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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)?

My attempt:

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?

tsaebeht
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2 Answers2

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The problem can be reformulated as follows: We have a bag with $N$ marbles, $R\geq1$ of them red and $B:=N-R$ of them blue. What is the expected number of draws until we draw a red marble?

During the process the number $b$ of blue marbles in the bag decreases, starting with $b=B$. Denote by $E(b)$ the expected number of additional draws when there are $b$ blue marbles left. Then $E(0)=1$, and $$E(b)=1+{b\over R+b}E(b-1)\qquad(b\geq1)\ .$$ One computes $E_1={R+2\over R+1}$, $E_2={R+3\over R+1}$. This leads to the conjecture that $$E(b)={R+b+1\over R+1}\ ,$$ which is then easily proved by induction. It follows that $$E(B)={N+1\over R+1}\ .$$

Christian Blatter
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Let $N=l\cdot b$ be the total number of squares, let $R$ be the number of squares initially with red markers, and let $X$ be the number of draws (without replacement) needed to obtain a red marker.

Now we have (see here)

$$E(X) = \sum_{k=0}^\infty k\ P(X = k) = \sum_{k=0}^\infty P(X \gt k),$$

and the probability that the first $k$ draws are not red is $$P(X \gt k) = \frac{{N-R} \choose {k}}{N \choose {k}} \quad (0 \le k \le N-R), $$ so $$E(X) = \sum_{k=0}^{N-R} \frac{{{N-R} \choose k}}{N \choose k} = \frac{N+1}{R+1}.$$

(I obtained the sum via Wolfram Alpha.)

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