Suppose that two balanced dice are tossed repeatedly and the sum of the two uppermost faces is determined on each toss. What is the probability that we obtain
a) a sum of 3 before we obtain a sum of 7?
b) a sum of 4 before we obtain a sum of 7?
Suppose that two balanced dice are tossed repeatedly and the sum of the two uppermost faces is determined on each toss. What is the probability that we obtain
a) a sum of 3 before we obtain a sum of 7?
b) a sum of 4 before we obtain a sum of 7?
Consider the probabilities $\displaystyle P(3) = \frac{2}{{36}}\,\& \,P(7) = \frac{6}{{36}}$.
In order for there to be a sum of three before a sum of seven happening on say the fifth toss we must have four sums of neither three nor seven and than a sum of three on the fifth toss. That probability is $\displaystyle \left( {\frac{{28}}{{36}}} \right)^4 \left( {\frac{2}{{36}}} \right)$.
Will this work $\displaystyle \sum\limits_{k = 1}^\infty {\left( {\frac{{28}}{{36}}} \right)^{k - 1} \left( {\frac{2}{{36}}} \right)} $?
a sum of $\displaystyle x$ before a sum of $\displaystyle y$ on the $\displaystyle n$th roll
$\displaystyle
\left[ {1 - \left( {\frac{{ - \left| {y - 7} \right| - \left| {x - 7} \right| + 12}}
{{36}}} \right)} \right]^{n - 1} \left[ {\frac{{ - \left| {x - 7} \right| + 6}}
{{36}}} \right]
$
[Math]
x \in Z\cap {\left[ {2,12} \right]}
[/tex]
[Math]
y \in Z\cap {\left[ {2,12} \right]}
[/tex]
[Math]
n \in Z^+\cup \{0\}
[/tex]
lol?