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

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- May 9th 2010, 05:26 AMsyke2Another dice question!
You have two dice What is the probability of rolling a "3" total before rolling a "7" total?

- May 9th 2010, 06:50 AMArchie Meade
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. - May 9th 2010, 07:19 AMsyke2
Awesome! Thank you very much!