2
$\begingroup$

I have this task about cards:

Consider choosing a card from a well-shuffled standard deck of 52 playing cards.

  • Suppose that, after the first extraction, the card is not reinserted in the deck. What is the probability that the second card is an ace?
  • Suppose that, after the first extraction, the card is reinserted in the deck. What is the probability that the second card is an ace?

For the first case, I've been thinking about $$\left(\frac{4*3}{52*51}\right) + \left(\frac{48*4}{52*51}\right)$$ Where I counted two examples: a)the ace was drawn already, and b) it wasn't. For the second part of the task, would it be $$\left(\frac{4*4}{52*52}\right) + \left(\frac{52*4}{52*52}\right)$$ I'm pretty sure that I'm assuming wrong, but I can't come up with anything else. Any help would be great!

$\endgroup$
  • 1
    $\begingroup$ Where does it say in the problem statement that the first card drawn was an ace? $\endgroup$ – David K Feb 5 '16 at 16:13
  • 3
    $\begingroup$ for both questions the answer is $4/52 = 1/13$ $\endgroup$ – Ant Feb 5 '16 at 16:16
  • $\begingroup$ @DavidK well it doesn't, that's why I counted it for both cases, since we do not know what was the first card. Could've been an ace, could've not been. $\endgroup$ – Jake Feb 5 '16 at 16:17
  • $\begingroup$ where do you have 52/52 from in the second case? $\endgroup$ – Scavenger23 Feb 5 '16 at 16:19
  • 4
    $\begingroup$ @Pauline mainly because of a symmetry argument. There is nothing special about the "first" card. You can very well say "look at the 13th card" or whatever, it does not make any difference; it would be different if you look at the first card and you see that it's NOT an ace (or you see that is IS an ace) but if you do not have any information, then probability does not change and it's still 1/13 for whatever card in the deck $\endgroup$ – Ant Feb 5 '16 at 16:24
1
$\begingroup$

As Ant said in the comments, in both cases, the answer is $\frac{4}{52} = \frac{1}{13}$. Your first equation simplifies to this: \begin{align*} \left(\frac{4\cdot 3}{52 \cdot 51}\right) + \left(\frac{48 \cdot 4}{52 \cdot 51}\right) &= \left(\frac{4}{52}\right)\left( \frac{3}{51} + \frac{48}{51}\right) \\ &= \left(\frac{4}{52}\right)(1) \\ &= \frac{1}{13} \end{align*} But this is actually overcomplicating things; drawing the first card doesn't affect the odds at all, regardless of whether or not it's replaced. We don't know anything about what the first card drawn was, so it doesn't add any information, and thus doesn't affect the probability that the second card is an ace.

$\endgroup$
1
$\begingroup$

The first question can be interpreted as:

If you give the cards a number then what is the probability that number $2$ is given to an ace?

The second question can be interpreted as:

If you give the cards a number then what is the probability that number $1$ is given to an ace?

(Here I mean the numbers $\{1,2,\dots,52\}$)

In the second cases the numbering takes place after the (irrelevant) extraction and reinserting of a card.

In both cases the answer is $\frac4{52}=\frac1{13}$

$\endgroup$
1
$\begingroup$

From a probabilistic point of view, extracting a card and placing it on same table of your deck without looking at it is like not extracting it at all: you have the same information you had before. More precisely given a probability space $(\Omega,\mathcal F,\mathbb P)$ your $\sigma$-algebra $\mathcal F$ is not changed, so your probability space is not changed.

Mathematically if $E_1=\{$Ace at the first extraction$\}$ and $E_2=\{$Ace at the second extraction$\}$ then $$P(E_2)=P(E_2\cap E_1)+P(E_2\cap E_1^c)$$ because $E_1$ and $E_1^c$ are a partition of $\mathcal F$.

From the definition of conditional probability $P(E_2|E_1)=\frac{P(E_2\cap E_1)}{P(E_1)}$, so $P(E_2\cap E_1)=P(E_2|E_1)P(E_1)$ and $P(E_2\cap E_1^c)=P(E_2|E_1^c)P(E_1^c)$.

Now, know the first card was an ace, means that you have 51 cards in your deck now and only 3 aces in that, so $P(E_2|E_1)=\frac 3 {51}$ and obviously $P(E_1)=\frac 4 {52}$. Similarly, knowing your first card wasn't an ace means you have 4 aces in a deck of 51 cards , so $P(E_2|E_1^c)=\frac 4 {51}$ and $P(E_1^c)=1-P(E_1)=\frac {48} {52}$. So $$P(E_2)=P(E_2\cap E_1)+P(E_2\cap E_1^c)\\=\frac 3 {51}\cdot\frac 4 {52}+\frac 4 {51}\cdot\frac {48} {52}\\=\frac 4 {52}[\frac 3 {51}+\frac {48} {51}]\\=\frac 4 {52}=\frac 1 {13} $$

$\endgroup$
0
$\begingroup$

For the first part,

$$\left(\frac{4*3}{52*51}\right) + \left(\frac{48*4}{52*51}\right) = \frac{4}{52} \left(\frac{3}{51} + \frac{48}{51}\right) = \frac{1}{13} \left(\frac{51}{51}\right) = \frac{1}{13} $$

which is the correct answer. The answer has to be $1/13$, because the answer must be the same if you ask the question with the word "king" instead of "ace", or "seven" instead of "ace", etc. There are exactly $13$ different words you can have where you wrote "ace", each producing the same non-zero probability.

The answer for the second one has to be $1/13$ for the same reason. But in this case,

$$ \left(\frac{52*4}{52*52}\right) = \frac{1}{13}, $$

so the term $\frac{4*4}{52*52}$ should not be added. It looks like you considered the case where the first card was an ace as a $4/52$ probability, but when you considered the "other" case you used a $52/52$ probability, so basically you counted twice the case where the first card is an ace.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.