<p>Three persons <i>A</i>, <i>B</i> and <i>C</i>, in order, cut a pack of cards replacing them after each cut on the condition that the first who cuts a spade shall win the prize. Find their respective chances.</p>
Step-by-Step Solution
Key Concept: Each player wins if they cut a spade on their turn while all previous players cut non-spades. The probability of cutting a spade is 13/52 = 1/4, and non-spade is 39/52 = 3/4. Sum the infinite geometric series for each player's winning probability.
<p><strong>Step 1:</strong> Identify basic probabilities. P(spade) = 13/52 = 1/4, P(non-spade) = 39/52 = 3/4.</p><p><strong>Step 2:</strong> Player A wins by cutting spade on turn 1, 4, 7, ... (every 3rd turn starting from 1).</p><p>P(A) = 1/4 + (3/4)³·(1/4) + (3/4)⁶·(1/4) + ... = (1/4)[1 + (27/64) + (27/64)² + ...]</p><p>P(A) = (1/4) · 1/(1 - 27/64) = (1/4) · 64/37 = 16/37</p><p><strong>Step 3:</strong> Player B wins by getting (non-spade, spade, non-spade, non-spade, spade, ...).</p><p>P(B) = (3/4)·(1/4) + (3/4)⁴·(1/4) + (3/4)⁷·(1/4) + ... = (3/16)[1 + (27/64) + (27/64)² + ...]</p><p>P(B) = (3/16) · 64/37 = 12/37</p><p><strong>Step 4:</strong> Player C wins when A and B fail in their turns.</p><p>P(C) = (3/4)²·(1/4) + (3/4)⁵·(1/4) + (3/4)⁸·(1/4) + ... = (9/64)[1 + (27/64) + (27/64)² + ...]</p><p>P(C) = (9/64) · 64/37 = 9/37</p><p><strong>Verification:</strong> 16/37 + 12/37 + 9/37 = 37/37 = 1 ✓</p><p>∴ <strong>P(A) = 16/37, P(B) = 12/37, P(C) = 9/37</strong></p>
Correct Answer: P(A)=16/37, P(B)=12/37, P(C)=9/37