IF I roll 10 dice, what is the probability that at least 5 of them are the same value? I understand that the probability of rolling 10 dice and getting the same number on each is 1/6^9 power which is 1/10,077,696.
SO what is the probability of me getting at least 5 dice of the same value if I roll 10, how do I account for the extra dice rolled that I don't really care the values of but still have a chance of helping me get the 5 out of 10?
Thanks!
 A: The probability of getting exactly five ones and the remaining five as other different numbers would be
$$\binom{10}{5}\left(\frac{1}{6}\right)^5\left(\frac{5}{6}\right)^5$$
Multiplying by $6$ would almost have given us the probability of having some number that we have exactly five of, however it overcounts the cases where we have two numbers, each of which occurring five times.
The probability then that we have some number that we have exactly five of us then $$6\binom{10}{5}\left(\frac16\right)^5\left(\frac56\right)^5-\binom{6}{2}\binom{10}{5}\left(\frac16\right)^{10}$$
If we are interested also in the cases where we have some number occurring more than five times, we would also want to add $6\binom{10}{6}\left(\frac{1}{6}\right)^6\left(\frac{5}{6}\right)^4,6\binom{10}{7}\left(\frac{1}{6}\right)^7\left(\frac{5}{6}\right)^3,\dots$ to account for the cases of having exactly six, exactly seven, etc... of a number appearing (noting that it is impossible to have strictly more than five of one number and five or more of another number simultaneously).
Combining the above into a more compact form, we have as a final probability:
$$\sum\limits_{k=5}^{10}\left[6\binom{10}{k}\left(\frac16\right)^k\left(\frac56\right)^{10-k}\right]-\binom{6}{2}\binom{10}{5}\left(\frac16\right)^{10}$$

Related reading: Binomial Distribution on Wikipedia.
A: For $i=1,\dots,6$ let $E_i$ denote the event that at least $5$ of the $10$ dice show value $i$.
Then you are looking for $$P(E_1\cup E_2\cup E_3\cup E_4\cup E_5\cup E_6)$$
With inclusion/exclusion and symmetry we find that this equals:$$6P(E_1)-\binom62P(E_1\cap E_2)$$
Can you find $P(E_1)$ and $P(E_1\cap E_2)$ yourself?
