Trouble understanding Number of Trials to First Success

Question

I'm having some trouble understanding how to calculate how many trials are needed before you expect to see a given event, I'm not sure what I've misunderstood yet.

I've followed these explanations so far:
https://www.cut-the-knot.org/Probability/LengthToFirstSuccess.shtml
https://www.geeksforgeeks.org/expected-number-of-trials-before-success/

Lets say we have a trial that produces an event $V$ with probability $0.3$, my understanding from the pages above is we expect to see event $V$ after $\frac{1}{0.3}$ trials, or $3.3$ trials.

But if we perform two trials, and look at the probability of seeing the event, don't we get 51%?
$$ 0.3^2+0.3*0.7+0.7*0.3 = 0.51 $$ or $$ 1 - (0.7^2) = 0.51 $$

So I would expect to expect $V$ to happen when the number of trials is as low as 2.

Can anyone show me what I've misunderstood here please?

Answer 1

Just because you “expect” a certain thing to happen, that does not make it an “expected value.” It’s just a quirk of mathematical English.

Expected value means the average value. Sure, more than half the time, it only takes 1 or 2 trials. Still, there is a $0.3· (0.7)^2=14.7$% chance that it will take $3$ trials, there is a $0.3· (0.7)^3\approx 10.3$% chance that it will take $4$ trials, and so on. The expected value considers all of these possibilities, and averages them together. It just so happens that the events of taking $3,4,5\ldots$ trials are significant enough to push the average up to $3.3$ trials.

Answer 2

I'm sure there are better ways to illustrate the difference between "The average number of trails until it happens is 2" and "in 2 trails, it happens for a chance higher than 50%". Here is how I see it.

Let $V_i=1$ if the $i$-th trial is a success, and $V_i=0$ if the $i$-th trial is a failure. The probability of success of each trial is $p$. Trials are independent. $R$ is the number of trials until the first success. $$ E\left[R\right]=\sum^{\infty}_{n=1}(1-p)^{n-1}pn=1/p. $$ The probability that there is at least one success in the first $n$ trials is $$ X_n=\mathbb{P}\left(\sum^n_{i=1}V_i\geq 1\right)=1-(1-p)^n. $$ You ask why in your example, $E\left[R\right]>2$ and yet $X_2>0.5$. It seems you presume for $n<E\left[R\right]$, $X_n$ should be smaller than $50\%$. This presumption is incorrect. Think about when $p\in(0.5,1)$, $X_1>0.5$ and yet $E[R]>1$ by definition.