Question

In the game of Blackball, at each turn Muggins draws a ball at random from a bag containing five white balls, three red balls and two black balls; after being recorded, the ball is replaced in the bag. A white ball earns him \(\$ 1\) whilst a red ball gets him \(\$ 2\); in either case he also has the option of leaving with his current winnings or of taking a further turn on the same basis. If he draws a black ball the game ends and he loses all he may have gained previously. Find an expression for Muggins' expected return if he adopts the strategy to drawing up to \(n\) balls if he has not been eliminated by then. Show that, as the entry fee to play is \$3, Muggins should be dissuaded from playing Blackball, but if that cannot be done what value of \(n\) would you advise him to adopt?

Short answer

Expected value after n turns is: \(5.5(1 - (0.8)^n)\). Avoid playing, but if forced, choose \(n = 1\).

Step-by-step solution

- Identify Probabilities

First, determine the probabilities of drawing each type of ball. There are 10 balls in total (5 white, 3 red, 2 black). The probability of drawing a white ball is \[ P(W) = \frac{5}{10} = 0.5 \] The probability of drawing a red ball is \[ P(R) = \frac{3}{10} = 0.3 \] The probability of drawing a black ball is \[ P(B) = \frac{2}{10} = 0.2 \]

- Calculate Expected Value After One Turn

Using the probabilities calculated, find the expected value (EV) after just one turn. The reward for a white ball is \$1, and for a red ball it's \$2. If a black ball is drawn, the game ends with a reward of \$0. The expected value of one turn can be calculated as follows: \[ EV_1 = P(W) \times 1 + P(R) \times 2 + P(B) \times 0 = 0.5 \times 1 + 0.3 \times 2 + 0.2 \times 0 = 0.5 + 0.6 = 1.1 \]

- Consider Multiple Turns

Now, consider the strategy to keep playing up to \(n\) turns unless a black ball is drawn. Let \(E_n\) represent the expected value of playing up to \(n\) turns. We can express \(E_n\) recursively: \[ E_n = P(W) \times (1 + E_{n-1}) + P(R) \times (2 + E_{n-1}) + P(B) \times 0 = 0.5 \times (1 + E_{n-1}) + 0.3 \times (2 + E_{n-1}) \] Simplify and substitute \(E_{n-1}\): \[ E_n = 0.5 + 0.5E_{n-1} + 0.6 + 0.3E_{n-1} = 1.1 + 0.8E_{n-1} \]

- Determine Expected Value When Drawing Up to n Balls

To find the general formula for \(E_n\), notice that this recursive relation suggests an infinite geometric series, which can be solved: \[ E_n = 1.1\left( \frac{1 - (0.8)^n}{1 - 0.8} \right) \] Simplifying further, we get: \[ E_n = 1.1 \times \frac{1 - (0.8)^n}{0.2} = 5.5 \times (1 - (0.8)^n) \] Therefore, \(E_n = 5.5 \times (1 - (0.8)^n)\)

- Evaluate Profitability Considering Entry Fee

Determine profitability considering the entry fee of \$3. Since \(E_n\) is the expected revenue from playing up to \(n\) turns, the net gain after paying the entry fee would be: \[ \text{Net Gain} = 5.5 \times (1 - (0.8)^n) - 3 \]

- Advise on Optimal Strategy

To maximize the expected return, calculate the value of \(n\) that makes the net gain maximum. Taking the derivative with respect to \(n\): \[ f(n) = 5.5 \times (1- (0.8)^n) -3 \] Find \(n\) where the net gain is highest while ensuring the return is positive. However, observe that \( f(\infty) = 2.5 \), which is negative, indicating all finite values of \(n\) also result in a negative net gain. Therefore, Muggins should be dissuaded from playing.

- If Forced to Play, Provide a Strategy

If Muggins must play, advise on an optimal \(n\) that minimizes loss or maximizes minor net gains. Realistically, smaller \(n\) values limit risk exposure. Evaluating smaller \(n\), such as \(n = 1\), significantly reduces potential losses. For \(n = 1\): \[ E_1 = 1.1 \text{ and Net Gain} = 1.1 -3 = -1.9 \]

Final Suggestion

Considering all options, Muggins should aim for \(n = 1\) to minimize losses if he must play.

Key concepts

Probability Distributions

In probability theory, a probability distribution assigns probabilities to different possible outcomes in an experiment. In our Blackball game, there are three possible outcomes when Muggins draws a ball: white, red, or black. The probability distribution is given as follows: the probability of drawing a white ball is 0.5, a red ball is 0.3, and a black ball is 0.2. These probabilities add up to 1, which confirms that all possible outcomes are accounted for. This distribution helps us determine the likelihood of various scenarios, crucial for calculating the expected return.

Decision Theory

Decision theory is about making optimal choices under uncertainty. In our context, Muggins must decide whether to draw another ball or quit after each turn. The decision hinges on the expected value (EV) of continuing. We calculated the EV for one turn as 1.1. To decide optimally, Muggins should consider the risk of drawing a black ball, which ends the game. Therefore, he evaluates the expected return from multiple draws using the recursive formula, deciding to continue only if the calculated EV justifies the risk taken each turn.

Recursive Equations

A recursive equation defines a sequence where the next term is calculated based on the previous ones. For Muggins’ game, the expected return after n turns, denoted as E_n, depends on the expected return after n-1 turns. The equation is given by:\[ E_n = 0.5 + 0.8E_{n-1} \]This means each term is built from the previous term, involving the probabilities and rewards of the game. Solving this recursive relation can quickly become complex, but it’s central to understanding how the expected value evolves with more turns.

Geometric Series

A geometric series is a series of terms each multiplied by a constant factor to get the next term. In Muggins’ game, solving the recursive equation reveals that the expected return forms a geometric series. We derive:\[ E_n = 5.5 \times (1 - (0.8)^n) \]This expression comes from summing the infinite series linked to the recursive equation, where each additional term contributes less due to the multiplicative factor of 0.8. Understanding geometric series allows us to predict long-term expected values and make informed decisions on how many turns to play.