A bag has $3$ black and $3$ white balls. A game starts with you extracting $3$ balls at random. If they are the same colour you win. Otherwise you return the two balls with the same colour to the bag, discard the other ball, and start again removing $3$ more balls at random. If in the end the bag contains $2$ balls you lose. What is the probability of winning the game?
  1. How many different pairs of students can a teacher select from a class of twenty?
  2. In a drawer there are six white socks and eight black socks. What is the probability that someone choosing two socks at random would get a matching pair?
  3. What is the probability of getting the following sequence when rolling a dice 4 times: 3, 1, 4, 1?

As the game progresses, you may want to keep track only of the ball-count for each color.

… and also ignore which color is which in your ball-count.

Try to write out the sequence of all possible ball-counts combinations for the first two rounds?

Can you identify rounds that guarantee winning or losing in the end?

You can compute either the winning probability or the losing probability. Which one do you think is easier in this case?

What are the possible ways to get to a round that guarantees losing in the end?

Let us denote with $(n,p)$ the state of the bag at the start of a round, where $n,p$ are the number of balls of first and second colour respectively, ignoring which is which. Initially we have $(3,3)$. At the start of round two we’ll have $(3,2)$. In round three we could either have $(2,2)$ or $(3,1).$ If it’s $(2,2)$ then we are guaranteed to lose the game. If it’s $(3,1)$ we win either in the current round or in the next round. It thus seems easier to derive the probability of losing, which is the probability of arriving at $(2,2),$ and then subtract it from $1$.

In round one there are 20 ways of drawing 3 balls (6 choose 3) and only 2 of them result in a win, so we progress to round 2 with probability $18/20$.

Round two starts in $(3,2)$ and there are 10 ways of drawing 3 balls (5 choose 3). These are: $WWW,$ $WWB,$ $WWB,$ $WBW,$ $WBW,$ $BWW,$ $BWW,$ $\underline{WBB},$ $\underline{BWB},$ $\underline{BBW},$ where the underlined ones lead to $(2,2).$ Note there are 2 ways of getting WBW for example. Hence the probability of going to $(2,2)$ in this round is $3/10$.

The overall probability of losing is $(18/20)(3/10) = 27/100,$ so the probability of winning is $73/100$.