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2001 Paper 3 Q12
D: 1700.0 B: 1518.2
probability expectation hypergeometric distribution combinatorics negative hypergeometric geometric distribution balls in bag summation

A bag contains \(b\) black balls and \(w\) white balls. Balls are drawn at random from the bag and when a white ball is drawn it is put aside.

  1. If the black balls drawn are also put aside, find an expression for the expected number of black balls that have been drawn when the last white ball is removed.
  2. If instead the black balls drawn are put back into the bag, prove that the expected number of times a black ball has been drawn when the first white ball is removed is \(b/w\,\). Hence write down, in the form of a sum, an expression for the expected number of times a black ball has been drawn when the last white ball is removed.

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1990 Paper 1 Q14
D: 1500.0 B: 1500.7
probability conditional expected value discrete probability distributions optimisation strategy sequential decision balls in bag

A bag contains 5 white balls, 3 red balls and 2 black balls. In the game of Blackball, a player draws a ball at random from the bag, looks at it and replaces it. If he has drawn a white ball, he scores one point, while for a red ball he scores two points, these scores being added to his total score before he drew the ball. If he has drawn a black ball, the game is over and his final score is zero. After drawing a red or white ball, he can either decide to stop, when his final score for the game is the total so far, or he may elect to draw another ball. The starting score is zero. Juggins' strategy is to continue drawing until either he draws a black ball (when of course he must stop, with final score zero), or until he has drawn three (non-black) balls, when he elects to stop. Find the probability that in any game he achieves a final score of zero by employing this strategy. Find also his expected final score. Muggins has so far scored \(N\) points, and is deciding whether to draw another ball. Find the expected score if another ball is drawn, and suggest a strategy to achieve the greatest possible average final score in each game.

Solution: The probability Juggin's has a non-zero score is the probability he never draws a black ball in his three goes. This is \((1-\frac15)^3 = \frac{64}{125}\). Let's consider the \(\frac{61}{125}\) probability world where he never draws a black ball. In this conditional probability space, he has \(\frac{5}{8}\) chances of pulling out white balls and \(\frac38\) or pulling out red. His expected score per pull is \(\frac58 \cdot 1 + \frac38 \cdot 2 = \frac{11}{8}\). Therefore his expected score in this universe is \(\frac{33}8\) and his expected score is \(\frac{33}{8} \cdot \frac{61}{125} = \frac{2013}{1000} = 2.013\) . The expected score after drawing another ball is \(( N + 1)\frac{5}{10} + (N+2) \frac{3}{10} + 0 \cdot \frac{2}{10} = \frac{8}{10}N + \frac{11}{10}\). A sensible strategy would be to only draw if \(\frac{8}{10}N + \frac{11}{10} > N \Rightarrow N < \frac{11}{2}\), ie keep drawing until \(N \geq 6\) or we bust out. [The expected score for this strategy is: \begin{array}{ccc} \text{score} & \text{route} & \text{count} & \text{prob} \\ \hline 6 & \text{6 1s} & 1 & \left ( \frac12 \right)^6 \\ 6 & \text{4 1s, 1 2} & 5 & 5 \cdot \left ( \frac12 \right)^4 \cdot \frac{3}{10} \\ 6 & \text{2 1s, 2 2s} & 6 & 6 \cdot \left ( \frac12 \right)^2 \cdot \left ( \frac{3}{10} \right)^2 \\ 6 & \text{3 2s} & 1 & 1 \cdot \left ( \frac{3}{10} \right)^3 \\ 7 & \text{5 1s, 1 2} & 1 &\left ( \frac12 \right)^5 \cdot \frac{3}{10} \\ 7 & \text{3 1s, 2 2s} & 4 & 4\cdot \left ( \frac12 \right)^3 \cdot \left ( \frac{3}{10} \right)^2 \\ 7 & \text{1 1, 3 2s} & 3 & 3\cdot \left ( \frac12 \right) \cdot \left ( \frac{3}{10} \right)^3 \\ \end{array} For an expected value of \(\frac{2171}{8000} \cdot 6 + \frac{759}{8000} \cdot 7 = \frac{18\,339}{8000} = 2.29 \quad (3\text{ s.f.})\)]

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