Question: What is the probability of rolling at least two 6's, when rolling 3 dice with two rolls? (with your first roll you keep dice only if they are 6's and roll the remainder for your second roll).

(6*6*6 = 216 = outcomes when rolling 3 dice) (6*6 = 36 = outcomes when rolling 2 dice)

1st roll: (3 dice)

A = 125/216 = 0 sixes

B = 75/216 = 1 six

C = 15/216 = 2 sixes

D = 1/216 = 3 sixes

Outcomes C and D fulfill requirement.

For outcome A: (first roll = no 6's, pick up all dice throw 3 dice again)

2nd roll:

a = 125/216 = 0 sixes

b = 75/216 = 1 six

c = 15/216 = 2 sixes

d = 1/216 = 3 sixes

Outcomes c and d fulfill requirements.

For outcome B: (first roll = one 6, pick up two non 6's and roll again)

2nd roll:

z = 25/36 = 0 sixes

y = 10/36 = 1 six

x = 1/36 = 2 sixes

Outcomes y and x fulfill requirements.

So would the formula below give me my answer?

C + D + Ac + Ad + By + Bx = X

What is the probability of rolling at least two 6's, when rolling 3 dice with two rolls?

If I substituted correctly and did the math correct the answer I got was 22.3%

Is this correct?

  • $\begingroup$ I'm guessing by your odds that you consider ordering important ? $\endgroup$
    – user451844
    Jul 29, 2017 at 22:20
  • $\begingroup$ I'm not sure if I know what you mean by ordering. $\endgroup$
    – Travis
    Jul 29, 2017 at 22:58
  • $\begingroup$ is (1,2,3) the same as (1,3,2) and (2,3,1) and (2,1,3) and (3,1,2) and (3,2,1) ? if so then it cuts down the number of distinct rolls down to 56. $\endgroup$
    – user451844
    Jul 29, 2017 at 23:00
  • $\begingroup$ Well let's say we had a blue green and a red die, if the red die shows a 1 and blue shows a 6 and green shows a 6, that would be different than if red showed a 6, green showed a 6, and blue showed a 1. So I guess ordering would matter. The numbers rolled are the same but show on different dice. Both rolls fulfill the two 6's requirement but are different rolls and should be counted as thus. I'm not sure if I'm correct on this, that's why I'm asking but does it makes sense? $\endgroup$
    – Travis
    Jul 29, 2017 at 23:09
  • $\begingroup$ yeah I'm just making sure, because the probability might change depending on conditions like this. of the 56 cases when order didn't matter a full 21 had at least one 6, but only 6 had more than one 6. etc. $\endgroup$
    – user451844
    Jul 29, 2017 at 23:19

2 Answers 2


An alternate approach would be to find the probability of the complementary event:

$\textbf{1)}$ The probability of getting no 6's is given by $\big(\frac{5}{6}\big)^3\cdot\big(\frac{5}{6}\big)^3=\big(\frac{5}{6}\big)^6$

$\textbf{2)}$ The probability of getting exactly one 6 is given by

$\hspace{.2 in}\big(\frac{5}{6}\big)^3\cdot3\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^2+3\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^2\cdot\big(\frac{5}{6}\big)^2=\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^5+\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^4$

Therefore the probability of getting at least two 6's is given by

$\hspace{.2 in} 1-\big(\frac{5}{6}\big)^6-\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^5-\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^4=\frac{5203}{23328}\approx.223$

  • $\begingroup$ From my point of view is your approach the most appropriate one. (+1) $\endgroup$ Jul 30, 2017 at 9:38

Your approach and your calculations are correct. Here is a variation based upon generating functions. We encode the roll of three dice with \begin{align*} (5+t)^3 \end{align*} marking an occurrence of $6$ with $t$ and collecting all other five possibilities with $5$. The probability to get $j$ sixes $0\leq j \leq 3$ in the first roll can be written as \begin{align*} [t^j](5+t)^3\cdot\frac{1}{6^3} \end{align*}

Encoding the second roll with $(5+u)^{3-j}$ we calculate \begin{align*} \sum_{{0\leq j,k\leq 3}\atop{j+k\geq 2}}&[t^j](5+t)^3[u^k](5+u)^{3-j}\cdot\frac{1}{6^{6-j}}\\ &=[t^0](5+t)^3\left([u^2]+[u^3]\right)(5+u)^3\cdot\frac{1}{6^6}\\ &\qquad+[t^1](5+t)^3\left([u^1]+[u^2]\right)(5+u)^2\cdot\frac{1}{6^5}\\ &\qquad+[t^2](5+t)^3\left([u^0]+[u^1]\right)(5+u)^1\cdot\frac{1}{6^4}\\ &\qquad+[t^3](5+t)^3\\ &=\binom{3}{0}5^3\left(\binom{3}{2}5^1+\binom{3}{3}5^0\right)\cdot\frac{1}{6^6}\\ &\qquad+\binom{3}{1}5^2\left(\binom{2}{1}5^1+\binom{2}{2}5^0\right)\cdot\frac{1}{6^5}\\ &\qquad+\binom{3}{2}5^1\left(\binom{1}{0}5^1+\binom{1}{1}5^0\right)\cdot\frac{1}{6^4}\\ &\qquad+\binom{3}{3}5^0\cdot\frac{1}{6^3}\\ &=\frac{125\cdot16}{6^6}+\frac{75\cdot11}{6^5}+\frac{15\cdot6}{6^4}+\frac{1}{6^3}\\ &=\frac{5\,203}{23\,328}\\ &\doteq 0.223 \end{align*}


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