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"A deck of cards is shuffled and 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."

Source : A First Course in Probability, Sheldon Ross, Chapter 3, Exercise 37

(My intention was not to be lazy and let the community do all the work for me, but I think that is what has been misunderstood seeing the 2 dislikes this question received. I just thought I should not clutter the question with more text than necessary)

What I've tried : Probability that the second half already contained 0,1,2 or 3 aces before the ace from the first half was added to it. Then, when the new ace was added to it from the first half, we could calculate the probabilities of drawing an ace considering each of these cases, and add them to get the answer. But I am not sure how do I calculate the P(second half contained 0/1/2/3 aces)..

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What have you tried so far? What relevant theory are you aware of? –  Brad Jan 6 '12 at 5:10
    
@Brad: I know Bayes' theorem, but I'm not sure how it is to be applied in this case. –  user118102114 Jan 6 '12 at 6:13

2 Answers 2

This is a nice application of Bayes' theorem/conditional probabilities. Well, not really Bayes' theorem since we aren't inverting any conditional probabilities, but we need to express a probability as a weighted sum of conditional probabilities.

Let $A$ be the event that we draw an ace the second time, and let $P(i)$ be the probability that there are $i$ aces originally in the second pile and $P(A|i)$ be the probability that we draw an ace conditional on the second pile starting with $i$ aces. We have

$$P(A)=\sum_{i=0}^{3} P(A|i)P(i)$$

It is easy to compute $P(A|i)=\frac{i+1}{27}$. Computing $P(i)$ is slightly more involved. We need to compute the number of possible half-decks with $i$ aces, which we will denote $N_i$, and then $P(i)=\frac{N_i}{\sum_{j=0}^{3}N_j}$.

If we have $i$ aces and $26-i$ other cards (out of the 48 non-aces), then there are $\binom{4}{i}\binom{48}{26-i}$ possible hands.

Computing the answer is now a matter of plugging into the formula. Note that true blue anil's answer is slightly incorrect (if you are trying to do a comparison between the responses), as when we computed our probabilities, we divided by $N_0+N_1+N_2+N_3$, and not by $\binom{52}{26}=N_0+N_1+N_2+N_3+N_4$. Also note that if one wants an exact numerical answer, the binomial coefficients can be expanded in terms of factorials, and after getting a common denominator, there will be significant cancellation.

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# of ways the 2nd half-deck could have 0,1,2,or 3 aces $=$

$$\binom{48}{26}+\binom41\binom{48}{25}+\binom42\binom{48}{24}+\binom43\binom{48}{23}$$

It now has 1 ace added to it, so

$$\begin{align*} \mathrm{P}[\text{ace drawn from it}] &= \frac1{27}\binom{52}{26}^{-1}\left(\binom{48}{26}+8\binom{48}{25}+18\binom{48}{24}+16\binom{48}{23}\right)\\\\ &\approx .10088 \end{align*}$$

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Since the original first half contained an ace, not all $\binom{52}{26}$ splits are actually possible. –  Brian M. Scott Jan 6 '12 at 9:51

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