How to calculate the expectation of a geometric random variable having parameter p

The Probability Distribution Function of a geometric variable is \[ E[x] = \underset{n}{\Sigma} np {(1-p)}^{n-1} : 0 \le n < \infty = \frac{1}{p}\]

Intuitively, this probability can be understood as follows. Let us consider the case of a coin toss and the probability of whether it will be a heads or tails is 0.5 and 0.5. If you are interested in the outcome being heads, 1/0.5 = 2 would mean that in the long run, we would need to wait for two tosses for the outcome to be heads. This does not imply however that the every-other toss is going to be favourable to you. But the outcomes match the expected number in the long run.

We will now do a simulation to see how the outcomes match up.

Probabilities 0.9 & 0.1

• In this case the probabilities of Failure and Success are 0.9 and 0.1 respectively.
• Each horizontal row of squares is a series of trials until a success is obtained.
• The red rectangles are the failures and the green rectangles at each row is a success.
• Observe how the mean number of trials needed stabilizes to \(\frac{1}{0.1} = 10\)

Probabilities 0.95 & 0.05

Code used for simulation and plots

    ClearAll[rect, geometric];

    rect[number_, yLocation_, height_] :=
        Piecewise[{{{Green, EdgeForm@Gray, Opacity@0.2, Rectangle[{0, yLocation}, {1, yLocation + height}]}, number == 1},
       {Table[{Red, EdgeForm@Gray, Opacity@0.2, Rectangle[{r, yLocation}, {r + 1, yLocation + height}]}, {r, 0, number - 2}]
         ~Join~{Green, EdgeForm@Gray, Opacity@0.2, Rectangle[{number - 1, yLocation}, {number, yLocation + height}]}, number > 1}}]

    accumulatingMean[list_List] := N@Accumulate[list]/Range[Length@list]

    geometric[p_, no_] :=
        Module[{data = Differences[Union @@ Position[RandomChoice[{p, 1 - p} -> {0, 1}, no], 1]], means},
        means = accumulatingMean[data];
        Graphics[{
        Table[rect[data[[n]], -n, 1], {n, 1, Length@data}],
        Line[Table[{means[[n]], -n}, {n, 1, Length@means}]]}
        , ImageSize -> 756
        , AspectRatio -> Length@data/Max@data
        , PlotRange -> {{0, 100}, All}
       ]
      ]

    Export[StringReplace[NotebookFileName[], ".nb" -> "_" <> ToString[#] <> ".svg"], geometric[#, 2000]] & /@ {0.9, 0.95}