In the video lectures from Harvard's Statistics 110:Probability course that can be found on iTunes and YouTube, I encountered this problem. I tried to summarize it here:

Suppose we are given a random two-card hand from a standard deck.

  1. What is the probability that both cards are aces given that we have at least one ace?

$$ P(both\ aces | have\ ace) = \frac{P(both\ aces, have\ ace)}{P(have\ ace)} $$

Since having at least one ace is implied if you have both aces, the intersection can be reduced to just $P(both\ aces)$

$$ P(both\ aces | have\ ace) = \frac{P(both\ aces)}{P(have\ ace)} $$

This is then just

$$ P(both\ aces | have\ ace) = \frac{4C2\ /\ 52C2}{1-48C2\ /\ 52C2}=\frac{1}{33} $$

  1. What is the probability that both cards are aces given that we have the ace of spades?

$$ P(both\ aces | have\ ace\ of\ spades)=\frac{P(both\ aces, have\ ace\ of\ spades)}{P(have\ ace\ of\ spades)} $$

$$ P(both\ aces | have\ ace\ of\ spades)=\frac{(3C1*1C1)\ /\ 52C2}{2!*\frac{51}{52}*\frac{1}{51}}=\frac{1}{17} $$

Now, somewhere along these examples I got lost...

The latter is obviously just the same as $\frac{3}{51}$, which makes a lot of sense (to me) that this would be the answer. If you are told that you have the ace of (say) spades, then you know that there are $3$ more aces and $51$ more cards.

But in the former example, the math seems fine (and I believe the lecturer wouldn't give this example if it was incorrect...), but I can't wrap my head around this.

How do I get some intuition for this problem?

  • 1
    $\begingroup$ Try answering this: My neighbor has two kids--you know that one of them is a boy. What's the probability that she has two boys. $\endgroup$
    – Steve S
    Aug 26, 2014 at 15:37
  • $\begingroup$ Thanks for including your attempt at the problem! Please add the [self-study] tag & read its wiki. $\endgroup$
    – Silverfish
    Jul 8, 2016 at 9:51

4 Answers 4


To aid the intuition, consider visualizing two events (sets of outcomes):

  1. The conditioning event, which is the information given.

  2. The conditioned event, whose probability you would like to find.

The conditional probability is found by dividing the chance of the second by the chance of the first.

There are $52 \times 51$ equally likely ways to deal two cards at random. A convenient way to visualize these deals is to lay them out in a table with rows (say) designating the first card dealt and columns the second card in the deal. Here is a part of this table, with ellipses ($\cdots$) designating the missing parts. Notice that because the two cards cannot be the same, no entries exist along the main diagonal of the table. The rows and columns are ordered from aces up through kings:

Figure 1

The questions focus on aces. The information "we have at least one ace" locates the pair within either the first four rows or the first four columns. In our mind we can visualize that schematically by coloring these rows and columns. I have colored them red, but where both aces appear I have colored them black:

Figure 2

There are $2\times 6 = 12$ pairs of all aces and $2\times (4\times 48) = 384$ other pairs with at least one ace, for a total of $12+384=396$ pairs on which you are conditioning, as represented by both the red and black areas. Because all such pairs are equally likely, the chance of the former is

$$\frac{12}{396} = \frac{1}{33}.$$

It is the black fraction of the red+black region.

The second question asserts "we have the ace of spades." This corresponds only to the very first row and column:

Figure 3

Now there are just $2\times 3 = 6$ such pairs with two aces and $2\times 48=96$ other pairs with the ace of spades, for a total of $96+6 = 102$ such pairs. Reasoning exactly as before, the chance of two aces is

$$\frac{6}{102} = \frac{1}{17}.$$

Again it is the black fraction of the red+black region.

For reference, the last figure includes the previous one shown in pink and gray. Comparing these regions reveals what happened: in going from the first question to the second, the number of pairs in the conditioning event (pink) dropped to about one-quarter of its original count (red), while the number of pairs in question dropped by only one-half (from gray to black, $12$ to $6$).

I have found such schematic figures to be helpful even--perhaps especially--when trying to understand more complicated concepts of probability, such as filtrations of sigma algebras.

  • $\begingroup$ Did you generate that first picture yourself? If so, how? $\endgroup$
    – Steve S
    Aug 27, 2014 at 5:23
  • $\begingroup$ By the way: +1 $\endgroup$
    – Steve S
    Aug 27, 2014 at 5:29
  • 1
    $\begingroup$ @Steve I used Mathematica, beginning with the playing card representations on the SE Mathematica site. I tabulated an outer product of an abbreviated list of the cards, where the "product" function combines a pair of randomly rotated card images to represent a two-card hand. $\endgroup$
    – whuber
    Aug 27, 2014 at 14:21
  • $\begingroup$ Unfortunately I don't use Mathematica, which is a shame, apparently, because that graphic really looks good (and definitely adds a lot to the post). $\endgroup$
    – Steve S
    Aug 28, 2014 at 12:00
  • $\begingroup$ @whuber To really round this out (as I am working on my own intuition of this problem), let's say I was gambling; when would asking the question 𝑃(π‘π‘œπ‘‘β„Ž π‘Žπ‘π‘’π‘ |β„Žπ‘Žπ‘£π‘’ π‘Žπ‘π‘’) be better than 𝑃(π‘π‘œπ‘‘β„Ž π‘Žπ‘π‘’π‘ |β„Žπ‘Žπ‘£π‘’ π‘Žπ‘π‘’ of spades)? $\endgroup$
    – 9301293
    Aug 27, 2021 at 16:28

A different way to set up a problem that leads to the second calculation is the following:

You draw two cards from the deck. What is the probability of two aces given that the first card you drew was an ace?

This phrasing makes it easier to contrast with the first calculation. The underlying chance of having picked two aces does not change, but the condition to have the first card as an ace is more restrictive than the condition if either being an ace. This means that in the conditional probability calculation the desired combination has to occur among fewer options, so it has a larger probability.

The two different phrasings (ace of spades versus first card as ace) are similar, because they break the symmetry / exchangeability between the aces: the suit or the order cannot be arbitrarily swapped.


How one can tell the answer is 3/51 without calculating?

If you took the ace of spades in the first place. I know which cards ar in the package. So there is stil 3 aces on 51 cards. so for the second one, you have 3/51 chances to have two aces.

And how to understand the difference between the two scenarios intuitively?

It's because "Have one ace" is included in "Have two aces". But "Have the ace of spades" is not included in "Have two aces". This is the difference.

In fact, if you have two ace, you have one but maybe not the ace of spades So it's not the same probability.

This answered was for a other post which has been moved on this one ..

  • $\begingroup$ I've mostly answered the second question : " And how to understand the difference between the two scenarios intuitively?" But i'm going to answer the first one $\endgroup$
    – el Josso
    Jul 8, 2016 at 14:38

At the beginning it was hard to me to have some intuition about.

One idea is to take the problem to the limit. In this case as Steve noted one identical problem is: My neighbor has two kids--you know that one of them is a boy. What's the probability that she has two boys.

Thje first idea is, ok, I have one boy, the other child has 1/2 chance to be a girl and 1/2 to be a boy, but in this case you are not taking all the information that gives you the fact (at least you have a boy) because it is implicit that this boy can be the youngest child being the oldest a girl or viceversa or both are boys which means that only one of the three possible outcomes is favorable.

As I said this is easier taking the problem to the limit...

Case1: Abstract case identical to "we have one ace"-> In this case imagines My neighbour has not 2 children but 27, and you know 26 are boys, the probability of this is almost zero. In this case it is clear that this information gives you a lot of information that the probabilistically speaking the remain child is a girl. To be precise, you will have one case with 27 boys, let's say a tuple (b,b,b,b,b,b...,b) and 27 cases with 1 girl and 26 boys (g,b,b,b...), (b,g,b,b,b...), so the probability of all boys is 1/27, in general it will be 1/(N+1)

case2: Concrete information. This would be identical to "We have the ace of spades" or "we have the first card being an ace". In this case imagine our neighbour have 26 children all boys and is pregnant with the 27th. What is the probability that the 27th will be a boy?

With case2 I am pretty sure we all can have a grasp of the intuition needed for this kind of not-so-obvious conditional probabilities problems.

If you want to become rich, you have to bet on the first case with 26 boys and a 27th because the lack of concrete information means a lot of probabilistic energy on the remain child while in the second case, the entropy is huge, we have not information to know where to bet.

I hope this is useful


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