## Equal Colours

#### Problem

A box contains a mixture of black and white discs, with more black than white discs. A game is played by taking one disc at a time, at random, and without replacement. If an equal number of each colour have been removed the game stops and the player wins.

It is found that the player has an equal chance of winning or losing.

If the box contains twelve discs in total, find the number of black discs.

#### Solution

Let the total number of blacks discs in the box be `b` and the number of white discs be `w`.

Clearly if `b`=`w`, the player would win every time; and for this problem we are given that `b``w`.

We shall express a win by the use of strings; for example, the sequence:

W, WW, WWB, WWBW, WWBWB, WWBWBB produces the winning string, WWBWBB, that contains three of each colour.

For each winning string, it will only contain an equal number of black and white discs for the first time when the final disc in the string is taken. For example, WBBW does not represent a valid winning string, as it contained an equal number of black and white discs after the second disc was taken. Therefore, given a winning string, we can safely invert the string to obtain a different winning string. For example, WWBWBB and BBWBWW are dual winning strings. Hence there are an equal number of winning strings that start with B and W.

As there are more black discs than white discs, any string starting with W must be a winning string; there will always be enough B's to eventually match the number of W's before the final disc in the box is taken.

P(1st disc W) = `w`/(`b`+`w`), hence P(winning) = 2`w`/(`b`+`w`), where `b``w`.

Therefore solving 2`w`/(`b`+`w`) = 1/2, leads to `b` = 3`w`.

As `b`+`w` = 3`w`+`w` = 4`w` = 12 `w` = 3. Hence there are 3 white discs and 9 black discs in the box.