Probability interpretation

Question

Supose we have a mouse on a labyrinth and we associate $p$ as the probability of the mouse escaping this labyrinth this turn, and $(1-p)$ for the probability of him not escaping that turn. Call $X$ the random variable associated with the number of turns that the mouse takes in order to escape. If I want him to take, at best, $4$ turns to escape the labyrinth then I would consider the possibilities of:

$$P(X \leq 4)= P(X = 4)+P(X = 3)+P(X = 2)+P(X = 1)$$

After some calculations I arrive at the probability:

$$ P(X \leq 4)= 1-(1-p)^4 $$

On the other hand:

$$ P(X \leq 4)=1-P(X > 4)$$

Which leads to:

$$ P(X > 4) = (1-p)^4$$

This gives us the probability of him failling more then $4$ times, given that it is the sum:

$$ P(X > 4) = P(X=5)+P(X = 6)+...=p(1-p)^4+p(1-p)^5+...=p \sum_{i=4}^n(1-p)^{i}=(1-p)^4$$

Can someone give me a intuitive interpretation for this? It seems a little odd that the summation would tend to $(1-p)^4$ when starting at $4$.

Answer 1

You have $P(X>4)=P(X=5)+...+P(X=n)+...=p\cdot\sum_{i=4}^{\infty}(1-p)^i$

This is a geometric series $a_1=(1-p)^4$ and $q=1-p$

By Geometric series formula, we have when $q<1$:

$lim_{n->\infty}\sum_{i=1}^{n}a_1 \cdot q^i = \frac{a_1}{1-q}=\frac{(1-p)^4}{1-(1-p)} = \frac{(1-p)^4}{p}$

But we had the multiplication with $p$ outside so we get $(1-p)^4$

It will work with any Geometric distribution, and for every value.

If you look for intuition, $(1-p)^4$ is the chance to "fail" in the first 4 times, so it leads to $P(X>4)$

Answer 2

Let $r\neq 1$. Then \begin{align*} \sum_{k=0}^{n-1} r^k = \frac{1-r^n}{1-r}, \end{align*} which for $r \in (0,1)$ tends to $\frac{1}{1-r}$ as $n$ tends to infinity.

Thus, if $p\in(0,1)$, then \begin{align*} p \sum_{k=4}^\infty (1-p)^k &= \left( p \sum_{k=0}^\infty (1-p)^k \right) - \left(p \sum_{k=0}^{4-1} (1-p)^k\right) \\ &= p\frac{1}{1-(1-p)} - p\frac{1-(1-p)^4}{1-(1-p)} \\ &= 1 - \left(1 - (1-p)^4\right) \\ &= (1-p)^4. \end{align*}

(In the above, the LHS of course corresponds to $\mathbb{P}(X>4) = \sum_{k=5}^\infty \mathbb{P}(X = k)$.)

Alternatively, one can re-write the LHS according to \begin{align*} p \sum_{k=4}^\infty (1-p)^k &= p \sum_{k=0}^\infty (1-p)^{k+4} \\ &= (1-p)^4 p \sum_{k=0}^\infty (1-p)^k, \end{align*} when the result is established by noting that the last term equals $\frac{1}{1-(1-p)} = \frac{1}{p}$.

Answer 3

You have that $\mathsf P(X=n)= p(1-p)^{n-1}$ and so $\mathsf P(X\gt n)=p\sum_{k=n}^\infty(1-p)^k$.

Thus $\mathsf P(X>0)= p\sum_{k=0}^\infty(1-p)^k$. However, this almost surely equals $1$. Theoretically, the mouse will escape sometime after the experiment starts.

$$\begin{align}\mathsf P(X>n)&=p\sum_{k=n}^n(1-p)^k\\&=(1-p)^n\cdot p\sum_{k=n}^\infty(1-p)^{k-n}\\&=(1-p)^n\cdot p\sum_{j=0}^\infty (1-p)^j\\&=(1-p)^n\cdot\mathsf P(X>0)\\&=(1-p)^n\end{align}$$

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In short...

The probability that the mouse escapes after $n$ failures is the probability of having $n$ consecutive failures times the probability the probability that it escapes sometime after that.

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And to check this intuition mathematically, we use the derivation for the closed form of a Geometric Series. $$\begin{align}\sum_{k=0}^\infty (1-p)^k &= (1-p)^0+(1-p)\sum_{k=1}^\infty (1-p)^{k-1}\\ &=1+(1-p)\sum_{k=0}^\infty(1-p)^k \\[2ex]\therefore\quad p\sum_{k=0}^\infty (1-p)^k &= 1 \end{align}$$