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We had this question arise in class today and I still don't understand the answer given. We were to assume that drawing cards are independent events. We were asked what the probability that the second card drawn is a queen if we take two from the deck. The answer given was 4/52, which seems counter-intuitive to me. How is the probability still 4/52 if there was a card drawn before it? What if the first card drawn was a queen?

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    $\begingroup$ If you are assuming that the drawing of cards are independent events, then it must be a drawing with replacement. $\endgroup$ – nathan.j.mcdougall May 18 '15 at 4:39
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    $\begingroup$ It's a red herring. The first and second cards are not independent, but it doesn't matter because you don't care about the first card at all. Taking two cards and looking only at the second is just a fancy way of drawing one card from the deck. $\endgroup$ – Robert Israel May 18 '15 at 4:48
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    $\begingroup$ "What if the first card drawn was a queen?" -- and what if the first card wasn't a queen? Now calculate the probability that either (first card is queen and second card is queen) or (first card is not queen and second card is queen). Not that this calculation helps with the intuition, but it should give you the correct answer. $\endgroup$ – Steve Jessop May 18 '15 at 9:39
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    $\begingroup$ Let $E_1,E_2,E_3,\dots,E_{11},E_{12},E_{13}$ be the events: second card drawn is A, 2, 3, . . ., jack, queen, king. As those $13$ events are mutually exclusive and exhaustive, their probabilities add up to one: $P(E_1)+P(E_2)+\cdots+P(E_{12})+P(E_{13})=1$. Now, if you can only manage to convince yourself that those probabilities are all equal, you can deduce that $P(E_{12})=\frac1{13}.$ Well, which do you think is more likely, getting a queen or getting a jack on the second draw? $\endgroup$ – bof May 18 '15 at 9:39
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    $\begingroup$ As with the Monty Hall problem, it's useful to extend the logic to the extreme. What if you take 51 cards and then the last card? What's the chance that the last card is a queen? In general, what's the chance p(N) that the N'th card is a queen? Obviously after 52 cards you must have had 4 queens to the chances better add up to 4.0. If the chance p(N) would decrease with N, p(1) would be the biggest and p(52) the smallest, so p(0) must be greater than 1/13 ! $\endgroup$ – MSalters May 18 '15 at 15:38

15 Answers 15

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There are two cases here:

Case 1: First card chosen is a queen

$$\frac{4}{52}*\frac{3}{51}=\frac{1}{221}$$

Case 2: First card chosen is not a queen.

$$\frac{48}{52}*\frac{4}{51}=\frac{16}{221}$$

Adding both the cases, we get $\frac{17}{221}$ = $\frac{4}{52}$ = $\frac{1}{13}$

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    $\begingroup$ Why on earth should one do two cases here? You could also amuse yourself in doing cases for the third card drawn, and the fourth, and so forth, but is there a point to it? $\endgroup$ – Marc van Leeuwen May 18 '15 at 13:22
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    $\begingroup$ @MarcvanLeeuwen: The point is, I think, to convince the OP that even if one does take the first card into account, the final probability is still the same. In other words, the analysis the OP would do is not wrong, even if it leads to a simpler end result that he had intuitively expected it to lead to. $\endgroup$ – Henning Makholm May 18 '15 at 13:59
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    $\begingroup$ @MarcvanLeeuwen: we do cases here because the question brought it up. $\endgroup$ – Joshua May 18 '15 at 20:31
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    $\begingroup$ @MarcvanLeeuwen The answer shows that there is no need to account for the different cases by showing that when you do account for both cases (which is allowed, just not required) the result is the same as when you don't account for both cases (the latter of which is implied rather than shown). $\endgroup$ – Jasper May 19 '15 at 15:55
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    $\begingroup$ Guys, don't be obnoxious. @A.J.'s answer is very effective in turning OP's mistake to the right direction. $\endgroup$ – André Chalella May 19 '15 at 15:55
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Think about it this way: Shuffle a deck of cards randomly. The probability of drawing a queen as your second card is the same as the probability that the second card in the deck is a queen, which is clearly 4/52.

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    $\begingroup$ What is "your probability"? I'm presuming you mean "the probability that the first card is a queen", but it isn't clear. Furthermore, you state "Your probability (of whatever) is the same as ...", but that feels like begging the question. Why are they the same. $\endgroup$ – jamesdlin May 18 '15 at 8:00
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    $\begingroup$ "Your probability" here means the probability the asker was asking about, namely the probability that when two cards are drawn from the deck in sequence, the second is a queen. The point of this answer is to get the reader to understand that in the absence of any information about the first card drawn, the second card drawn has an equal chance of being any card in the deck, just as in a shuffled deck the second card in the deck has an equal chance of being any card. $\endgroup$ – Hammerite May 18 '15 at 11:06
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    $\begingroup$ This is a good way of looking at it. The probability of any card in the deck being a queen is 4/52. The probability over a high enough number of runs doesn't change after drawing any number of other cards first. If you add the probabilities of all the cards in the deck you get 4 which is obviously correct because there are four queens in the pack! What is confusing is that when you draw a card and look at it, you forget that there was the other possibility. $\endgroup$ – CJ Dennis May 18 '15 at 12:11
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A slightly more intuitive way of looking at this:

The probability that the second card is a queen should be the same as the probability that the second card is an ace, and the same as the probability that the second card is a 2 etc. There are $ 13 $ possibilities for the card number/letter, so the answer is $ \frac{1}{13} $

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    $\begingroup$ This is the cleverest answer I have seen to this question. Entirely skips the debate around 1st, second card, etc. Rather changes the question altogether by thinking along a different dimension. Off the beaten path, and brilliant. $\endgroup$ – Amrinder Arora May 18 '15 at 17:42
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    $\begingroup$ I don't consider this intuitive at all. If the first card is an ace, then it is less likely that the 2nd card is an ace versus any other card. $\endgroup$ – Tony Ennis May 18 '15 at 22:55
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    $\begingroup$ @TonyEnnis Look at this: there's same number of aces (namely four) as queens, and shuffling makes no preference to one or the other. So the probability of getting each is the same. Similary the probability of getting any chosen value (4, 10 or King alike) at any pre-selected position (be it second or eleventh, or first or fifty-second) is the same. So for 13 possible values it is... what Cliff said. $\endgroup$ – CiaPan May 19 '15 at 8:44
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    $\begingroup$ @Tony - If the first card is an ace, then it is less likely that the 2nd card is an ace. From a card player's perspective, that's true. But, what happens when the first card is not an ace? Then, it is more likely that the second card is an ace. 1/13th of the time the probability will go down, while 12/13ths of the time the probability will go up. In the end, the numbers cancel out, and we're back to 1/13th. A.J.'s answer has the actual values. $\endgroup$ – J.R. May 19 '15 at 20:42
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You can draw a pair of cards by drawing the first card, then drawing the second card. Let's call these cards A and B. You're interested in the probability that card B is a queen.

Now consider a different experiment: draw a pair of cards as before, but this time call the first one card B, and the second one card A. I claim that these two experiments are identical. The reason is that for any two cards X,Y, the probability to draw X then Y is the same as the probability to draw Y then X.

The second experiment makes it clear that the probability that card B is a queen is 4/52, since there are 4 queens out of 52 cards.

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  • $\begingroup$ This is good. Does this assume that we don't look at the "other" card? $\endgroup$ – Tony Ennis May 18 '15 at 22:59
  • $\begingroup$ You can look at the other card if you want. $\endgroup$ – Yuval Filmus May 18 '15 at 23:46
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    $\begingroup$ @TonyEnnis What you have drawn does not depend on whether you look at it (or at anything else)... $\endgroup$ – CiaPan May 19 '15 at 8:53
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Probablities are a hard thing to wrap your mind around. Let's try a more feeling-based approach:

  1. Take a deck of cards
  2. Shuffle the cards
  3. Flip over the second card of the deck and check if it is a queen, while leaving the first card on the deck

Perhaps you can see how the chance is the exact same ($\frac{4}{52}$) if you had picked the top card instead of the second to top. Chance, after all, does not play favorites between the two cards.

Then, consider the following set of actions:

  1. Take a deck of cards
  2. Shuffle the cards
  3. Draw the first card
  4. Draw the second card and check if it is a queen

Now, you should be able to see that the chances for the two cases are the exact same. After you shuffle the deck, it doesn't matter if you draw two cards and check the second or you just check the second card without looking at the first; if the second card is a queen, it's a queen, if it's not, it's not.

The flip side of the feeling-based approach is the math to back it up. Let's start with the simple case: you have 52 cards, and want a queen on the second draw. (This has also been done by other answers, but I'll repeat it here.)

You draw a queen on the second draw if:

  • You draw a queen on the first draw and one on the second $$\frac{4}{52}*\frac{3}{51}=\frac{12}{2652}=\frac{1}{221}$$
  • You draw something other than a queen on the first draw and a queen on the second $$\frac{48}{52}*\frac{4}{51}=\frac{192}{2652}=\frac{16}{221}$$

So in total, the chance is:

$$\frac{1}{221}+\frac{16}{221}=\frac{17}{221}=\frac{4}{52}$$

Now, let's up the ante a little bit. Rather than wanting to know something about queens in a complete deck, I want to know about the more general case. I have a pile of $n$ shuffled cards. In that pile, I know there are $p$ cards that I "like". What I want to know is: what is the chance I draw a card I like.

For the first card, it's simple. The chance simply is $\frac{p}{n}$.

For the second card, we once again have two options:

  • I like both the first and second card $$\frac{p}{n}*\frac{p-1}{n-1}=\frac{p^2-p}{n^2-n}$$
  • I like the second card, but not the first $$\frac{n - p}{n}*\frac{p}{n-1}=\frac{pn-p^2}{n^2-n}$$

Adding the two, you get:

$$\frac{p^2-p}{n^2-n}+\frac{pn-p^2}{n^2-n}=\frac{p^2+pn-p^2-p}{n^2-n}=\frac{pn-p}{n^2-n}$$

Moving things around a bit more:

$$\frac{pn-p}{n^2-n}=\frac{p(n-1)}{n(n-1)}=\frac{p}{n}$$

Which is the same as the chances for the first card. So, now I can say that no matter the deck size or the amount of cards that represent "success", it doesn't matter if I look to the first or second card to determine success. (Of course, if I look at the second card, it's important that I don't care what the first card is at all.)

I could actually repeat the experiment for each different card in the deck, and then I could draw the conclusion that in general: it doesn't matter if I look at the first or second card, the chances for the card to be a specific one are equal.

The next step could be to proof that the the other cards (third, fourth, etc) have the same chance as well, but I'll leave that as an exercise for the reader.

Disclaimer: my proof probably isn't elegant, optimal or nice, but I do believe it to be correct.

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  • $\begingroup$ As pointed out by user CiaPan, a simpler way to proceed after "Moving things around a bit more:" is $\dfrac{pn-p}{n^2 - n} = \dfrac{p(n-1)}{n(n-1)} = \dfrac{p}{n}$. $\endgroup$ – epimorphic May 19 '15 at 13:38
  • $\begingroup$ To CiaPan: Simplifications are probably best communicated in the comments, given the prohibition against "drastic" edits to non-CW posts authored by others and given that one of the purposes of comments is to "suggest improvements". If you are to make an edit, please respect the structure and flow of the post. Don't add colors if the original post doesn't have them, and don't sign the edit with your name. $\endgroup$ – epimorphic May 19 '15 at 13:39
  • $\begingroup$ @epimorphic, CiaPan Thanks. I was really struggling with the proof there myself and this is much simpler. $\endgroup$ – Jasper May 19 '15 at 15:19
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It seems to me that the problem here is not distinguishing between the prior probability and the conditional probability.

If we observed that the first card drawn was a queen, then that would give us a lowered probability of the second card being a queen of $3/51$.
However, this is not the same probability as the one asked of in the problem, but the conditional probability given the first card being a queen.

If we instead had had the first card not be a queen, then the conditional probability would have been heightened to $4/51$.
The prior probability could then be calculated using the probability of both cases as $$\frac{4}{52}\cdot\frac{3}{51}+\frac{48}{52}\cdot\frac{4}{51}=\frac{4}{52}$$

However, the problem can be simplified as what is important isn't the drawing of the card from the deck, but the observation of the value of the card.

As the value of the first card is never considered in the problem, it can simply be considered as not being drawn at all, which brings us to a second point;
If instead of drawing two cards from the deck, you simply fanned out the cards and drew the second card from the top, you would still get the same probabilities as if you chose the top card.

In fact, you would get the same probabilities even if you chose the tenth card or even a card at random.
There is nothing special about the top card of the deck unless you choose to do something with it.

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Consider a more complete event:

When drawing all 52 cards, what is the probability that the second card is a queen?

The reason of changing the original question to this is because this is an expanded process, in other words, if you do draw all cards you certainly drop the first two ones. To solve the new problem is not difficult. One way is to consider every card of the 52 has an equal chance be at the second place. So the probability is 4/52.

It is equal to the same probability that the first card, the 3rd card or the 27th card etc. is a queen, or any other type of card.

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  • $\begingroup$ If the first card you draw is a queen than the second has a different probability. $\endgroup$ – hownowbrowncow May 18 '15 at 13:06
  • $\begingroup$ @hownowbrowncow - how, now? $\endgroup$ – user165601 May 18 '15 at 14:13
  • $\begingroup$ @GalacticCowboy because then there would be 3 queens within the deck :D $\endgroup$ – hownowbrowncow May 18 '15 at 14:21
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    $\begingroup$ @David The author didn't specify that one would replace the drawn card back into the deck and therefore, I would assume, one would not. Thus changing the probability with each additional drawing of a card (the second draw would also have 51 cards within the deck, btw). Perhaps there is replacement and perfect shuffling after every draw then it's 4/52 all the way down the line. $\endgroup$ – hownowbrowncow May 18 '15 at 14:27
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    $\begingroup$ @hownowbrowncow - A.J.'s answer demonstrates that, even if you consider the first card, the overall probability of card #2 being a queen is still the same. $\endgroup$ – user165601 May 18 '15 at 14:31
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2 things to consider. What if the problem stated draw all cards and what is the probability that the last card drawn is a queen then what would your answer be? You might think it is close to 0 because in 51 cards drawn, there is a very good chance that all queens would be already drawn by then. However, the answer should still be 4/52 because there are 4 queens in the deck and each has an equal chance of being the last card drawn just as all the other cards have.

Also, if you notice on TV when they have Texas Hold 'em, they "waste" the top card when drawing community cards (turn, river...) so if that changed the probability, it would likely be disallowed.

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put the first card back in the deck; now you've only drawn one card.
the chance it's queen 4/52

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Imagine you buy a ticket in a lottery, where a fraction $p$ of the tickets will win. (From the comments I gather that there are some people who are confused by all kinds of newfangled games with different rules, such as being able to compose the number on one's own ticket; I therefore emphasise this is just an old fashioned lottery with a fixed set of tickets established beforehand, all distinct and all sold before the drawing; drawing then determines a subset of the tickets as winners, with the size of that subset being $p$ times the total number of tickets.) What is your chance of winning? It would be hard to argue that it is anything other then $p$.

But now let us add that before buying a ticket, you had to wait in line, and you happen to notice that the person before you also bought a ticket for the same lottery. Of course you don't know whether her ticket will win, but what is the chance that your ticket will win? It is still $p$. But what if her ticket actually wins? Well with that new information your chances will no doubt be less, but you don't have that information. If her ticket actually looses, that will slightly improve your odds, but you don't know that either. In fact you can be sure that there are hundreds of people who also bought tickets, but as long as you know nothing about their results (and you won't until the winning lots are drawn), it will not affect your chances the slightest bit.

The situation you describe is entirely similar to this.

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  • $\begingroup$ Why would my chances of winning a lottery be less if I know the person in front of me in line already won? Many lotteries share the jackpot if there are multiple winners. In that case, my winnings will be reduced but not my chance of winning. Also if that person loses it will not increase my chance of winning at all. The only exception is if I choose the same numbers as some other person in a lottery (such as 5, 10, 15, 20, 25, and 30). Then if I know that person won then my chances of winning are likely affected (depending on the type of lottery). You didn't say a fixed fraction will win. $\endgroup$ – David May 18 '15 at 11:38
  • $\begingroup$ It wouldn't be p if the lottery/drawing is biased (which some are). For example, if people put their name on a piece of paper and place it into a container to be drawn. It would depend if they fold it and how and also when they put it in the container since they may not be mixed well before being drawn. I've actually seen this happen in real life where some last minute entry won because they were not mixed well and they just drew her name from the top. $\endgroup$ – David May 18 '15 at 11:51
  • $\begingroup$ @David "Imagine you buy a ticket in a lottery, where a fraction p of the tickets will win." <-> "You didn't say a fixed fraction will win." Please elaborate. $\endgroup$ – Alexander May 18 '15 at 12:18
  • $\begingroup$ @David: I'm using a very simple lottery model, where all tickets are sold, and then a fraction $p$ is drawn as winning lots. Obviously if I know another ticket that wins, that will decrease the chances that my ticket too is among the winners (in an extreme example there is only one winning ticket, and knowing another ticket wins ensures tha mine won't; but even if there are say 2 wins out of 100, then $\frac1{99}$ is worse than $p=\frac2{100}$). I don't know what kind of lottery you are thinking of that would not see my chances diminished by knowing a winner that is not my ticket. $\endgroup$ – Marc van Leeuwen May 18 '15 at 12:55
  • $\begingroup$ @David: Re-reading what you wrote, I really cannot see what you are worrying about, you must have a strange idea of a lottery. It is not about writing anything on tickets, or putting them in a container, and there cannot be "identical" tickets. Tickets are simply a fixed known collection, say numbers $1$ to $n$, and when all are sold then some $k$ of these numbers are "drawn" (from a separate copy of those numbers) to be winning numbers. The chance of winning with one ticket is $p=\frac kn$. Also I don't see what complaining about biased-ness in real-world lotteries has to do with this. $\endgroup$ – Marc van Leeuwen May 18 '15 at 13:10
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Can I convince you of this statement:

If I draw two cards, the probability that the first card is a queen is the same as the probability that the second card is a queen.

If you believe this, then your question has the same probability as the following the situation:

What is the probability that, if I draw two cards, the first card is a queen?

It is clear now that the second card pull in this case has no effect on the first, just as how the first card pull has no effect on the second in your case.

So the problem breaks down into:

What is the probability that, if I draw a card from a standard deck, the card is a queen?

which of course is $\frac{4}{52} = \frac{1}{13}$.

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Let us find out, the $E$ be the event of drawing a queen in the second draw. Let $X_1$ be the event of drawing a queen at the first draw and $X_2$ is a card other than queen in the first draw. So the total probability is $P(E|X_1)+P(E|X_2)= 4/52*3/51+48/52*4/51=4/52$.

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Key word is independent. Probability of nth card being a queen is 4/52. The question is to teach independent.

Which means that the chances don't change because of the first draw . (whatever card is drawn , it should be put back in the stack) otherwise the second draw is dependent on the first draw.

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    $\begingroup$ How is independence used in this question or answer? $\endgroup$ – JiK May 19 '15 at 10:49
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This is basically the Monty Hall problem -- as long as you are talking in advance of the draw, the probability is unchanged, once you draw the odds adjust.

You have two goats and a car, if you open two doors, what is the chance that there is a goat behind the second door? What is the chance it is a car? 2/3 and 1/3 respectively.

After you open the first door, and find that it is a car, you can say that there is zero chance that there is a car behind the second door. But at that point, they are no longer independent events, which is why in the Monty Hall problem you always switch.

In this case you are effectively saying that the second card is a queen, and you are doing this before you know what the first card was. The chance that that is correct is 1/13. It (the chance you were correct) will remain 1 in 13 until either the second card is revealed or the position of all queens have been determined, at which point it is not a matter of probability.

In both cases the odds of you having identified the card is unchanged, what changes is the current odds of that card being a queen (ie is this one of the 1 in 13 times where you were correct).

This is where people go wrong with the Monty Hall problem, they confuse the current odds of the car being behind the door with the odds that they correctly identified the door with the car. In this instance the confusion is between having identified the card vs the card actually being.

Here's another way of looking at it. Your odds of correctly predicting card 2 is 1 in 13. So take 4 decks of cards, divide them into 13 rows by suit (throwing away the 3 extra suits). Then pick a value for the 2nd card in each row. Now start randomly revealing cards until each row has been determined -- do you expect this random reveal to result in you having correctly guessed more than the expected 1 row? Is it any different from just revealing the 2nd card of each row? Now, suppose that someone who knows what the cards are does the reveal, revealing each card until there are only 2 cards left per row -- when you do the final reveal for each row, how many rows would you expect to have guessed correctly?

Your chance of predicting the card in advance is 1 in 13 for each row, and it remains 1 in 13 -- revealing cards doesn't change the odds of your having guessed correctly. Your guess is in the past and is now fixed, forever unchanging. Either it was right or it wasn't. What changes (assuming random reveal) is my current odds of guessing whether you were right or not. As with the related Deal Or No Deal game, I get additional information with each random reveal.

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    $\begingroup$ You start by saying that this is basically the Monty Hall problem, and then you explain that the Monty Hall problem is different from this. I don't understand your explanation; in the Monty Hall problem, no doors with a car are opened first. $\endgroup$ – JiK May 19 '15 at 10:52
  • $\begingroup$ @JiK: The surface elements, cards vs goats, and the number of items, 3 vs 52 are different, but the basic "probability of original guess remains unchanged despite wrong guesses being removed" is the same. $\endgroup$ – jmoreno Nov 5 '17 at 12:43
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I would think it would be slightly below 1/13 since 1/13 is the probability for the first pick being a queen then the second round should account for the possibility of there only being 3 queens in the deck.

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    $\begingroup$ As noted, the same logic holds for a jack, so it's chance also would have to be < 1/13 because there might be only 3 jacks left. That means the chance of drawing any second card are <1.0 - it literally doesn't add up. $\endgroup$ – MSalters May 18 '15 at 15:34

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