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Perhaps its easiest to first ask for the probability $P_0$ that the card does not move at all ($k=0$). One way to achieve this is to cut immediately below and above that card, with probability $1/2\times 1/2=1/4$. But any symmetric pair of cuts will do, for example you could cut at two cards below and two cards above, with probability $1/4\times 1/4$, or three cards above and three cards below, with probability $2^{-3}\times 2^{-3}$, and so on. Summing the geometric series gives

$$P_0 = \sum_{n=1}^\infty \frac{1}{4^n}=\frac{1/4}{1-1/4}=\frac{1}{3}$$

For the general case you first start with a cut just below the card and a second cut $k+1$ cards above, with probability $1/4\times 2^{-k}$. Then extending this symmetrically at both sides and summing the geometric series gives you the additional factor $1/3$, and you arrive at the general formula $P_k=\frac{1}{3} \times 2^{-k}$.

We are assuming here an infinite set of cards, to sum the geometric series to infinity. For a small number of cards the formula is not exact, as noted by Kevin in a comment.

Perhaps its easiest to first ask for the probability $P_0$ that the card does not move at all ($k=0$). One way to achieve this is to cut immediately below and above that card, with probability $1/2\times 1/2=1/4$. But any symmetric pair of cuts will do, for example you could cut at two cards below and two cards above, with probability $1/4\times 1/4$, or three cards above and three cards below, with probability $2^{-3}\times 2^{-3}$, and so on. Summing the geometric series gives

$$P_0 = \sum_{n=1}^\infty \frac{1}{4^n}=\frac{1/4}{1-1/4}=\frac{1}{3}$$

For the general case you first start with a cut just below the card and a second cut $k+1$ cards above, with probability $1/4\times 2^{-k}$. Then extending this symmetrically at both sides and summing the geometric series gives you the additional factor $1/3$, and you arrive at the general formula $P_k=\frac{1}{3} \times 2^{-k}$.

We are assuming here an infinite set of cards, to sum the geometric series to infinity. For a small number of cards the formula is not exact, as noted by Kevin in a comment.

I could try to generalize this to arbitrary cutpoint probability $p\in(0,1)$. Then

$$P_0=p^2+p^2(1-p)^2+p^2(1-p)^4=\frac{p^2}{1-(1-p)^2}=\frac{p}{2-p},$$

and hence

$$P_k=(1-p)^kP_0=\frac{p(1-p)^k}{2-p}.$$

Perhaps its easiest to first ask for the probability $P_0$ that the card does not move at all ($k=0$). One way to achieve this is to cut immediately below and above that card, with probability $1/2\times 1/2=1/4$. But any symmetric pair of cuts will do, for example you could cut at two cards below and two cards above, with probability $1/4\times 1/4$, or three cards above and three cards below, with probability $2^{-3}\times 2^{-3}$, and so on. Summing the geometric series gives

$$P_0 = \sum_{n=1}^\infty \frac{1}{4^n}=\frac{1/4}{1-1/4}=\frac{1}{3}$$

For the general case you first start with a cut just below the card and a second cut $k+1$ cards above, with probability $1/4\times 2^{-k}$. Then extending this symmetrically at both sides and summing the geometric series gives you the additional factor $1/3$, and you arrive at the general formula $P_k=\frac{1}{3} \times 2^{-k}$.

Perhaps its easiest to first ask for the probability $P_0$ that the card does not move at all ($k=0$). One way to achieve this is to cut immediately below and above that card, with probability $1/2\times 1/2=1/4$. But any symmetric pair of cuts will do, for example you could cut at two cards below and two cards above, with probability $1/4\times 1/4$, or three cards above and three cards below, with probability $2^{-3}\times 2^{-3}$, and so on. Summing the geometric series gives

$$P_0 = \sum_{n=1}^\infty \frac{1}{4^n}=\frac{1/4}{1-1/4}=\frac{1}{3}$$

For the general case you first start with a cut just below the card and a second cut $k+1$ cards above, with probability $1/4\times 2^{-k}$. Then extending this symmetrically at both sides and summing the geometric series gives you the additional factor $1/3$, and you arrive at the general formula $P_k=\frac{1}{3} \times 2^{-k}$.

We are assuming here an infinite set of cards, to sum the geometric series to infinity. For a small number of cards the formula is not exact, as noted by Kevin in a comment.

Perhaps its easiest to first ask for the probability $P_0$ that the card does not move at all ($k=0$). One way to achieve this is to cut immediately below and above that card, with probability $1/2\times 1/2=1/4$. But any symmetric pair of cuts will do, for example you could cut at two cards below and two cards above, with probability $1/4\times 1/4$, or three cards above and three cards below, with probability $2^{-3}\times 2^{-3}$, and so on. Summing the geometric series gives

$$P_0 = \sum_{n=1}^\infty \frac{1}{4^n}=\frac{1/4}{1-1/4}=\frac{1}{3}$$

in agreement with the general formula $P_k=\frac{1}{3} 2^{-k}$.

Perhaps its easiest to first ask for the probability $P_0$ that the card does not move at all ($k=0$). One way to achieve this is to cut immediately below and above that card, with probability $1/2\times 1/2=1/4$. But any symmetric pair of cuts will do, for example you could cut at two cards below and two cards above, with probability $1/4\times 1/4$, or three cards above and three cards below, with probability $2^{-3}\times 2^{-3}$, and so on. Summing the geometric series gives

$$P_0 = \sum_{n=1}^\infty \frac{1}{4^n}=\frac{1/4}{1-1/4}=\frac{1}{3}$$

For the general case you first start with a cut just below the card and a second cut $k+1$ cards above, with probability $1/4\times 2^{-k}$. Then extending this symmetrically at both sides and summing the geometric series gives you the additional factor $1/3$, and you arrive at the general formula $P_k=\frac{1}{3} \times 2^{-k}$.