7\left(\frac{5}{6}\right)^3, $$ since the probability of success is $1/6$.

Then there are six numbers to choose one from which could be the "good" kind of which you want $7$. Hence you multiply the above expression by $\binom{6}{1}$: $$ \binom{6}{1}\binom{10}{7}\left(\frac{1}{6}\right)^7\left(\frac{5}{6}\right)^3. $$

Note that this gives the exact probability $\frac{625}{419904}\approx\frac{1}{672}$, consistent with the approximate value given by Wolframalpha.

Solution 2:

To be short: The result is $p\approx \frac{1}{672}$

You can use the second formula in the Wikipedia article and sum the probabilities for every face of 1-6: http://en.wikipedia.org/wiki/Liar%27s_dice#Basic_dice_odds

Note that Wolframalpha is also able to get this: http://www.wolframalpha.com/input/?i=10+dice (Press on more under probabilities)

Solution 3:

Calculate the probability of exactly 7 sixes.
Calculate the probability of exactly 8 sixes.
Calculate the probability of exactly 9 sixes.
Calculate the probability of exactly 10 sixes.
Add them all up, and multiply by 6.