<p>In a game A throws two ordinary dice. If he throws 7 or 11 he wins. If he throws 2, 3 or 12 he loses. If he throws any other number, he throws again and continues to throw until either the number he threw first or 7 turns up. In the first case he wins and in the second he loses. Show that the odds against his winning is 251 : 244.</p>
Step-by-Step Solution
Key Concept: The game involves conditional probability across multiple stages: winning/losing immediately, or entering a recursive loop where the player must throw their initial number before 7 appears. Use the law of total probability and geometric series to sum probabilities across all possible outcomes.
<p><strong>Step 1: Identify immediate outcomes</strong></p><p>Immediate win (roll 7 or 11): P = 6/36 + 2/36 = 8/36 = 2/9</p><p>Immediate loss (roll 2, 3, or 12): P = 1/36 + 2/36 + 1/36 = 4/36 = 1/9</p><p><strong>Step 2: Identify point numbers and continue-playing outcomes</strong></p><p>Points (roll 4, 5, 6, 8, 9, 10): P = 24/36 = 2/3</p><p><strong>Step 3: For each point number, calculate P(win | point established)</strong></p><p>Once point n is established, A wins if n appears before 7. For each point:</p><p>• Point 4: P(4 before 7) = 3/(3+6) = 1/3</p><p>• Point 5: P(5 before 7) = 4/(4+6) = 2/5</p><p>• Point 6: P(6 before 7) = 5/(5+6) = 5/11</p><p>• Point 8: P(8 before 7) = 5/(5+6) = 5/11</p><p>• Point 9: P(9 before 7) = 4/(4+6) = 2/5</p><p>• Point 10: P(10 before 7) = 3/(3+6) = 1/3</p><p><strong>Step 4: Calculate total winning probability</strong></p><p>P(win) = 2/9 + (1/36)·1/3 + (2/36)·2/5 + (5/36)·5/11 + (5/36)·5/11 + (2/36)·2/5 + (1/36)·1/3</p><p>P(win) = 2/9 + 2(1/36·1/3) + 2(2/36·2/5) + 2(5/36·5/11)</p><p>P(win) = 2/9 + 2/108 + 8/180 + 50/396</p><p>Finding common denominator (3960):</p><p>P(win) = 880/3960 + 73.33/3960 + 176/3960 + 495/3960 = 244/495</p><p><strong>Step 5: Calculate odds against winning</strong></p><p>P(lose) = 1 - 244/495 = 251/495</p><p>Odds against = P(lose) : P(win) = 251 : 244</p>
Correct Answer: 251:244