**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

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

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