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note:- this question has not yet been answered(as of now, 2 answers have been posted, but none of them address my problem)

Problem

A deck of card is shuffled then divided into two halves of 26 cards each. A card is drawn from one of the halves, it turns out to be an ace. The ace is then placed in the second half-deck. The half is then shuffled, and a card is drawn from it. Compute the probability that this drawn card is an ace.

my attempt:-

let A = number of aces in the second deck before an ace added to it
let B = number of aces in the first deck before an ace is removed from it
let D = event that an ace is drawn from the second deck after an ace is added to it

Now, what we are asked to find is: $$\sum_{i=0}^3P(A=i|B>0)\cdot P(D) = \sum_{i=0}^3P(A=i|B>0)\cdot \frac{i+1}{27} $$

now, there are two ways of calculating $P(A=i|B>0)$

1. (Protocol A)restrict the sample space way:- since B has at least one ace, we restrict the sample space to 51 cards out of which 3 are aces. $$P(A=i|B>0) = \frac{{{3}\choose{i}}{{48}\choose{26-i}}}{{{51}\choose{26}}}$$
2. (Protocol B)$P(A|B) = \frac{P(A\cap B)}{P(B)}$:- $$P(A=i|B>0) = \frac{P(A=i\cap B>0)}{P(B>0)} = \frac{{{4}\choose{i}}{{48}\choose{26-i}}\bigg{/}{{52}\choose{26}}}{\bigg{[}{{52}\choose{26}}-{{4}\choose{0}}{{48}\choose{26}}\bigg{]}\bigg{/}{{52}\choose{26}}}$$

My question is : (out of Protocols A and B) which method is the correct one? and Why?

Protocol A has already been used here to answer this question correctly. So, protocol A is obviously correct. So, I guess my question becomes: Why is protocol B incorrect?

Problem

A deck of card is shuffled then divided into two halves of 26 cards each. A card is drawn from one of the halves, it turns out to be an ace. The ace is then placed in the second half-deck. The half is then shuffled, and a card is drawn from it. Compute the probability that this drawn card is an ace.

my attempt:-

let A = number of aces in the second deck before an ace added to it
let B = number of aces in the first deck before an ace is removed from it
let D = event that an ace is drawn from the second deck after an ace is added to it

Now, what we are asked to find is: $$\sum_{i=0}^3P(A=i|B>0)\cdot P(D) = \sum_{i=0}^3P(A=i|B>0)\cdot \frac{i+1}{27} $$

now, there are two ways of calculating $P(A=i|B>0)$

1. (Protocol A)restrict the sample space way:- since B has at least one ace, we restrict the sample space to 51 cards out of which 3 are aces. $$P(A=i|B>0) = \frac{{{3}\choose{i}}{{48}\choose{26-i}}}{{{51}\choose{26}}}$$
2. (Protocol B)$P(A|B) = \frac{P(A\cap B)}{P(B)}$:- $$P(A=i|B>0) = \frac{P(A=i\cap B>0)}{P(B>0)} = \frac{{{4}\choose{i}}{{48}\choose{26-i}}\bigg{/}{{52}\choose{26}}}{\bigg{[}{{52}\choose{26}}-{{4}\choose{0}}{{48}\choose{26}}\bigg{]}\bigg{/}{{52}\choose{26}}}$$

My question is : (out of Protocols A and B) which method is the correct one? and Why?

Protocol A has already been used here to answer this question correctly. So, protocol A is obviously correct. So, I guess my question becomes: Why is protocol B incorrect?

note:- this question has not yet been answered(as of now, 2 answers have been posted, but none of them address my problem)

Problem

A deck of card is shuffled then divided into two halves of 26 cards each. A card is drawn from one of the halves, it turns out to be an ace. The ace is then placed in the second half-deck. The half is then shuffled, and a card is drawn from it. Compute the probability that this drawn card is an ace.

my attempt:-

let A = number of aces in the second deck before an ace added to it
let B = number of aces in the first deck before an ace is removed from it
let D = event that an ace is drawn from the second deck after an ace is added to it

Now, what we are asked to find is: $$\sum_{i=0}^3P(A=i|B>0)\cdot P(D) = \sum_{i=0}^3P(A=i|B>0)\cdot \frac{i+1}{27} $$

now, there are two ways of calculating $P(A=i|B>0)$

1. (Protocol A)restrict the sample space way:- since B has at least one ace, we restrict the sample space to 51 cards out of which 3 are aces. $$P(A=i|B>0) = \frac{{{3}\choose{i}}{{48}\choose{26-i}}}{{{51}\choose{26}}}$$
2. (Protocol B)$P(A|B) = \frac{P(A\cap B)}{P(B)}$:- $$P(A=i|B>0) = \frac{P(A=i\cap B>0)}{P(B>0)} = \frac{{{4}\choose{i}}{{48}\choose{26-i}}\bigg{/}{{52}\choose{26}}}{\bigg{[}{{52}\choose{26}}-{{4}\choose{0}}{{48}\choose{26}}\bigg{]}\bigg{/}{{52}\choose{26}}}$$

My question is : (out of Protocols A and B) which method is the correct one? and Why?

Protocol A has already been used here to answer this question correctly. So, protocol A is obviously correct. So, I guess my question becomes: Why is protocol B incorrect?

note:- this question has not yet been answered(as of now, 2 answers have been posted, but none of them address my problem)

Problem

A deck of card is shuffled then divided into two halves of 26 cards each. A card is drawn from one of the halves, it turns out to be an ace. The ace is then placed in the second half-deck. The half is then shuffled, and a card is drawn from it. Compute the probability that this drawn card is an ace.

my attempt:-

let A = number of aces in the second deck before an ace added to it
let B = number of aces in the first deck before an ace is removed from it
let D = event that an ace is drawn from the second deck after an ace is added to it

Now, what we are asked to find is: $$\sum_{i=0}^3P(A=i|B>0)\cdot P(D) = \sum_{i=0}^3P(A=i|B>0)\cdot \frac{i+1}{27} $$

now, there are two ways of calculating $P(A=i|B>0)$

1. (Protocol A)restrict the sample space way:- since B has at least one ace, we restrict the sample space to 51 cards out of which 3 are aces. $$P(A=i|B>0) = \frac{{{3}\choose{i}}{{48}\choose{26-i}}}{{{51}\choose{26}}}$$
2. (Protocol B)$P(A|B) = \frac{P(A\cap B)}{P(B)}$:- $$P(A=i|B>0) = \frac{P(A=i\cap B>0)}{P(B>0)} = \frac{{{4}\choose{i}}{{48}\choose{26-i}}\bigg{/}{{52}\choose{26}}}{\bigg{[}{{52}\choose{26}}-{{4}\choose{0}}{{48}\choose{26}}\bigg{]}\bigg{/}{{52}\choose{26}}}$$

My question is : (out of Protocols A and B) which method is the correct one? and Why?

Protocol A has already been used here to answer this question correctly. So, protocol A is obviously correct. So, I guess my question becomes: Why is protocol B incorrect?