Not an answer to your question but an alternative that might interest you.

The ace that was drawn from the first half has probability $\frac1{27}$ to become the last drawn card.

All other cards have equal probability to become the last drawn card, so if $p$ denotes this probability then:$$51p+\frac1{27}=1$$ so that:$$p=\frac1{51}\frac{26}{27}$$

Three of those cards are aces so the probability that the last drawn card is an ace equals:$$\frac1{27}+3p=\frac1{27}+\frac3{51}\frac{26}{27}=\frac{43}{459}$$

Not an answer to your question but an alternative that might interest you.

The ace that was drawn from the first half has probability $\frac1{27}$ to become the last drawn card.

All other cards have equal probability to become the last drawn card, so if $p$ denotes this probability then:$$51p+\frac1{27}=1$$This makes clear that:$$p=\frac1{51}\frac{26}{27}$$

Three of those cards are aces so the probability that the last drawn card is an ace equals:$$\frac1{27}+3p=\frac1{27}+\frac3{51}\frac{26}{27}=\frac{43}{459}$$