Question:
So the question is, in a box, there are a number of cards:
- x cards black from both sides
- y cards whote from both sides
- z cards having one side black and one white
So if a person draws a card from the box and show you one side which is black, what is the probability that other side is also black.
My Approach:
I personally think that since we have seen one side which is black, the cards white from bpth sides are no longer a part of our sample space. Thus total sample space would be x+z and probability that other side is also black will be x/(x+z).
Objection:
However, my friend objects that and says it is an application of Bayes Theorem and I should follow the formula to calculate the probability.
Counter Argument:
I think that it would be a case of Bayes Theorem only if we would not know the color of one side then we have probabilities of One side black and other side black given the one side is black for the application of Bayes Theorem.
Any suggestions?
As discussed in the comments above, the probability will depend heavily on what strategy your friend uses in determining which face to show.
If after drawing a card at random he then chooses which side to show at random, if it happen to be that the side he shows you is black then the probability that the other side is black will be:
$$\frac{2x}{2x+z}$$
If after drawing a card at random he then chooses to show you a black side if at least one of the sides is black, the probability that the other side is black will be:
$$\frac{x}{x+z}$$
The calculation for the second case, you already performed correctly. The idea being, we know that having seen a black card it was not one of the $y$ cards who have both sides as white. By removing those from those to consider, each remaining card is equally likely to have been chosen. That leaves $x$ "good" results out of $x+z$ total equally likely results.
The calculation for the first case, imagine for a moment that each side has a "front" and a "back." By treating each side separately, we recognize that each of the sides across all cards are equally likely to have been selected. $2x$ of the faces have an opposite face as also black. This is out of the $2x+z$ total equally likely black faces that we could have been shown.