What is the probability of rolling a total of 7 with two dice at least once in 10 rolls?

1 Answer
Nov 27, 2017

P("at least one 7 in 10 rolls of 2 dice")~~83.85%P(at least one 7 in 10 rolls of 2 dice)83.85%

Explanation:

When rolling 2 dice there are 36 possible outcomes.
[to see this imagine one die is red and the other green; there are 6 possible outcomes for the red die and for each of these red outcomes there are 6 possible green outcomes].

Of the 36 possible outcomes 6 have a total of 7:
{color(red)1+color(green)6,color(red)2+color(green)5,color(red)3+color(green)4,color(red)4+color(green)3,color(red)5+color(green)2,color(red)6+color(green)1}{1+6,2+5,3+4,4+3,5+2,6+1}

That is 3030 out of 3636 outcomes will not be a 7 total.
3/36=5/6336=56

We will not get a total of 77 on the first roll 5/656 of the time.

Of the 5/656 of the time we did not get a 77 on the first roll,
we will not get a 77 on the second roll 5/656 of the time.
That is 5/6xx5/6=(5/6)^256×56=(56)2 of the time we will not get a total of 77 on either of he first two rolls.

Continuing with this reasoning, we see that we will not get a total of 77 on any of the first 1010 rolls (5/6)^10(56)10 of the time.

With the help of a calculator we find that we will not get a total of 77 on any of the first 1010 rolls approximately 16.15%16.15% of the time.

This implies that we will get a total of 77 on at least one of the first 1010 rolls 100%-16.15%=83.85%100%16.15%=83.85% of the time.