3 problems found
A bag contains \(b\) balls, \(r\) of them red and the rest white. In a game the player must remove balls one at a time from the bag (without replacement). She may remove as many balls as she wishes, but if she removes any red ball, she loses and gets no reward at all. If she does not remove a red ball, she is rewarded with \pounds 1 for each white ball she has removed. If she removes \(n\) white balls on her first \(n\) draws, calculate her expected gain on the next draw and show that %her expected total reward would be the same as before it is zero if \(\ds n = {b-r \over r+1}\,\). Hence, or otherwise, show that she will maximise her expected total reward if she aims to remove \(n\) balls, where \[ n = \mbox{ the integer part of } \ds {b + 1 \over r + 1}\;. \] With this value of \(n\), show that in the case \(r=1\) and \(b\) even, her expected total reward is \(\pounds {1 \over 4}b\,\), and find her expected total reward in the case \(r=1\) and \(b\) odd.
You play the following game. You throw a six-sided fair die repeatedly. You may choose to stop after any throw, except that you must stop if you throw a 1. Your score is the number obtained on your last throw. Determine the strategy that you should adopt in order to maximize your expected score, explaining your reasoning carefully.
Solution: Once you have thrown, all previous throws are irrelevant so the only thing which can affect your decision is the current throw. Therefore the strategy must consist of a list of states we re-throw from, and a list of states we stick on. It must also be the case that if we stick on \(k\) we stick on \(k+1\) (otherwise we can improve our strategy by switching those two values around). Therefore we can form a table of our expected score: \begin{array}{c|c|c} \text{stop on} & \text{possible outcomes} & \E[\text{score}] \\ \hline \geq 2 & \{1,2,3,4,5,6\} & \frac{21}{6} = 3.5 \\ \geq 3 & \{1,3,4,5,6\} & \frac{19}{5} = 3.8 \\ \geq 4 & \{1,4,5,6\} & \frac{16}{4} = 4 \\ \geq 5 & \{1,5,6\} & \frac{12}{3} = 4 \\ =6 & \{1,6\} & \frac{7}{2} = 3.5 \end{array} Therefore the optimal strategy is to stop on \(4\) or higher. If we cared about variance we might look at the variance of the two best strategies, \(4\) or higher has a variance of \(\frac{1+16+25+36}{4} - 16 = 3.5\) and \(5\) or higher has a variance of \(\frac{1+25+36}3 - 16 = \frac{14}3 > 3.5\) so \(4\) or higher is probably better in most scenarios.
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 \(\frac{2171}{8000} \cdot 6 + \frac{759}{8000} \cdot 7 = \frac{18\,339}{8000} = 2.29 \quad (3\text{ s.f.})\)]