Lucky Dip

Problem

A bag contains one red and one blue disc. In a game a player pays £2 to play and takes a disc at random. If the disc is blue then it is returned to the bag, one extra red disc is added, and another disc is taken. This game continues until a red disc is taken.

Once the game stops the player receives winnings equal in pounds to the number of red discs that were in the bag on that particular turn.

Find the exact value of the expected return on this game.

Solution

Let $X$ be the number of red discs in the bag when play stops.

\begin{align}P(X = 1) &= \dfrac{1}{2}\\P(X = 2) &= \dfrac{1}{2} \times \dfrac{2}{3}\\P(X = 3) &= \dfrac{1}{2} \times \dfrac{2}{3} \times \dfrac{3}{4}\\...\\P(X = k) &= \dfrac{1}{2} \times \dfrac{2}{3} \times ... \times \dfrac{1}{k} = \dfrac{k}{(k+1)!}\\\therefore E(X) &= \dfrac{1}{2!} \times 1 + \dfrac{2}{3!} \times 2 + \dfrac{3}{4!} \times 3 \times ...\\&= \dfrac{1^2}{2!} + \dfrac{2^2}{3!} + \dfrac{3^2}{4!} + ...\end{align}

To evaluate this series we need to consider the general term, $\dfrac{k^2}{(k+1)!}$, but before this we will evaluate a different series:

$$\dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + \dfrac{4}{5!} + ...$$

As $\dfrac{k}{(k+1)!} = \dfrac{k + 1 - 1}{(k+1)!} = \dfrac{k + 1}{(k+1)!} - \dfrac{1}{(k+1)!} = \dfrac{1}{k!} - \dfrac{1}{(k+1)!}$, we can write the original series becomes a telescoping series:

\begin{align}\dfrac{1}{2!} + \dfrac{2}{3!} + ... &= \left( \dfrac{1}{1!} - \dfrac{1}{2!} \right) + \left( \dfrac{1}{2!} - \dfrac{1}{3!} \right) + \left( \dfrac{1}{3!} - \dfrac{1}{4!} \right) + ...\\&= 1\end{align}

Now we write, $\dfrac{k^2}{(k+1)!} = \dfrac{k(k + 1) - k}{(k + 1)!} = \dfrac{k(k + 1)}{(k + 1)!} - \dfrac{k}{(k + 1)!} = \dfrac{1}{(k - 1)!} - \dfrac{k}{(k + 1)!}$.

\begin{align}\therefore E(X) &= \dfrac{1^2}{2!} + \dfrac{2^2}{3!} + \dfrac{3^2}{4!} + ...\\&= \left(\dfrac{1}{0!} - \dfrac{1}{2!} \right) + \left(\dfrac{1}{1!} - \dfrac{1}{3!} \right) + \left(\dfrac{1}{2!} - \dfrac{1}{4!} \right) + ...\\&= \left(1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!} + ... \right) - \left(\dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + ... \right)\\&= e - 1\end{align}

Hence the return on each game will be $2 - (e - 1) = 3 - e \approx 28$ pence in the "bankers" favour.

Problem ID: 276 (21 Apr 2006)     Difficulty: 4 Star

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