Geometric distribution question for $p=0.5$

Question

Wiki says that geometric distribution $p(k)=p \cdot (1-p)^k$ describes a probability of Bernoulli trials required to get success.

My question is: Let us assume that we're tossing a coin. A success is when a face occurs. What is the probability that only one toss is required to get a face. According to geometric distribution $p(1)=0.25$. My intuition somehow says it should be $0.5$. Where I misunderstand the whole thing.

Answer 1

Your intuition is correct here.

There are two sort of geometric distributions.

Start with experiments (Bernoulli trials) that are independent and end up succesfully with chance $p$.

Let $T$ stand for the number of trials that are made until the first success occurs. Then:

$P\left[T=k\right]=p\left(1-p\right)^{k-1}$ for $k=1,2,...$

Let $F$ stand for the number of failures. Then $F$=$T-1$ so:

$P\left[F=k\right]=P\left[T=k+1\right]=\left(1-p\right)^{k}p$ for $k=0,1,...$

In the case you mention we are dealing with $T$.