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
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
In brackets is an infinite geometric series...
for which the sum is
hence the probability is
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.