Which of the two methods is correct and why?

Question

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?

Answer 1

I am having difficulty reading either your "protocol B" or quasi's answer and understanding where you are getting your numbers or what events you think you are using. To address this, I will go through a similar derivation below explaining my thoughts to organize myself and share it with you after it is done. Hopefully then we can see what went wrong.

We are drawing a card from the first half, seeing it is an ace, and then placing it in the second half. We then shuffle the second half and draw again and ask what the probability is that the next card drawn is an ace.

Let us define some events. $X$ is the event we drew an ace from the first deck. $Y$ is the event we drew an ace from the second deck. $A_i$ is the event that we began with $i$ aces in the first deck (thus making it $4-i+1$ aces in the second deck after the swap).

$\Pr(Y\mid X) = \dfrac{\Pr(Y\cap X)}{\Pr(X)}=\dfrac{\Pr(Y\cap X \cap (A_1\cup A_2\cup A_3\cup A_4))}{\Pr(X)}$

$=\dfrac{\Pr(Y\cap X\cap A_1) + \Pr(Y\cap X\cap A_2)+\dots + \Pr(Y\cap X\cap A_4)}{\Pr(X)}$

$=\dfrac{\Pr(Y\mid X\cap A_1)\Pr(X\mid A_1)\Pr(A_1) + \dots + \Pr(Y\mid X\cap A_4)\Pr(X\mid A_4)\Pr(A_4)}{\Pr(X)}$

$\dfrac{\frac{4}{27}\cdot \frac{1}{26}\cdot \binom{4}{1}\binom{48}{25}/\binom{52}{26} + \dots + \frac{1}{27}\cdot \frac{4}{26}\cdot \binom{4}{4}\binom{48}{22}/\binom{52}{26}}{\frac{4}{52}}$

$=\frac{43}{459}\approx 0.09368\dots$

Looking at portions of this calculation in greater detail, here we use that $A_0,A_1,\dots,A_4$ partitions the sample space into disjoint events and that $A_0\cap X=\emptyset$ so may be ignored. We then use the law of total probability to split the numerator up into separate probabilities. We then use the product rule to split each of these up further.

Now, as for $\Pr(Y\mid X\cap A_k)$ we have $4-k+1$ aces out of $27$ cards in the second deck at the time of drawing from the second deck, so the probability here is $\frac{4-k+1}{27}$. $\Pr(X\mid A_k)$ is us drawing one of $k$ aces from the first half deck with $26$ cards in it, and $\Pr(A_k)$ is a straightforward hypergeometric probability $\binom{4}{k}\binom{48}{26-k}/\binom{52}{26}$. In related posts there are claims about using $\binom{52}{26} - \binom{48}{26}$ as the denominator, but here we are looking purely at $\Pr(A_k)$ and are not conditioning on there being an ace available to draw at this point. Alternatively phrased, we could also have had $\Pr(X\mid A_0)\Pr(A_0)$ appearing in the calculations but this would have equated to zero.

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As to what went wrong in your calculation? Looking it over, I see now that you had been conditioning not on having successfully selected an ace in the first deck, but rather you were conditioning on it being possible to select an ace in the first deck whether or not one of the aces available to select happened to be the top card of the deck.

Answer 2

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:$$\frac1{27}+3p=\frac1{27}+\frac3{51}\frac{26}{27}=\frac{43}{459}$$