# In a deck of cards, if the second card picked is a heart, what is the probability that the first card picked was a heart?

Assume its a deck of 8 cards with 2 cards of each suit.

My analysis is:

A = First card is heart

B = Second card is heart

P(A) = 1/4

P(B) = 1/4

P(B|A) = 1/7

P(A|B) = P(A) * P(B|A) / P(B)

= 1/4 * 1/7 / 1/4

= 1/7

Is this correct? I tried to create an example to do with Bayes theorem, but I'm not sure if I'm understanding it correctly. Is this a well-posed problem? I notice that P(A|B) is the same as P(B|A). That seems weird.

• Not weird, more like (after a while) obvious. All orders are equally likely. So there is an easier way to do it. – André Nicolas Sep 26 '13 at 10:11
• @AndréNicolas Yeah I mean it is kind of obvious. I sort of said that to bait an explanation as to why. Can you elaborate? – mr real lyfe Sep 26 '13 at 10:20

The best way to understand Bayes theorem is by construction of trees. As problems based on Bayes theorem have a time sequence involved, we can state the sequence as the levels in this tree.

For this problem, we can construct a tree as follows;

All the probabilities in each box are event probabilities in that particular case.

P(A - first card is heart) = 2/8 = 1/4 (directly) P(B - second card is heart) = 2/8 * 1/7 + 6/8 * 2/7 = 14/56 = 1/4

P(B|A) = P(B and A)/P(A) = (1/7 * 2/8)/(2/8) = 1/7 P(A|B) = P(A and B)/P(B) = (2/8 * 1/7)/(1/4) = 1/7

I think you are just surprised at the results, to a very well formed Bayes problem. The reason for P(B|A) = P(A|B) is because P(A) = P(B)

You're totally correct. To see why $P(A|B)$ and $P(B|A)$ are the same, just notice that the order of the cards totally doesn't matter. Suppose you picked the two cards at the same time and put one on the table in front of you to the left, and one to the right.

Let $A$ be "the left card is a heart"

Let $B$ be "the right card is a heart"

Then clearly $P(A|B)=P(B|A)$.