# Why is this counting way wrong?

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.

This is how I count based on the hint given in this thread Question regarding Conditional Probability with a deck of card.
Assuming the 27-half deck will contains from 1, 2, 3, 4 aces, then I conditioned on how we can distribute the a 52 cards into 2 decks, each has 26 cards. Let $A$ denote the event that the card chosen from this 27 half-deck is ace, and $D_i$ denotes this second half deck (before adding an ace) contains $i$ aces. $$P(A \mid D_0)P(D_0) + P(A \mid D_1)P(D_1) + P(A \mid D_2)P(D_2) + P(A \mid D_3)P(D_3)$$ where $$P(A \mid D_0)P(D_0) = \dfrac{1}{27} \cdot \dfrac{\binom{48}{26}}{\binom{52}{26}} = \dfrac{46}{22491}$$ $$P(A \mid D_1)P(D_1) = \dfrac{2}{27} \cdot \dfrac{\binom{4}{1}\binom{48}{25}}{\binom{52}{26}} = \dfrac{416}{22491}$$ $$P(A \mid D_2)P(D_2) = \dfrac{3}{27} \cdot \dfrac{\binom{4}{2}\binom{48}{24}}{\binom{52}{26}} = \dfrac{325}{7497}$$ $$P(A \mid D_3)P(D_3) = \dfrac{4}{27} \cdot \dfrac{\binom{4}{3}\binom{48}{23}}{\binom{52}{26}} = \dfrac{832}{22491}$$

Thus, $$P(A) = \dfrac{46}{22491} + \dfrac{416}{22491} + \dfrac{325}{7497} + \dfrac{832}{22491} = \dfrac{2269}{22491}$$

The correct answer is $\dfrac{43}{459}$. My answer is off by 0.007 percent, but I couldn't figure out why it was wrong. I understand the way in the book, but I just want to see how can I count it in a different way. Any idea?

Update
Following the hints given by Didier Piau and TonyK, I got: $$\dfrac{1}{27} \cdot \dfrac{\binom{48}{26}}{\binom{51}{26}} = \dfrac{92}{22491}$$ $$\dfrac{2}{27} \cdot \dfrac{\binom{3}{1}\binom{48}{25}}{\binom{51}{26}} = \dfrac{208}{7497}$$ $$\dfrac{3}{27} \cdot \dfrac{\binom{3}{2}\binom{48}{24}}{\binom{51}{26}} = \dfrac{325}{7497}$$ $$\dfrac{4}{27} \cdot \dfrac{\binom{3}{3}\binom{48}{23}}{\binom{51}{26}} = \dfrac{416}{22491}$$

$$P(A) = \dfrac{92}{22491} + \dfrac{208}{7497} + \dfrac{325}{7497} + \dfrac{416}{22491} = \dfrac{43}{459}$$

• The arithmetic is correct. – TonyK Oct 9 '11 at 11:57

Updated to add: The second part of my answer (Protocol $B$) was, I think, correct. But the first half failed to take Didier Pau's correction into account: it should be $\binom{51}{26}$ in the denominator, not $\binom{52}{26}$. Also, I think, $\binom{4}{i}$ in the numerator should be $\binom{3}{i}$.
End of update

The reason for the discrepancy is quite subtle. Suppose we do this (call it Protocol $A$):

1. Remove an Ace from the deck
2. Deal 26 of the remaining cards at random
3. Add the Ace that we removed to these 26 cards
4. Pick one of the resulting 27 cards at random

Then I think your calculations are correct. But that's not quite what happened, if we take the natural interpretation of the question. There is nothing special about Aces in this context, so we can recast it as follows (Protocol $B$):

1. Divide the deck into two halves
2. Select a card at random from the first half of the deck; suppose its rank is $R$
3. Add the selected card to the second half of the deck
4. Pick one of the resulting 27 cards at random

Now the question asks the probability that the chosen card has rank $R$. The situation has changed slightly from Protocol $A$, because the fact that a card of rank $R$ was chosen from the first half of the deck increases the prior probability that the first half of the deck contained more such cards. So in this situation, the probability of choosing an $R$ is slightly less.

Edited to add: I don't know how the book did it, but I got the same answer, like this:

If you pick the card that was added (with probability $1/27$), the probability that its rank is $R$ is $1$.
If you pick another card (with probability $26/27$), the probability that its rank is $R$ is $3/51$.
So the probability of picking an $R$ is $$\frac{1}{27} \times 1 + \frac{26}{27} \times \frac{3}{51} = \frac{43}{459}$$

• Your solution is identical with the one in the book ;). Thanks for the explanation though, I still couldn't find the right solution for the counting method I gave. – Chan Oct 10 '11 at 0:38

The error lies in the denominators ${52\choose 26}$. These count the number of ways of choosing $26$ cards from $52$ to make your first deck. But this is not what you should count since you know there is at least one ace in the other deck. Hence you should discard all the splits where the $4$ aces landed in the first deck.

• Aren't you going to do the sums for us? – TonyK Oct 9 '11 at 11:58
• @TonyK, What do you mean? – Did Oct 9 '11 at 12:22
• I mean: What answer do you get? ("sums" = "arithmetic".) – TonyK Oct 9 '11 at 12:29