# dice

• Jan 30th 2010, 01:34 AM
stressedout
dice
suppose that 2 balanced dice are tossed repeatedly and the sum of the 2 uppermost faces is determined on each toss. what is the porbabilty that we obtain a sum of 3 before we obtain a sum of 7.

my working:

since obtain 3 and 7 are independent,
P( getting sum of 3) = 2/6 x 1/6
P ( getings sum of 7) = 3/6 x 1/6

so the answer should be 2/6 x 1/6 x 2/6 x 1/6

but my book says taht it is 1/4...
• Jan 30th 2010, 02:20 AM
h2osprey
Quote:

Originally Posted by stressedout
suppose that 2 balanced dice are tossed repeatedly and the sum of the 2 uppermost faces is determined on each toss. what is the porbabilty that we obtain a sum of 3 before we obtain a sum of 7.

my working:

since obtain 3 and 7 are independent,
P( getting sum of 3) = 2/6 x 1/6
P ( getings sum of 7) = 3/6 x 1/6

so the answer should be 2/6 x 1/6 x 2/6 x 1/6

but my book says taht it is 1/4...

This one's actually a little tricky. What you seem to be calculating is the probability that you'll get a sum of 3 on the first throw, and a sum of 7 on the second throw (or vice versa). What you actually want is the probability that you attain a sum of 3 before a sum of 7.

To begin with, determine the respective probabilities:

$P$(sum of 3 on any roll of the dice) $= \frac{2}{36}$

$P$(sum of 7 on any roll of the dice) $= \frac{6}{36}$

(You should be able to obtain these easily by counting the outcomes).

Now, we want the probability of attaining sum of 3 before sum of 7. This is equivalent to the probability that, when you obtain the first sum of 3, your previous rolls have not had sums of 7 (or 3, obviously). In this sense we have what resembles a geometric random variable:

$P$(first sum of 3 obtained on the $x^{th}$ roll, without obtaining any sums of 7 before the $x^{th}$ roll) $= (\frac{28}{36})^{x-1}\frac{2}{36}$.

This is of course for only some value of $x$. Summing over all possible values of $x$, we get:

$P$(first sum of 3 obtained without obtaining sums of 7 before) $= \sum_{x=1}^\infty (\frac{28}{36})^{x-1}\frac{2}{36}
= \frac{\frac{2}{36}}{1 - \frac{28}{36}}
= \frac{1}{4}$
• Jan 30th 2010, 02:48 AM
stressedout
supposed it is to get a sum of 4 before a sum of 7, the method would be the same just that instead of 2/36 in the final equation, it will be 3/ 36 divide by 8/ 36 right?
• Jan 30th 2010, 10:11 AM
h2osprey
Quote:

Originally Posted by stressedout
supposed it is to get a sum of 4 before a sum of 7, the method would be the same just that instead of 2/36 in the final equation, it will be 3/ 36 divide by 8/ 36 right?

You've got the right idea with $\frac{3}{36}$, but remember you have to adjust the other denominator accordingly.

Probability of not rolling a sum of 7 or 4 is $1 - \frac{3}{36} - \frac{6}{36} = \frac{27}{36}$

Hence you want to divide $\frac{\frac{3}{36}}{\frac{9}{36}}$.