Suppose I have a Checkerboard of length l and breadth b. Two people, A and B decide to play a game. The rules of the game are as follows:
-A is blindfolded
-B arranged the checkerboard as follows: On every grid, he will place either a blue marker OR a red marker. It is guaranteed that no grid will be left empty by B.
-Every turn, A picks a non-empty square at random and B will tell A the colour of the marker he picked up.
-If the colour of the marker is red, the game will end. Else, A will repeat the step of choosing a non-empty square
-Assume that the game will always end. That means, that there is always atleast 1 red marker on the checkerboard.
Question: Given the dimensions of the board, the number of red markers R and the number of blue markers B, what is the expected value of the number of tries taken by A to complete the game (R+B=l*b)?
Assume Fail is picking up a Blue marker and Pass is picking up a Red Marker. What I tried doing was to form a series of attempts for the expected value. In the sense, Fail+Fail,Pass+Fail,Fail,Pass... etc. Consider the case Fail,Fail,Pass. The probability for failing (picking up a blue marker) is B/(l*b) for the first case. Similarly picking up a blue marker in the second turn is with probability B/(l *b-1) and finally for the third term, which is picking up a red marker will be R/(l *b-1). All three terms will be multiplied together and multiplied with the number of turn taken, which is 3. I'm not sure of this series. I think I haven't really understood how to work such problems out. I'm familiar with the idea of expectation value and it's definition which is ΣP(i)*xi where xi is your value and P(i) is it's probability. Somehow I have a feeling my solution to this answer is terribly wrong.
How do I go about solving this problem?