Probability of rolling a 10 and an 11 before rolling a 7?

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Question: What is the probability of rolling a 10 and an 11 before rolling a 7? Solution: As stated, must roll a 10 and 11 before rolling a 7 P(10)=3/36 P(11)=2/36 P(7)= 6/36 Two options: a) 10 first or b) 11 first a) P that 10 is rolled before 7 is rolled: P(10)/P(7)= 3/6 then 11 rolled before 7 is rolled: P(11)/P(7)= 2/6 These are two sequential, independent events: Therefore: P(10 then 11 before 7)= 3/6*2/6=6/36 b) P that 11 is rolled before 7 is rolled: P(11)/P(7)= 2/6 then 10 rolled before 7 is rolled: P(10)/P(7)= 3/6 Therefore P(11 then 10 before 7)= 2/6*3/6=6/36 Sum a) plus b) = 12/36 = 33.33% Don't believe it? Get out the dice. Record 10-7, 10-11-7, 11-7, 11-10-7 and 7-7 and ignore all other rolls. Number of sequences of 10-11-7 plus 11-10-7 should be ~1/3 of 7-7 POST YOUR OPINION IF YOU HAVE BETTER SOLUTION

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Question:
What is the probability of rolling a 10 and an 11 before rolling a 7?

Solution:
As stated, must roll a 10 and 11 before rolling a 7

P(10)=3/36 P(11)=2/36 P(7)= 6/36
Two options: a) 10 first or b) 11 first
a) P that 10 is rolled before 7 is rolled:
P(10)/P(7)= 3/6
then 11 rolled before 7 is rolled:
P(11)/P(7)= 2/6

These are two sequential, independent events:
Therefore: P(10 then 11 before 7)= 3/6*2/6=6/36
b) P that 11 is rolled before 7 is rolled:
P(11)/P(7)= 2/6
then 10 rolled before 7 is rolled:
P(10)/P(7)= 3/6
Therefore P(11 then 10 before 7)= 2/6*3/6=6/36

Sum a) plus b) = 12/36 = 33.33%

Don't believe it? Get out the dice. Record 10-7, 10-11-7, 11-7, 11-10-7 and 7-7 and ignore all other rolls. Number of sequences of 10-11-7 plus 11-10-7 should be ~1/3 of 7-7

POST YOUR OPINION IF YOU HAVE BETTER SOLUTION