A bag has 20 blue balls and 14 red balls. Each time you randomly take two balls out. (Assume each ball in the bag has equal probability of being taken). You do not put these two balls back. Instead, if both balls have the same color, you add a blue ball to the bag; if they have different colors, you add a red ball to the bag. Assume that you have an unlimited supply of blue and red balls, if you keep on repeating this process, what will be the color of the last ball left in the bag?
Solution: Let (B, R) represent the number of blue balls and red balls in the bag. We can take a look what will happen after two balls are taken out.
Both balls are blue: (B, R) (B - 1, R)
Both balls are red: (B, R) (B + 1, R - 2)
One red and one blue: (B, R) (B - 1, R)
Notice that R either stays the same or decreases by 2, so the number of red balls will never become odd if we begin with 14 red balls. We also know that the total number of balls decreases by one each time until only one ball is left. Combining the information we have, the last ball must be a blue one.
My question is couldn't there technically not be one "last ball in the bag"? We know that the number of red balls is always even, but couldn't we get to a point where there are 2 red balls and 1 blue, and then we remove a red and a blue and now have 2 red and 0 blue? Am I missing something, or is the wording of the question just not accounting for the scenarios when there is never a "last ball"?
The problem says uses this phrase:
What they really mean is this:
This means the "$2$ red, $0$ blue" situation never arises, because you would not be done at that point. There are still at least two balls, so the instructions demand that you repeat the process, drawing the two red balls and leaving one blue ball.