# If we have a beta-geometric distribution X with pdf P(X=k|alpha, beta)=(beta (alpha+1,k+beta))/(beta(alpha,beta))

Probability of events for beta-geometric distribution
If we have a beta-geometric distribution X with pdf $P\left(X=k\mid \alpha ,\beta \right)=\frac{Beta\left(\alpha +1,k+\beta \right)}{Beta\left(\alpha ,\beta \right)}$.
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Reagan Tanner
Step 1
To avoid confusion you should not use $\beta$ for both the parameter and the beta function. I suggest using B for the beta function.
Further $\mathsf{P}\left(k;\alpha ,\beta \right)$ is a probability density function. X is a continuous random variable realised within the support of [0;1]. It certainly is not equal to its own pdf.
Step 2
So, the claims you make at the very beginning of your chain are complete absurdities.
So the reasoning that follows from there is invalid because it is based on an unjustifiable premise.