You have two dice What is the probability of rolling a "3" total before rolling a "7" total?

Hi syke2,

presuming you mean the probability of throwing the 2 dice, repeatedly if necessary,

and you get a total of 3 before ever getting a total of 7, then

there are 2 ways to get a total of 3

1,2 and 2,1

there are 6 ways to get 7

1,6 and 6,1

2,5 and 5,2

3,4 and 4,3

Hence the probability of getting neither a total of 3 or 7 is \(\displaystyle \frac{36-8}{36}=\frac{28}{36}\)

Hence the probability of getting a total of 3 before a total of 7

(not a total of 3 followed immediately by a total of 7) is

a total of 3 on the first 2 throws

or neither total on the first 2 throws and a total of 3 on the next 2

or neither total on the first 4 throws and a total of 3 on the next 2

etc...

This probability is

\(\displaystyle \frac{2}{36}+\left(\frac{28}{36}\right)\frac{2}{36}+\left(\frac{28}{36}\right)^2\frac{2}{36}+\left(\frac{28}{36}\right)^3\frac{2}{36}+........\)

\(\displaystyle =\frac{2}{36}\left(1+\frac{28}{36}+\left(\frac{28}{36}\right)^2+\left(\frac{28}{36}\right)^3+.....\right)\)

In brackets is an infinite geometric series... \(\displaystyle a=1,\ r=\frac{28}{36}\)

for which the sum is \(\displaystyle \frac{a}{1-r}=\frac{1}{1-\frac{28}{36}}=\frac{1}{\frac{36-28}{36}}=\frac{1}{\left(\frac{8}{36}\right)}=\frac{36}{8}\)

hence the probability is

\(\displaystyle \frac{2}{36}\left(\frac{36}{8}\right)=\frac{1}{4}\)

Or...

there are 8 chances of getting a total of 3 or 7,

there are 2 chances of getting a total of 3.

Hence if we got a total of 3 rather than a total of 7, that's 2 chances in 8.