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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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  • $\begingroup$ @Brad: I know Bayes' theorem, but I'm not sure how it is to be applied in this case. $\endgroup$ Jan 6, 2012 at 6:13

4 Answers 4

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Here are two solutions:
Let $A$ be the event an ace is drawn from the second half
Let $I$ be the event that the interchanged card is selected

$P(A) = P(A|I)\cdot P(I) + P(A|I^c)\cdot P(I^c)$
$P(A) = 1\cdot\dfrac{1}{27} + \dfrac{3}{51}\cdot\dfrac{26}{27}$
$P(A) = \dfrac{43}{459} = 0.094$

Note that $P(A|I^c) = \dfrac{3}{51}$ because there are 3 aces to choose from and all 51 other cards are equally likely to be in the second half

Second Solution:
Let $A$ be the event an ace is drawn from the second half
Let $N_i$ be the event that $i$ aces are in the second half with $0 \leq i \leq 3$

$P(A) = \sum_{i=0}^3 P(A|N_i)\cdot P(N_i)$
$P(A) = P(A|N_0)\cdot P(N_0) + P(A|N_1)\cdot P(N_1) + P(A|N_2)\cdot P(N_2) + P(A|N_3)\cdot P(N_3)$ $P(A) = \dfrac{1}{27}\cdot P(N_0) + \dfrac{2}{27}\cdot P(N_1) + \dfrac{3}{27}\cdot P(N_2) + \dfrac{4}{27}\cdot P(N_3)$

Note that there are only 3 aces to choose from and 51 cards total, since one ace is in the first half

$P(N_0) = \dfrac{\dbinom{3}{0}\dbinom{48}{26}}{\dbinom{51}{26}}$
$P(N_1) = \dfrac{\dbinom{3}{1}\dbinom{48}{25}}{\dbinom{51}{26}}$
$P(N_2) = \dfrac{\dbinom{3}{2}\dbinom{48}{24}}{\dbinom{51}{26}}$
$P(N_3) = \dfrac{\dbinom{3}{3}\dbinom{48}{23}}{\dbinom{51}{26}}$

$P(A) = \dfrac{43}{459} = 0.094$

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    $\begingroup$ I have downvoted this answer, as (1) the lack of appropriate formatting using MathJax makes it very hard to read and (2) the question already has several answers, and I don't see how this answer adds much that hasn't already been said. $\endgroup$
    – Xander Henderson
    Jun 9, 2018 at 23:18
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    $\begingroup$ All the other answers are incorrect. The probability is 0.094, not 0.10678 $\endgroup$
    – nimo956
    Jun 9, 2018 at 23:29
  • $\begingroup$ I cleaned up the formatting. $\endgroup$
    – nimo956
    Jun 9, 2018 at 23:47
  • $\begingroup$ First solution needs formality. It seems a bit vauge why your solution works :/ $\endgroup$ Sep 24, 2023 at 8:00
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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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When solving a word problem the student should always remember that they are looking for a mathematical model to match the words. This question looks like it was set up by magician taking a break from his usual job of shuffling a deck and tricking his audience.

You can approach this problem by setting up an equivalent word problem that sides steps the shuffling. Then you break out the $27$ mutually exclusive "examine card" events of the sample space:

Answer to "A First Course in Probability:, Sheldon Ross
$\quad\quad\quad\quad$ Chapter 3, Exercise 37:

$$ (\frac{1}{27}\cdot 1) + \sum_{k=2}^{27} \,(\frac{1}{27}\cdot \frac{3}{51}) = \frac{1}{27} + \frac{26}{9\cdot51} = \frac{43}{459} $$

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Let $A$ be the event that an ace is drawn from the first half of the deck.

Let $X$ be the count of aces in the second half-deck before the additional ace is added from the first deck.

$$\mathsf P(X=x\mid A) ~=~ \dfrac{\dbinom{4}{x}\dbinom{48}{26-x}\mathbf 1_{x\in\{0,1,2,3\}}}{\dbinom{52}{26}-\dbinom{48}{26}}$$

Let $B$ be the event of drawing an ace after adding one and shuffling.   There are now $X+1$ aces among $27$ cards.

$$\mathsf P(B\mid X=x, A)~=~\dfrac{x+1}{27}$$

By the Law of Total Probability.

$$\begin{align}\mathsf P(B\mid A)~=~& \sum_{x=0}^3 \mathsf P(B\mid X=x, A)\mathsf P(X=x\mid A) \\[1ex] ~=~& \dfrac{1\cdot \dbinom{3}{0}\dbinom{48}{25}+2\cdot \dbinom{3}{1}\dbinom{48}{24}+3\cdot\dbinom{3}{2}\dbinom{48}{24}+4\cdot \dbinom{3}{3}\dbinom{48}{23}}{27~\left(\dbinom{52}{26}-\dbinom{48}{26}\right)}\end{align}$$

Remark: Since we are given that event $A$ has happened, the probability that we seek is indeed conditional on this event: $\mathsf P(B\mid A)$

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  • $\begingroup$ this post seems to contain a lot of trivial mistakes. $\endgroup$
    – abhishek
    Aug 15, 2020 at 14:37

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