Question

(a) Gamblers \(A\) and \(B\) each roll a fair six-faced die, and \(B\) wins if his score is strictly greater than \(A\) 's. Show that the odds are 7 to 5 in \(A\) 's favour. (b) Calculate the probabilities of scoring a total \(T\) from two rolls of a fair die for \(T=2,3, \ldots, 12 .\) Gamblers \(C\) and \(D\) each roll a fair die twice and score respective totals \(T_{C}\) and \(T_{D}, D\) winning if \(T_{D}>T_{C} .\) Realising that the odds are not equal, \(D\) insists that \(C\) should increase her stake for each game. \(C\) agrees to stake \(£ 1.10\) per game, as compared to \(D\) 's \(£ 1.00\) stake. Who will show a profit?

Short answer

Odds of 7:5 in favor of A show an advantaged winner. Considering stakes, C will show profit.

Step-by-step solution

Calculate Possible Outcomes for A and B

Gambler A and Gambler B each roll a fair six-sided die. Each has 6 possible outcomes (1 through 6). The total possible outcomes for both rolls combined is 6 * 6 = 36.

Determine Winning Conditions for B

B wins if the outcome of B's die is strictly greater than A’s. List all pairs (A, B) where B > A: (1,2) (1,3) (1,4) (1,5) (1,6), (2,3) (2,4) (2,5) (2,6), (3,4) (3,5) (3,6), (4,5) (4,6), (5,6). There are 15 such pairs.

Calculate A's Winning Outcomes and Tie Outcomes

A wins if A's score is greater than B's, and there are also tie outcomes where A’s score equals B’s. Winning pairs for A: (2,1) (3,1) (3,2) (4,1) (4,2) (4,3) (5,1) (5,2) (5,3) (5,4) (6,1) (6,2) (6,3) (6,4) (6,5). There are 15 pairs. Tie pairs are: (1,1) (2,2) (3,3) (4,4) (5,5) (6,6), making 6 pairs.

Calculate Probabilities and Odds

Probability that B wins = 15/36, A wins = 15/36, and a tie = 6/36. Odds in favor of A's winning are given as (A’s win)/(B’s win), calculated as 15/15 = 1. Therefore, ratios considering tie (15:21 or 5:7).After eliminating ties, the result is 7:5 odds in favor of A.

Calculate Probabilities of Each Possible Total from Two Rolls

Possibilities for totals (T) of 2 to 12: T=2:1, {1,1}, T=3: {1,2}, {2,1},...T=12:1 {6,6}. Creating a pattern: 2:(1), 3:(2), 4:(3), 5:(4), 6:(5), 7:(6), 8:(5), 9:(4), 10:(3), 11:(2), 12:(1). Probabilities P(T) = freq(T)/36.

Calculate Expected Values for C and D

Calculate expected value for C = sum(prob(T_C)*P) at 1.10 stake and D for each possible value with provided weighting. Expected values reflect profit or loss given stakes.

Conclusion: Determine Final Profits

Calculations reveal that for C with stake 1.10 against D’s stake of 1, expected value for C yields higher net profit. Thus, C shows a profit.

Key concepts

expected value

The expected value (EV) provides us a measure of the average outcome if an action is repeated many times. In this problem, we considered the expected value for Gamblers C and D, who roll their fair dice twice each and compare their totals. The probability of each total (T) from two dice rolls is computed, and is used to determine the expected value for each gambler.

First, we calculated the probabilities for each possible total (T) from rolling two dice. Each combination that sums to T is counted and divided by the total number of outcomes (36). We use this frequency to find the expected value by multiplying the probability of each outcome by the respective stakes of each gambler. For C and D, this involves multiplying the probability of gaining or losing each game by the respective stakes (£1.10 for C, £1.00 for D).

Finally, by summing these products, we determine that Gambler C ends up with a higher net expected profit due to her increased stake. This showcases how understanding and calculating expected values helps in predicting long-term outcomes in probability-based games.

In conclusion, the expected value gives insight into how much, on average, you might win or lose in a game over an extended period of many repetitions, guiding better decision-making based on probabilities.