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