Geometric distrubution and expectancy I have following problem:
We throw a fair dice time after time.What is time expectancy to get first time 66  i.e two six in the row for the first time. I probably should use this http://en.wikipedia.org/wiki/Expected_value#Discrete_distribution_taking_only_non-negative_integer_values though I don't understand how.                                                                                                                                                                                                                                                                                                                                                                                                     
 A: Let $x$ be the expected number of tosses to get one six. Then $x = \frac{5}{6}(1+x)+ \frac{1}{6}(1)$. So $x=6$. Let $y$ be the expected number of throws to two consecutive sixes. Then $y = \frac{1}{6}(x+1)+ \frac{5}{6}(x+1+y)$. So $y = 6(x+1)$. Plugging in $x=6$, we get $y = 42$. 
A: Define a live n-sequence to be a sequence of $n$ dice throws in which no double 6 occurs. Define:
$r_n$ = number of live $n$-sequences that don't end in 6
$s_n$ = number of live $n$-sequences that do end in 6
Then we get the recurrence relations:
$r_{n+1} = 5r_{n} + 5s_{n}$
$s_{n+1} = r_n$
with initial conditions $r_0 = 1, s_0 = 0$. Substituting the second relation into the first gives  
$s_{n+2} = 5s_{n+1} + 5s_{n}$
The solutions of this are of the form $s_n = A\alpha^n + B\beta^n$, where $\alpha$ and $\beta$ are the roots of the equation $x^2 - 5x - 5 = 0$:  
$\alpha = (5 + 3\sqrt5)/2, \beta = (5 - 3\sqrt5)/2$
The probability that we need exactly $n$ throws is $\frac{1}{6} s_{n-1}/6^{n-1} = \frac{s_{n-1}}{6^n}$, so the expected number of throws is  
$\sum_{n=1}^\infty n\frac{s_{n-1}}{6^n} = \frac{A}{\alpha}\sum_{n=1}^\infty nu^n + \frac{B}{\beta}\sum_{n=1}^\infty nv^n$
where $u = \frac{\alpha}{6}$ and $v = \frac{\beta}{6}$. This is  
$\frac{A}{\alpha}\frac{u}{(1-u)^2} + \frac{B}{\beta}\frac{v}{(1-v)^2}$
From the initial terms $s_0=0, s_1=1$, we get $A = \frac{1}{3\sqrt 5}, B = -A$. So we get  
$\frac{A}{6}(\frac{1}{(1-u)^2} - \frac{1}{(1-v)^2})$, which my shaky algebra simplifies to...wait for it...   
$42$
A: The probability that you get a 6 after $n+1$ throws is $$ \frac{5^n}{6^n} \frac{1}{6}$$ that is getting $n$ times something else than a $6$ then getting a $6$. Therefore, the probability to throw $n$ times the dice and then to get two $6$ in a row is $$ \frac{5^n}{6^n} \frac{1}{6^2}.$$ The expectancy is thus $$\frac{1}{6^2} \sum_{n=0}^{\infty}\frac{5^n}{6^n}.$$
Does it answer your question?
