On the intuition behind a conditional probability problem.

Question

This is a very similar question to this one. But notice the subtle difference that the event that I define $B$ is that I am dealt at least an ace.

Suppose I get dealt 2 random cards from a standard deck of poker (52 cards).

Let the event $A$ be that both cards are aces, let $B$ be the probability that I am dealt at least an ace and let $C$ be the probability that I am dealt the ace of spades.

We have

$P(A| C) = \frac{3}{51}= \frac{1}{17} $ and $$P(A|B) = \frac{P(A,B)}{P(B)} = \frac{P(A)}{P(B)} = \frac{\frac{\binom{4}{2}}{\binom{52}{2}}}{1 - \frac{\binom{48}{2}}{\binom{52}{2}}} = \frac{1}{33}$$

I do not intuitively uderstand how $P(A|B) < P(A|C)$. Surely the probability of having been dealt 2 aces given that I am dealt at least an ace should be higher than the probability of having been dealt 2 aces given that I am dealt the ace of spades?

Could I get some help in understanding where my intuition is failing? Is there an other way to approach this problem?

Answer 1

Maybe this can help: think of a deck of $4$ cards containing spade ace, another ace and two non-aces and ask the same question. Then:

$P\left(\text{two aces}\mid\text{ace of spades}\right)=\frac{P\left(\text{two aces}\right)}{P\left(\text{ace of spades}\right)}$

$P\left(\text{two aces}\mid\text{at least one ace}\right)=\frac{P\left(\text{two aces}\right)}{P\left(\text{at least one ace}\right)}$

Note that $P\left(\text{at least one ace}\right)>P\left(\text{ace of spades}\right)$

Answer 2

By exactly specifying the card, you increase the probabily of both cards to be aces (specifying the card is stronger than demanding that just one of them is an ace). Maybe the following example will help:

We have a three closed boxes $A$, $B$ and $C$. One box contains $100$ blue balls, one box contains $100$ red balls, and one box contains $1$ red ball and $99$ blue balls. Now I tell you that box $A$ contains at least one red ball, and let you pick a ball at random from box $B$. If the ball you pick is red, which box would you guess to contain $100$ red balls?

Answer 3

Be careful about the meaning of the events $A$, $B$, and $C$. The event $B$ is "of the two cards dealt, at least one of them is an ace". And I think you intended the event $C$ to be "the first card is an ace". As you've calculated above, event $A$ is contained in both events $B$ and $C$, so the conditional probabilities reduce to $$ P(A|B) = { P(A)\over P(B) } $$ and $$P(A|C) = { P(A) \over P(C) }. $$ Now notice that event $B$ has higher probability than event $C$ (since if $C$ is true, then $B$ is true). This means that in the first calculation you have a larger denominator than in the second.

Answer 4

For me attaching some numbers to the problem makes it more intuitive.

In all, there are $52*51-48*47=396$ possible ways in which you can draw at least 1 ace from the deck. There are $1*51+51*1=102$ ways in which you can draw the Ace of Spades from the deck. So it's about $\frac14$ as likely to draw specifically the Ace of Spades. The deviation from exactly $\frac14$ comes from the 12 ways in which is possible to draw 2 aces. ($102*4-12=396$)

Now, out of the 396 ways you can draw at least one ace from the deck, there are only 12 ways to draw 2 aces. However, out of the 102 ways in which you can specifically draw the Ace of Spaces, there are 6 total ways in which you can draw two aces.

So drawing a specific Ace from the deck is only about a fourth as likely, but drawing a specific ace includes half the ways in which you can draw two aces. This meshes perfect with your calculated probabilities, since $\frac1{33}$ is almost half of $\frac1{17}$.

Answer 5

I like pictures.

$A$ is the event that both cards are aces, B is the event that at least one card is an ace and $C$ is the event that one of the cards is the ace of spades. Since, $ C \subset B$ and $A \subset B$, the size of these events can be visualized as follows

$$P(A|B) = { P(A)\over P(B) }$$ and $$P(A|C) = { P(A \cap C) \over P(C) }.$$