**Question:** A coin is randomly selected from a group of ten coins, the nth coin having a probability ^{n}⁄_{10} of coming up heads. The coin is then repeatedly flipped until a head appears. Let N denote the number of flips necessary. What is the probability distribution of N? Is N a geometric random variable? When would N be a geometric random variable; that is, what would have to be done differently?

**Analytical Solution**

We will solve for both the probability as well as the expected value. We can solve this using conditional expectation. One could follow an approach as outlined below.

E[N] = E[ E[ N|Coin=n ] ]

⇒ E[N] = Summation( E[ N|Coin=n ] ] P{ Coin=n } )

⇒ E[N] = Summation( ^{1}⁄_{i⁄10} ^{1}⁄_{10} ) ≅ 2.92897

Next part

P{ N=t } = P{ N=t | Coin=n } P{ Coin=n }

⇒ P{ N=t } = Summation over n : ^{n}⁄_{10} (1-^{n}⁄_{10})^{(t-1)} ^{1}⁄_{10}

And no, N would not be a geometric variable unless of course all the coins have the same probability or the same coin is deliberately chosen every single time (that would be a lot of bias ;))

**Simulation Solution**

The plots are self explanatory and the code required for them is provided below in the code section.

**Code**

Module[{probabilities = Divide[Range@10, 10]}, Labeled[Histogram[ Mean /@ Partition[(RandomVariate[GeometricDistribution[#]] + 1) & /@ RandomChoice[probabilities, 1000000], 100], {0.05}, Frame -> True, FrameLabel -> Style["\[Mu](N) \[Congruent] E(N)", 15, Lighter@Red], ImageSize -> 788], Style["Means calculated for batches of size 15", Lighter@Red, 20], Top] ] Module[{probabilities = Divide[Range@10, 10], data}, data = KeySort@ Counts[(RandomVariate[GeometricDistribution[#]] + 1) & /@ RandomChoice[probabilities, 10000]]; Labeled[BarChart[data, ChartLabels -> Placed[{Keys@data, Rotate[#, Divide[\[Pi], 2]] & /@ (Values@data)}, {Below, Above}], Frame -> True, ImageSize -> 788], Style["Counts of the number of turns taken for a pass", Lighter@Red, 20], Top] ]

End of the post 🙂

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