geometric distribution

Assuming we want to know the probability of succeeding at least once within the first \[k\] tries (or the probability of success within the first \[k\] tries), i.e. \[P(X\le k)\], the formula is given as \[1-q^{k}\] where \[q=1-p\], or the probability of failure.

The derivation of the formula is quite straightforward. If \[k\ge 1\], \[P(X\le k)=P(X=1)+P(X=2)+\cdots+P(X=k)\], \[P(X\le k)=\sum_{i=1}^{k}\left( (1-p)^{i-1}p \right)=p\sum_{i=0}^{k-1}q^{i}\], assuming \[q=1-p\]. Now, the finite geometric series tells us that \[\sum_{i=0}^{k=1}q^{i}=1+q+q^{2}+\cdots+q^{k-1}=\frac{1-q^{k}}{1-q}\], thus \[p\sum_{i=1}^{k-1}q^{i}=p \left( \frac{1-q^{k}}{1-q} \right)=1-q^{k}\].

The expected value of a geometrically distributed random variable \[X\], \[E(X)\], is the mean number of trials required to get the first success, and is given as \[E(X)=\frac{1}{p}\].

For example, when rolling a six-sided dice, the average number of rolls needed to land on a one will be \[E(X)=\frac{1}{\left( \frac{1}{6} \right)}=6\].

The variance, or expected value of the squared deviation from the mean of a random variable, is given as \[\text{Var}(X)=\frac{1-p}{p^{2}}\].

Converting this to standard deviation, it quantifies the unpredictability of the number of trials to get a the first success. For instance, let's say you have a 20% chance of hitting a shot. \[p=0.2\implies \text{Var}(X)\approx4.47\], thus it would not be surprising if you took 9 or 10 shots to hit the target, despite being usually able to hit it on the fifth shot.