I am currently learning about conditional probability from "The Book of R" by Tilman Davies and I'm having trouble understanding a conditional probability problem. The problem is as follows:

You randomly draw a card, and after replacing it, you draw another. Let A be the event that the card is a club; let B be the event that the card is red. What is Pr(A|B)? That is, what is the probability the second card is a club, given the first one was a red card? Are the two events independent?

This assumes your standard 52 card deck. I understand that Pr(B) = 26/52, however where I get stuck is when the question asks about clubs which are black and not red. This means that it is impossible to pull a club out of the new sample space (now 26) because clubs are black and not red. The answer key to the book says the final value is 13/52 but I have absolutely no idea how that conclusion was reached. I have always struggled with understanding probability so any help anyone could provide I would greatly appreciate.

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    $\begingroup$ What makes you think the sample space has been reduced to only red cards? $\endgroup$ – Ruben van Bergen Jun 6 '17 at 7:16
  • $\begingroup$ I thought that was what was the word "given" always implied in a conditional probability word problem, or that's how I interpreted it at least. $\endgroup$ – user164161 Jun 6 '17 at 12:11
  • $\begingroup$ As Glen_b pointed out, there is some poor wording in the second sentence of the problem, which I missed at first and might have thrown you off. However, it seems that A is meant to be the event "The first card drawn is a club" while B is the event "The second card drawn is red". So yes, you are given the fact that A occurred (i.e. the first card was indeed a club), but since this first card was put back in the deck, this tells you nothing about the probability of the second card being red. $\endgroup$ – Ruben van Bergen Jun 6 '17 at 12:49
  • $\begingroup$ Here's my question. Since we know the first event occurred, shouldn't it limit the overall sample space size to the number of possible outcomes for that first event (in this case 26 for red cards)? That's how I interpreted and obviously I'm wrong but to be honest I have no idea why I'm wrong. And why would the sample space be reduced when lets just say we were asking for black cards for the club suit and not red? This is beyond confusing and why I've always had problems with probability. It's not that I hate it, it just sometimes makes no sense at all. $\endgroup$ – user164161 Jun 6 '17 at 13:35
  • $\begingroup$ Here is a thought experiment for you. Suppose that instead of replacing the first card in the deck and drawing again, you were to take a different (shuffled) deck and draw the second card from it. (1) Intuitively, wouldn't you consider separate draws from separate decks to be statistically independent? Could you demonstrate that they are independent? (2) How, if at all, would using a second deck change the probabilities in the question? $\endgroup$ – whuber Jun 6 '17 at 15:17

Wow, that's poor notation$^\dagger$ they have in the question; it doesn't distinguish which draw the events apply to properly.

Before you draw any cards you have 26 red cards, 13 clubs (and 13 spades). After the first draw is red, they say you replace it (presumably shuffling again), so you still have 26 red cards, 13 clubs and 13 spades. At the second draw there are 52 cards, of which 13 are clubs; the chance of drawing a club is $\frac{13}{52}$. (Since the first-drawn card was replaced it changes nothing; you're exactly in the situation you started with)

$\dagger$ A better notation would be: Let $R_i$ be the event "a red card was drawn at the $i$th card draw" and let $C_i$ be the event "a club was drawn at the $i$th card draw; then the problem is "What is $P(C_2|R_1)$?" which is useful to distinguish from various other possible questions. Ambiguous notation is a recipe for confusion.

In reply to comment:

The actual sample space for two draws is the $52\times 52$-element set consisting of all possible pairs of cards from $(A♡,A♡)$ to $(K♠,K♠)$. When you draw a red card on the first draw, you restrict attention to the $26\times 52$ set that has only red cards on the first draw.

Given that, the events with a club on the second draw are the $26 \times 13$ cases that have a red on the first draw and a club on the second draw, and the conditional probability is $\frac{26\times 13}{26\times 52}$. Note that the first draw simply cancels out of numerator and denominator, because the two draws are independent; you can simply ignore it, because it contributes no information.

[Notation that ignores that there are two draws encourages you to make the error of thinking the sample space only has 52 elements.]

  • $\begingroup$ Ok, I still really don't understand. From how I learned conditional probability when an event is given to occur I though this limited the number of samples you have. I understand that overall the card being put back brings the total sample space S size back to 52, but wouldn't the final probability be 13/26 due to the fact that red occurred and that sub-sample space has 26 possible outcomes? This is why I've always had trouble with probability because these explanations are about as conceptually clear as mud to be honest. $\endgroup$ – user164161 Jun 6 '17 at 13:28
  • $\begingroup$ See the edit at the end of my answer. $\endgroup$ – Glen_b Jun 6 '17 at 14:59
  • $\begingroup$ This explanation helps out a lot! How I figured it out is that since the first draw is a red it doesn't "limit the outcome choices" of the second event if that makes any intuitive sense. It also appears that I wasn't computing the sample space size properly since I assume the whole sample space was 52 and not 52 X 52 representing the pairs of draws. Now it makes sense why you said the notation was confusing because I wouldn't have guessed that. Thanks again for the help and this points me in the right direction. $\endgroup$ – user164161 Jun 6 '17 at 15:50
  • $\begingroup$ Also I assume that your calculation is based on the conditional probability formula? $\endgroup$ – user164161 Jun 6 '17 at 16:02
  • $\begingroup$ I'm not sure which conditional probability formula you mean. Are you referring to $P(A|B) = \frac{P(AB)}{P(B)}$? My initial answer is based on reasoning about independence and then applying elementary counting arguments. The discussion about the 52 x 52 sample space at the end can be seen as applying that formula. $\endgroup$ – Glen_b Jun 6 '17 at 22:48

You randomly draw a card, and after replacing it, you draw another. Let A be the event that the card is a club; let B be the event that the card is red. What is Pr(A|B)? That is, what is the probability the second card is a club, given the first one was a red card? Are the two events independent?

Let's interpret this paragraph for you. I believe the problem you are having is one of English. Specifically, the word given is not quite what you think it is. You ought to interpret the word given to only mean that if I tell you that A has happened, now what are the chances that B will happen.

  1. You have a deck of 52 cards
  2. You pull one out, it is a red card, you put it back in. (the text specifically says, "and after replacing it")
  3. You pull out a new card (from the same 52 card deck)
  4. What are the odds this new card is a club?

In some cases, the first activity might limit the sample size for the second event. In this particular case, it does not.


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