Why is the sum to infinity of a geometric distribution equal to 1?

Question

I know that the geometric distribution follows the rules of a geometric progression thus by using the sum to infinity formula (which I know its proof and is really convinced by it), $$\frac {a}{1-r}$$

We can easily arrive at this:

$\frac {p}{1-(1-p)} =\frac {p}{1-1+p} = \frac pp = 1$

But the thing is, when my teacher asked my class what we would expect the sum to be no one hesitated to say 1. They all knew it by logic. However I couldn't understand how they arrived at that.

One guessing was that maybe the sum is 1 because "P", probability of getting my success at first trial will be added to small fractions of it, but I still couldn't arrive at anything, after all, these fractions are going to keep decreasing to very tiny numbers so I couldn't get my head around the fact that they'll end up as the same value as P and thus their sum will be 1.

Can someone explain to me the logic or intuition behind this?

Answer 1

Try looking at it this way - the geometric distribution measures the probability of achieving some number of failures in a series Bernoulli trials before getting the first success. So it's the number of tails you flip on a coin until you get your first heads. So if you sum up the probabilities of $0, 1, 2, \ldots,$ up to $n-1$ successes, you get the total probabilities of getting a heads before you've flipped $n$ tails. And that's $\sum_{k=0}^{n-1} p(1-p)^k = p + p(1-p) + \ldots + p(1-p)^{n-1} = 1 - (1-p)^n$, if you do the algebra.

And so, what if you take that probability away from 1? Then that's $(1-p)^n$, the probability that it takes you at least $n$ failures before your first success, i.e. the probability that you've flipped the coin $n$ times and still haven't gotten heads.

So what does it mean to look at that value as $n \rightarrow \infty$? It's the probability that you've been flipping that coin forever, never getting heads even once. Just and endless sequence of tails, for all eternity. And if you know that the probability of getting heads on any flip is $p > 0$, then that seems ridiculous - surely, somewhere in that series of flips, there must have been at least one time you flipped heads. It's impossible! Or at least, it's an event with zero probability - meaning that the probability that you flipped heads at least once, i.e. the probability of one of the events from the geometric distribution happening, must be 1.

Answer 2

They are probably using the fact that the sum of the probabilities of all possible outcomes must equal $1$. This is true for any [discrete] random variable. They are probably not doing the computation $p + p(1-p) + p(1-p)^2 + \cdots = \frac{p}{1-(1-p)} = 1$ in their heads instantaneously.