Radioactive decay, falling pencils and the uncertainty principle

[Happiness]

Consider an ensemble of identically prepared pencils balanced on its tip. If a pencil is precisely vertical, $x=0$, and precisely at rest, $p=0$, then it will never fall. But some pencils in the ensemble would fall, because according to the uncertainty principle, the standard deviation $\sigma_x$ of position of the ensemble and the standard deviation $\sigma_p$ of momentum of the ensemble is related by $\sigma_x\sigma_p\geq\hbar/2$. From this inequality, it can be shown that the mean time $\tau$ of fall for an ensemble is at most 4s. Also, the smaller the product $\sigma_x\sigma_p$, the longer the mean time $\tau$.

If we set up a large number of pencils and wait sufficiently long, the set of pencils still standing would have a big value of $\tau$. This would imply that for this set of pencils, $\sigma_x\sigma_p$ must be smaller than $\hbar/2$, violating the uncertainty principle.

Next consider an ensemble of identically prepared radioactive uranium. Suppose its half life is 4s. If we started off with 100 uranium atoms, then after 8s, we would have roughly 25 atoms left. For this set of 25 atoms, the mean time $\tau$ of decay is greater than 8s. This set of undecayed uranium atoms must somehow be inherently more stable than those 75 that have already decayed. Then the half life of these 25 remaining uranium atoms must be more than 4s. But this contradicts experimental observations. Why is this so?

It seems that even though the mean time $\tau$ of decay is greater than 8s for the set of 25 remaining uranium atoms, they are not more stable than those 75 that have already decayed. Similarly, even though the mean time $\tau$ of fall is greater for the set of remaining pencils, their $\sigma_x\sigma_p$ is still greater than $\hbar/2$. But why?

 

[Nugatory]

Those $\sigma{x}$ and $\sigma{p}$ values and decay half-lives are statistical properties of the ensemble, not properties of anyone pencil or decaying atom.

If I prepare an ensemble of coins and start flipping them...
After one toss, half the coins will have come up heads. After two tosses, one-fourth of the coins will have come up heads both times. After three tosses, one-eight will have come up heads all three times, and after $N$ tosses one in $2^{N}$ will have come up heads all $N$ times. It doesn't follow that that coin is any more likely to come up heads than the others, and likewise there's nothing special about the one in eight uranium atoms that go three or more half-lives before they decay.

 

[Nugatory]

A note for all posters:

The quantum mechanical uncertainty principle applies to the decay of radioactive atoms, but it is not at all clear that it can applied to a classical metastable system like a pencil balanced on its point, at least not in the straightforward way it's being done here. This point has already been made in post #11 of https://www.physicsforums.com/threads/is-the-uncertainty-principle-an-ontological-statement.877360/