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You have a standard deck of cards and randomly take one card away without looking at it and set it aside. What is the probability that you draw the Jack of Hearts from the pile now containing 51 cards?

I'm confused by this question because if the card you removed from the pile was the Jack of Hearts then the probability would be zero so I'm not sure how to calculate it.

Edit: I asked this question about a year ago because I was struggling to get an intuitive understanding of an important concept in probability, and the comments and answers were really helpful for me (especially the one about "no new information being added so the probability doesn't change").

J. Zimmerman
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    Hint: $P(A)=P(A|B)P(B)+P(A|\bar B)P(\bar B)$. – Arnaud Mortier May 22 '18 at 01:39
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    Removing the top card has given you no new information. It is the exact same as the Jack of hearts being the top card on the deck before you burned one card. – Doug M May 22 '18 at 01:40
  • So even if you removed the Jack of hearts originally the probability would still be 1/52 even though the probability of drawing it from that pile is actually zero? – J. Zimmerman May 22 '18 at 01:46
  • I'm struggling to understand this – J. Zimmerman May 22 '18 at 01:48
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    If you know what card you removed, that is new information and new information changes your estimate of the probability. If you don't know what card you removed then there is no new information so your estimate of the probability does not change. – lulu May 22 '18 at 01:54
  • If you removed the jack of clubs first the probability would be 0. But you only have a $\frac 1{52}$ chance of this happening. You have $\frac {51}{52}$ chance of that not happening and, if not, the probability will by $\frac 1{51}$. Putting that all together the probability of 2nd card is JH is $\frac 1{52}\times 0 + \frac {51}{52}\frac 1{51} = \frac 1{52}$. – fleablood May 22 '18 at 02:09
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    @geocalc33 Consider this tweak to the scenario: Before you remove the card, I take a peek at the second card while you weren't looking. What's the chance I'll see the Jack of Hearts there? Clearly, it's 1/52. So how can you putting the top card somewhere else change the identity of the card I've already looked at? Clearly it can't. – Monty Harder May 23 '18 at 19:48

6 Answers6

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The Hard Way

The probability that the first card is not the Jack of Hearts is $\frac {51}{52}$ so the probability that the first card is not the Jack of Hearts and the second card is the Jack of Hearts is $\frac {51}{52}\times \frac 1{51}$.
The probability that the first card is the Jack of Hearts is $\frac 1{52}$ so the probability that the first card is the Jack of Hearts and the second card is the Jack of Hearts is $\frac 1{52}\times 0$.

So the total probability that the second card is Jack of Hearts is:
The probability that the second card is after the first card is not +
The probability that the second card is after the first card already was $$= \frac {51}{52}\times \frac 1{51} + \frac 1{52}\times 0 = \frac 1{52} + 0 = \frac 1{52}$$

That was the hard way.

The Easy Way

The probability that any specific card is any specific value is $\frac 1{52}$. It doesn't matter if it is the first card, the last card, or the 13th card. So the probability that the second card is the Jack of Hearts is $\frac 1{52}$. Picking the first card and not looking at, just going directly to the second card, putting the second card in an envelope and setting the rest of the cards on fire, won't make any difference; all that matters is the second card has a one in $52$ chance of being the Jack of Hearts.

Any thing else just wouldn't make any sense.

The thing is throwing in red herrings like "what about the first card?" doesn't change things and if you actually do try to take everything into account, the result, albeit complicated, will come out to be the same.

SQB
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fleablood
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    Are you sure? That first card reminds me of the Monty Hall Problem – Mawg May 23 '18 at 12:00
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    @Mawg it would be like the Monty Hall Problem if you picked the first card, and someone else picked a random non-Jack-of-Hearts card and turned it over, then you had a choice of keeping that first card, or drawing another, to try to pick the Jack of Hearts. That's much more complicated than this scenario, in which you always just discard the first one and turn the second, with no decision making. – Max Williams May 23 '18 at 12:23
  • I am positive and it's not in the least bit like the monty hall problem. Think about it. If someone asks you what is the probability that the first card is JH you'd say $\frac 1{52}$. If someone said what is the probability of the first card is JH but first I'm going to dance the maranga you'd still say $\frac 1{52}$ because dancing the maranga has nothing to do with anything. Then he says what is probability of 1st card being JH but first I'm going to touch the second card. Suddenly by magic the probability changes? – fleablood May 23 '18 at 14:52
  • In the monty hall problem you are given information that changes things. And as I said, even if we do this the hard way and calculate the math it comes out to be the same. The probability that the 4th is the JH is $\frac 1{52}0+\frac{51}{52}\frac 1{51} 0+ (1- (\frac 1{52}+\frac {51}{52}\frac{50}{51}))\frac {1}{50}0 + ( 1- (\frac 1{52}+\frac {51}{52}\frac{1}{51} + \frac {51}{52}\frac{50}{51}\frac{1}{50})\frac 1{49}$... i.e. the probability that that the first three are not time the probability that the fourth is given that the first three are not... which is.... $\frac 1{52}$. – fleablood May 23 '18 at 15:03
  • This would be the monty hall problem if you chose the second and then he showed you what the first card was. Because he didn't show you what the first card was itd be like the monty hall problem if Monty Hall didn't show you a goat, but instead pointed to another door and said "Hey, you know that's a door over there, did you think about it" but did show you what was behind the door. – fleablood May 23 '18 at 15:10
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I'm confused by this question because if the card you removed from the pile was the Jack of Hearts then the probability would be zero so I'm not sure how to calculate it.

Well, to go the long way, you need to use the Law of Total Probability.   Letting $X$ be the card you took away, and $Y$ the card you subsequently draw from the remaining deck.

$$\mathsf P(Y=\mathrm J\heartsuit) = \mathsf P(Y=\mathrm J\heartsuit\mid X\neq \mathrm J\heartsuit)~\mathsf P(X\neq \mathrm J\heartsuit)+0\cdot\mathsf P(X=\mathrm J\heartsuit)$$

Well, now, $\mathsf P(X\neq\mathrm J\heartsuit) = 51/52$ is the probability that the card taken is one from the 51 not-jack-of-hearts.

Also $\mathsf P(Y=\mathrm J\heartsuit\mid X\neq \mathrm J\heartsuit)$ is the probability that the subsequent selection is the Jack of Hearts when given that that card is among the 51 remaining cards.

Alternatively, the short way is to consider : When given a shuffled deck of 52 standard cards, what is the probability that the second from the top is the Jack of Hearts?

Graham Kemp
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  • Really like the way you put it in the second solution: The probability that any specific card is at a certain point in the deck is always 1/52. – rubikscube09 Aug 19 '19 at 15:14
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The probability is 1/52.

It doesn't matter that one card was moved somewhere else.

NicNic8
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I think the reason you're getting a bit confused is that you're conflating two different probabilities:

  1. Your level of belief that you will choose the Jack of Hearts.
  2. Whether or not you will choose the Jack of Hearts.

The first case depends entirely on your knowledge of the state of the deck, and won't be altered until you learn something new about the deck. If you take a random card from the deck and don't look at it, you've learned nothing and your level of belief remains at $1/52$.

The second case depends entirely on the actual, true state of your deck and can be altered if the deck is altered: if the random card you discard is the Jack of Hearts, the true chance of you subsequently randomly selecting that card is zero, otherwise it is $1/51$ as you said.

The logic using in the second case, however, can be taken a step further: The probability of choosing the Jack of Hearts is one if the next card you choose to draw is the Jack of Hearts, and zero otherwise. This indicates to me that the question is probably asking about the first case as it poses a more interesting problem.

Philip C
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Suppose you don't remove the first card from the pile, but just remember its position, and choose the second card from any of remaining $51$ positions.
This is equivalent to the initial problem, isn't it?

Now, that in turn is equivalent to randomly chosing two positions in a pile and taking one card from the second chosen position.

But in all possible ordered pairs of distinct numbers from the set $\{1,2,\dots 52\}$, each number has exactly the same probability of appearing as the second one in a pair.
This implies you can drop choosing the first position at all — just choose randomly the 'second' position and see if it is the Jack of Hearts there.

This, however, is a simple random selection of a single element from the set, consequently the probability of taking the specific one is $1$ over the number of elements: $\frac 1{52}$.

CiaPan
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Through extension: You randomly remove 1 card and then 50 cards from a shuffled deck and put them all in an envelope. What is the probability the card left behind is the Jack of Hearts?

Paul Evans
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