# probability or rolling 1 1 2 with six dice

What is the probability of rolling at least two ones and at least one two (order not important) with six 6-sided die?

Equivalently, to win a game, I need two 1s and a 2, (1,1,2). What is the probability of my rolling at least this with 6 dice.

There are $6^6=46656$ possible rolls. I believe that there are 6896 'winning' rolls among these as I wrote some code to check every roll. This makes the probability around 0.1478

I'm not sure how to get this without the help of a computer though.

• What have you tried? Also, since you have six dice, is unclear whether, for instance, $(1,1,2,3,1,2)$ is valid. Is it valid? – Arthur Jun 18 '17 at 20:29
• yes, 1,1,2,x,x,x where x could be any value – mjmdavis Jun 18 '17 at 20:38
• Common practice is to say at least two 1's and one 2 in your case and exactly " ... " otherwise. – infinitylord Jun 18 '17 at 20:44
• @infinitylord like that? – mjmdavis Jun 18 '17 at 20:48
• If you want positive responses, I recommend stating the question as: "What is the probability of rolling at least two ones and at least one two (order not important) with six 6-sided die" and then (and this part is really huge for this site) show your attempt at the problem. – infinitylord Jun 18 '17 at 21:01

Approach via inclusion-exclusion and De Morgan's laws.

Let $A$ be the event that you roll at least two $1$'s. Let $B$ be the event that you roll at least one $2$.

We are attempting to calculate then $Pr(A\cap B)$

$Pr(A\cap B)=1-Pr((A\cap B)^c)=1-Pr(A^c\cup B^c) = 1-Pr(A^c)-Pr(B^c)+Pr(A^c\cap B^c)$

We calculate each term on the right now that they are in simpler form.

$Pr(A^c)$ refers to the probability of strictly fewer than two $1$'s occurring, i.e. at most one $1$ occurring. Either exactly one $1$ occurs or no $1$'s occur. In the case of exactly one $1$, pick its location and then pick each remaining digit for a total of $6\cdot 5^5$ possibilities. In the case of no $1$'s, pick each digit for a total of $5^6$ possibilities. Taking the ratio of this compared to the $6^6$ equally likely dice rolls, we calculate $Pr(A^c)=\frac{6\cdot 5^5+5^6}{6^6}$.

$Pr(B^c)$ refers to the probability of strictly fewer than one $2$ occurring, i.e. no $2$'s. This occurs with probability $Pr(B^c)=\frac{5^6}{6^6}$

$Pr(A^c\cap B^c)$ refers now to the probability of at most one $1$ and no $2$'s, which similarly to before we break into cases for either exactly one $1$ or no $1$'s. For exactly one $1$, first pick the location, then pick each remaining digit for $6\cdot 4^5$ possibilities and for no $1$'s pick each digit for $4^6$ possibilities for a probability of $Pr(A^c\cap B^c)=\frac{6\cdot 4^5+4^6}{6^6}$

This gives a final probability of:

$$Pr(A\cap B) = 1-\frac{6\cdot 5^5+5^6}{6^6}-\frac{5^6}{6^6}+\frac{6\cdot 4^5+4^6}{6^6}$$

$$=1-\frac{39760}{46656} = \frac{431}{2916}\approx 0.147805$$

You can consider the possibility to find probability directly without inclusion-exclusion principle. This is a longer way, but there are those who like it. Consider all possibilities to roll at least two ones and at least one two with six 6-sided die. We can have exactly two $1$ and one $2$. Denote this event by $A_{21}$. Or we can have exactly two $1$ and two $2$: event $A_{22}$. There are $10$ disjoint events total: $A_{21}$, $A_{22}$, $A_{23}$, $A_{24}$, $A_{31}$, $A_{32}$, $A_{33}$, $A_{41}$, $A_{42}$, $A_{51}$.

Find and then add their probabilities. The probability of $A_{21}$ is $$\mathbb P(A_{21})=\frac{\binom{6}{2}\binom{6-2}{1}4^{6-2-1}}{6^6}=\frac{15\cdot 4\cdot 4^3}{6^6}.$$ By first multiplier $\binom{6}{2}=15$ we choose the rolls which result in $1$, next multiplier $\binom{6-2}{1}=4$ we choose the roll among the rest which results in $2$, and the last multiplier $4^3$ is number of possible outcomes in the rest $6-2-1=3$ rolls.

Similar, $$\mathbb P(A_{22})=\frac{\binom{6}{2}\binom{4}{2}4^{2}}{6^6}=\frac{15\cdot 6\cdot 16}{6^6},$$ $$\mathbb P(A_{23})=\frac{15\cdot 4\cdot 4}{6^6},\quad \mathbb P(A_{24})=\frac{15}{6^6},\quad \mathbb P(A_{31})=\frac{20\cdot 3\cdot 16}{6^6},$$ $$\mathbb P(A_{32})=\frac{20\cdot 3\cdot 4}{6^6}, \quad \mathbb P(A_{33})=\frac{20}{6^6}, \quad \mathbb P(A_{41})=\frac{15\cdot 2\cdot 4}{6^6},$$ $$\mathbb P(A_{42})=\frac{15}{6^6}, \quad \mathbb P(A_{51})=\frac{6}{6^6}.$$ The total probability of rolling at least two ones and at least one two with six 6-sided die is equal to the sum of probabilities above: $$\frac{6896}{6^6}=0,147805213.$$

As I understood your question, you are asking about probability that between $6$ values will be at least two $1$'s and one $2$ (don't looking at other values and order).

Well, so the probability of $(1,1,2,x,x,x)$ (excluding order and remarking that there may be other $1$'s and $2$'s between $x$'s) is exactly the number of sets of six digits from $1$ to $6$ that start with $1,1,2$ divided by total number of sets of six digits from $1$ to $6$. There are as many (disordered) sets of six digits from $1$ to $6$ that start with $1,1,2$ as just (disordered) sets of three digits from $1$ to $6$.

Thus the answer is $P = \frac{(3^3)/3!}{(6^6)/6!} = \frac{5}{72}$.

• Hasek I'm not sure if this result is correct. – mjmdavis Jun 18 '17 at 23:07
• The line "There are as many (disordered) sets of six digits from 1 to 6 that start with 1,1,2 as just (disordered) sets of three digits from 1 to 6." is incorrect. After all, what does it mean for a "disordered set to start with something"? Further, your calculations don't appear to make much sense. What would $3^3/3!$, your supposed numerator, represent? Notice that $3!$ is even but $3^3$ is odd and so $3^3/3!$ is not an integer, so it could not possibly represent the answer to a counting problem. Similarly for the denominator. – JMoravitz Jun 18 '17 at 23:28
• Yes, I'm definitely wrong, sorry. – Hasek Jun 19 '17 at 6:44