Probability of two consecutive successes before two consecutive failures

Question

An infinite sequence of independent trials is to be performed. Each trial results in a success with probability $p$ and a failure with probability $1-p$. What is the probability that we see two successes in a row before we see two consecutive failures?

So far I have $$p^2+(1-p)p^2+(1-p)p^3+(1-p)^2p^3+ (1-p)^2p^4+\cdots$$

But I am not sure how to simplify this and get a proper answer in a single term.

Answer 1

Solution 1

Let's denote

• $P_{S}$ probability that the current trial is success and previous trial is not success (i.e., either the current trial is the first trial or the previous trial is failure);
• $P_{F}$ probability that the current trial is failure and previous trial is not failure (i.e., either the current trial is the first trial or the previous trial is success)
• $P_{SS}$ probability of 2 consecutive successes before 2 consecutive failures.

Then

• $P_{SS}=pP_S$
• $P_S=p+pP_F$
• $P_F=(1-p)+(1-p)P_S$

and finally after solving the above system $$P_{SS}=\frac{p^2(2-p)}{1-p(1-p)}$$

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Solution 2

Given the first trial is success, the probability of 2 consecutive successes before 2 consecutive failures is $$P(SS|S_1)=p+p^2(1-p)+p^3(1-p)^2+\cdots=\frac{p}{1-p(1-p)}$$

Given the first trial is failure, the probability of 2 consecutive successes before 2 consecutive failures is $$P(SS|F_1)=p^2+p^3(1-p)+p^4(1-p)^2+\cdots=\frac{p^2}{1-p(1-p)}$$

The total probability $$P(SS)=pP(SS|S_1)+(1-p)P(SS|F_1)=\frac{p^2}{1-p(1-p)}+\frac{p^2(1-p)}{1-p(1-p)}=\frac{p^2(2-p)}{1-p(1-p)}$$

Answer 2

Possibilities are starting with a success $$ \color{#C00}{(SF)^n}\color{#090}{SS}=\frac{\color{#090}{p^2}}{\color{#C00}{1-p(1-p)}} $$ and starting with a failure $$ F\color{#C00}{(SF)^n}\color{#090}{SS}=\frac{(1-p)\color{#090}{p^2}}{\color{#C00}{1-p(1-p)}} $$ giving a total of $$ \frac{(2-p)p^2}{1-p(1-p)} $$