A bag contains three cards – one is red on both sides, the second is white on both sides, and the third is red on one side and white on the other. Amy draws a card without looking and keeps it on the table. If the card is red face up, what is the probability that it’s also red on this hidden side?
Intuition tells the probability to be 1/2. The argument goes like this: if the side up is red, there are two equal possibilities for the hidden side – red or white. Therefore, it’s 1/2. A slightly different version of the same logic estimates that once the person sees it red, it shuts the options white-white card, leaving only two red-red and red-white. The card must be one out of two.
Conditional Probability
Let’s investigate the problem using conditional probability (the Bayes’ rule).
P(RR|Ru) represents the required probability, or the card is RR given R is up.
P(Ru|RR) = probability of red up given RR is the selected card
P(RR) = Prior probability of choosing the RR card
P(Ru|RW) = probability of red up given RW is the picked card
P(RW) = Prior probability of selecting the RW card
P(Ru|RR) must be 1 as RR will always show red up
P(RR) = 1/3, as there are three cards to choose from
P(Ru|RW) = 1/2, there is a 50:50 chance for red to show up from an RW card
P(RW) = 1/3
