Okay, letâ€™s do this slightly differently.

Put the first coin in at $t=0$, the next at $t=0.5$, then at $t=0.75$, etc. By $t=1$, $\aleph_0$ coins will be disposed of. Letâ€™s write $\omega$ for $\aleph_0$; itâ€™s easier.

So by $t=1$, $\omega$ coins disappear. Now speed things up twice as fast. Then by $t=2$, $\omega^2$ coins will have disappeared.

But now we can adjust the speed in such a way that the next $\omega^2$ coins will disappear by $t=2.25$, the next after by $t=2.375$, and so on â€¦ so this time $\omega^3$ coins disappear by $t=2.5$. Now carry on adjusting the speed so that $\omega^4$ coins disappear by $t=2.75$, $\omega^5$ by $t=2.875$, and so on.

Result: by $t=3$, $\omega^{\omega}$ coins will be gone. But $\omega^{\omega}$ is uncountable! For $\omega^{\omega}>2^{\omega}=\left|\frak P(\mathbb N)\right|$ and we know that the power set of the natural numbers is uncountable. Since the gambler only has a countable amount of coins to get rid of, he might be able to do that in three seconds!

What do you think?