I am having trouble with a probability question regarding to G-distribution.

Background: Let's assume I have a disc that has two sides, one is black and one is white. Note: the disc has biased probability: the probability to get black is $p$ and the probability to get white is $1-p$. when u toss the disc time after time, the number of tosses ($Z$) until u get black at the first time distribute geometrically to: $Z$~$G(p)$ . Note: formula for G-distribution: $P(Z=k)=p(1-p)^{k-1}$.

Questions: 1.let's toss the disc time after time until we get two black color in a row (for the first time). 1.a Find the probability function of the random variable: $T =${number of tosses until getting 2 black colors in a row for the first time}. Make sure there normalization properties exists.

this is what i have achieved so far: i know that i have to shoe that the sum of the probabilities of the experiment should equals to $1$.

$P(T=k)=((k-1)*(1-p)^{k-2}*p)*p$ the arguments in the brackets demonstrating one success of getting black color in $k-1$ tries. the otter $p$ present success in the $k$-ith. time. let's show normalization: $$\frac{p^2}{(1-p)^{2}}\sum_{k=1}^{\infty}(k-1)(1-p)^{k-2}$$ We know that $$\frac{1}{1-x}\sum_{n=0}^{\infty}x^n=> \frac {d}{dx}=\frac{1}{(1-x)^2}=\sum_{n=1}^{\infty}n \cdot x^{n-1}$$

we multiply both sides with $x^{2}$ and then we will get :$\frac {(1-p)^2}{(1-(1-p)^2}\frac{p^2}{(1-p)^2}=1$. this is my answer but my professor told me its wrong and i don't understand why. if anyone can please help me understand what's wrong or give me a hint or a solution to this problem will be grateful.

  • $\begingroup$ You have computed the probabilities for a different random variable, $T’$, which is the number of tosses it takes to get two black colors (not necessarily in a row). $\endgroup$ – Mike Earnest Aug 15 '18 at 12:26
  • $\begingroup$ I just edited my post. if u can please help. $\endgroup$ – Tomer Levy Aug 15 '18 at 13:23
  • $\begingroup$ Your expression for $P(T=k)$ is incorrect. $(k-1)(1-p)^{k-1}p^2$ is the probability that the first $k$ flips will contain exactly two blacks, with the second occurring at flip $k$. What you instead want to calculate is the probability that the first $k$ flips contain any number of black flips, as long as the only place where two consecutive black flips occur is in the last two flips. This is harder to write down a formula for. $\endgroup$ – Mike Earnest Aug 15 '18 at 13:28
  • $\begingroup$ Am I understanding the problem correctly? In the below sequence of flips, is $T=10$? $$ W,B,W,W,B,W,B,W,B,B $$ $\endgroup$ – Mike Earnest Aug 15 '18 at 14:09

Let $a_k=P(T=k)$. You can show that $$ a_k =(1-p)a_{k-1}+p(1-p)a_{k-2},\qquad k>2 $$ Why? There are two ways to have a sequence of $k>2$ coin flips whose first occurrence of $BB$ is at flips $k-1$ and $k$. Either you start with a white flip $W$, and then follow it up with a sequence which gets its first $BB$ at flips $k-2$ and $k-1$, or you start with $BW$ and get a sequence of $k-2$ flips ending in $BB$. You cannot start with $BB$, since that would imply $T=2$, and we are assuming $k>2$.

Combined with the base cases $a_1=0$, $a_2=p^2$, the above is a linear recurrence which can be solved using the methods in this article.

  • $\begingroup$ so i tried to solve this linear recurrence, but i am getting solutions with roots in the process to solve this recurrence ( i also tried to use a variable to avoid roots, but no luck) and other stuff that are very not easy to solve. i really don't think my professor aimed for this solution. maybe there is another way you have in mind? $\endgroup$ – Tomer Levy Aug 16 '18 at 7:17

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