Similar questions:
Chance of 7 of a kind with 10 dice
Probability of getting exactly $k$ of a kind in $n$ rolls of $m$-sided dice, where $k\leq n/2$
Probability was never my thing, so please bear with me.
I've reviewed the threads above to the best of my ability, but I still wonder how to go about finding a match of 3 from 7 dice.
At least three match, but no more (two sets of three is okay, a set of three and a set of four is not):
(a) : $ \frac{6 \binom{7}{3} 5^4}{6^7} $
In the other discussions, this wasn't desired since it would allow for a second triple to occur, or even a quadruple. Odds of a quadruple with the remaining 4 dice:
(b) : $(1/5)^4 $
Then, the probability that from rolling 7 dice that there is at least three that match, and no more than three, would be:
(c) : $ \frac{6 \binom{7}{3} * 5^4}{6^7}- (1/5)^4 $
Exactly two sets of three:
(d) : $ \frac{6 \binom{7}{3} \binom{4}{3} \binom{1}{1}}{ 6^7} $
Maybe? My thought process was that if $\binom{7}{3}$ will give me a set of three, then with the remaining 4, I could pick 3 yielding $\binom{4}{3}$ with 1 leftover. I realize this is probably wrong. Why? What would be the proper way to go about this?
Exactly one set of three:
Then to find the probability that there is one and only one set of three from 7 dice, we could take the probability of one or more sets of three (c) and subtract the probability of exactly two sets (d), for:
$ \frac{6 \binom{7}{3} 5^4}{ 6^7} - (1/5)^4 - \frac{6 \binom{7}{3} \binom{4}{3} \binom{1}{1} }{ 6^7} $
(e) : $ \left(\frac{6 \binom{7}{3}}{ 6^7}\right) \left( 5^4 - \binom{4}{3} \binom{1}{1} \right) - (1/5)^4 $
Is this at all on the right path?
Thank you!
PS. Sorry about the syntax, but I couldn't figure out how to make the standard nCr() symbol with MathJaX.
\binom{n}{r}
. $\endgroup$