10^{-27}$ kg, a human of $100$ kg, if constituted only by hydrogen atoms, would contain approximately a number $N$ of them given by: $$N\approx {100\, {\rm kg}\over 1.67 \times 10^{-27}\, {\rm kg}}\approx ≈ 4.2 \times 10^{28}.
\label{eq3}$$

Each of these atoms will have to individually re-locate in a specific region of space, to reconstitute the entire body structure, with no errors. Therefore, the (estimated) probability ${\cal P}_{\rm body}$ of self-teleportation of the overall body structure will be given by the product of the self-teleportation probabilities ${\cal P}$ of every single atom contained in that structure. If the body would be formed only by two atoms, that is, $N= 2$, the probability would be ${\cal P}_{\rm body} \approx {\cal P}^2\approx 10^{-45\times 2} = 10^{-90}$. With three atoms, the probability would become: ${\cal P}_{\rm body} \approx {\cal P}^3\approx 10^{-45\times 3} = 10^{-135}$. Therefore, with $N= 10^{28}$ atoms, we obtain: $${\cal P}_{\rm body} \approx {\cal P}^{10^{28}}\approx 10^{-4.5\times 10^{29}}. 
\label{eq4}$$

Let us reflect for a moment on the amazing infinitesimality of this number. To write it in decimal, non-scientific notation, we must use more than $10^{29}$ zeros, i.e., more than one hundred billion billion billion zeros! If we write with a printer on paper 