Probability of 3 of a kind with 7 dice

Question

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.

Answer 1

You are on the right track with $6^7 = 279936$ as the denominator. To find how many cases have three but no more matching, I would start by looking at the four partitions of 7 into up to 6 parts where the largest is 3: 3+3+1, 3+2+2, 3+2+1+1, 3+1+1+1+1. You can then work out each systematically, taking account both the numbers that appear and the order they appear in.

I think the first (which has two sets of three) is $$\frac{6!}{2!\;1!\;3!} \times \frac{7!}{3!\;3!\;1!} = 8400$$ which is ten times what you have in (c). The others (just one three-of-a-kind) are $$\frac{6!}{1!\;2!\;3!} \times \frac{7!}{3!\;2!\;2!} + \frac{6!}{1!\;1!\;2!\;2!} \times \frac{7!}{3!\;2!\;1!\;1!} + \frac{6!}{1!\;4!\;1!} \times \frac{7!}{3!\;1!\;1!\;1!\;1!} = 113400$$.

So I get about 0.405 for the probability of exactly one three-of-a-kind (but no four or more) and about 0.435 for the probability of the one or more threes-of-a-kind (but no four or more)

Answer 2

Carrying around the $6^7$ is just complicating your life. Instead, just count how many distinct rolls have exactly one triple; the end probability will be that count, divided by the total number of possible rolls (namely, $6^7$).

There is also the issue of distinguishable and non-distinguishable dice. But let's assume you roll a single dice in sequence and write down the results.

Proceeding along the lines you have, and hoping I'm not making a mistake:

1. At least 3-of-a-kind of x, with the other four dice being anything except x. This one you compute correctly: pick what is x (six possibilities), then pick which three dice fall come up x ($\binom{7}{3}$ ways), then choose any of five possible outcomes for each of the remaining four dice, $5^4$. So you have $\binom{6}{1}\binom{7}{3}5^4$.

   Edit. However, this overcounts! (Thanks to Henry for spotting it) If a roll has two different three-of-a-kind, then I count that roll twice; once when I'm counting the lower three-of-a-kind, and again when I'm counting the higher one. So we need to subtract those rolls that have two three-of-a-kinds. We select the two ranks by using $\binom{6}{2}$, then select the dice that show the lower rank with $\binom{7}{3}$, and then select the dice, out of the remaining $4$, that have the higher rank three-of-a-kind with $\binom{4}{3}$. Then the remaining die can be any of the remaining $4$ ranks; so we overcounted by $\binom{6}{2}\binom{7}{3}\binom{4}{3}4$. So the real number of rolls that have at least one three-of-a-kind is $$\binom{6}{1}\binom{7}{3}5^4 - \binom{6}{2}\binom{7}{3}\binom{4}{3}4.$$

2. Discard the case in which we get 3-of-a-kind of x, but the other four dice are equal to y, with x$\neq$ y. There are $\binom{6}{1}$ ways of picking x, and then you need to pick what the other $4$ are; there are $\binom{5}{1}$ ways of doing it; then pick the three dice that will come up x, $\binom{7}{3}$ ways. So you have $\binom{6}{1}\binom{5}{1}\binom{7}{3}$ outcomes that you don't want.

3. Discard the case in which you get two sets of 3-of-a-kind. We already figured out, when we computed the overcount, that there are $$4\binom{6}{2}\binom{7}{3}\binom{4}{3}$$ rolls with two 3-of-a-kind. So we must subtract this total again.

So: the total number of rolls in which the largest number of dice that come up the same is $3$, and there is one and only one set that does this, is: $$\binom{6}{1}\binom{7}{3}5^4 - \binom{6}{1}\binom{7}{3}\binom{5}{1} - 2\times\left(\binom{6}{2}\binom{7}{3}\binom{4}{3}4\right) = 113400.$$

Given that there are $6^7$ sequential rolls possible, you get a probability of $$\frac{\mbox{number of rolls with exactly one three-of-a-kind, no 4-of-a-kind}}{6^7} = \frac{113400}{6^7} \approx 0.4050925926.$$