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