Throw 10 dice, probability of getting 6 identical numbers?

Question

I got this result when I threw a set of 10 dice (btw. for the first with this set):

What is a probability of getting 6 identical numbers (any) with 10 dice, with no particular order?

Answer 1

If you count only throws that have exactly 6 identical dice, the probability is exactly $\frac{787500}{6^{10}}$, which is around 1.3%, broken down as follows:

    Pattern     Rolls out of 6^10    Percent chance

    AAAAAABBBB    6300               0.0104      
    AAAAAABBBC  100800               0.167   
    AAAAAABBCC   75600               0.125   
    AAAAAABBCD  453600               0.75    
    AAAAAABCDE  151200               0.25   

The left column is the pattern of numbers in the roll. The pattern depicted in your question, with 555555 33 44, would be AAAAAABBCC.

The middle column counts the number of ways of rolling each pattern out of $6^{10}$ possible different rolls. The right-hand column is the percentage of total rolls that show each pattern: it's $100\cdot m\cdot 6^{-10}$, where $m$ is the middle column.

If you also want to count rolls that have 7 to 10 identical dice, add $\frac{97056}{6^{10}} \approx 0.16\%$ for a total of exactly $\frac{884556}{6^{10}}\approx 1.46\%$; these rolls are exceedingly rare.

This article explains how I calculated these probabilities, and there is a form at the bottom of the page that invokes a program that tabulates them. The complete tabulation for 10 dice is here.

Answer 2

You want to see 6 "successes" out of 10 trials, where the probability of one success is 1/6. For a fixed number, that likelihood is provided by using the Binomial distribution: $$P(X=6) = { 10 \choose 6}\left( \frac{1}{6} \right)^6 \left( 1- \frac{1}{6} \right)^{10-6} = \frac{10!}{6! 4!}\left( \frac{1}{6} \right)^6 \left(\frac{5}{6} \right)^{4}=0.002171$$ Since you actually asked about any dice, we have to multiply the answer by 6, thus getting $0.013 = 1.3\%$.

Answer 3

We count the total number of favorable outcomes,i.e., outcomes where there are six dies that land in the same number, and divide by the total number of outcomes.

EDIT: As Hurkyl correctly pointed out, I was assuming at least 6 equal dice. Correction (multiplied by $5^4$ ):

You have 6 possible equal values that the dice can take; {$1,2,3,4,5,6$}, and you have $10P6$ ways of selecting the dice that will take the equal value. But if you want exactly $6$ equal dice, then each of the remaining four dice can have 5 different value, so you multiply by $5^4$ . And there $6^{10}$ total possible outcomes, since each die can show any of the numbers in {$1,2,3,4,5,6$}. Then, putting it together, we get:

$\frac{6{ 10 \choose 6}5^4}{6^{10}}$