Exercise 10 Ch.3

Total number of permutations to deal 13 cards:

\[ P(52,13) = \frac{52!}{(52-13)!} \]

Number of permutations to deal an ace last:

\[ P(12,12) \times P(13,1) = \frac{12!}{(12-12)!} \times \frac{13!}{(13-1)!} \]

\[ P(52,13) = \frac{52!}{39!} \]

\[ P(12,12) \times P(13,1) = \frac{12!}{0!} \times \frac{13!}{12!} \]

\[ Probability = \frac{P(12,12) \times P(13,1)}{P(52,13)} \]

\[ P(52,13) = \frac{52!}{39!} = \frac{52 \times 51 \times 50 \times 49 \times 48 \times 47 \times 46 \times 45 \times 44 \times 43 \times 42 \times 41 \times 40 \times 39!}{39!} \]

\[ P(12,12) \times P(13,1) = \frac{12!}{0!} \times \frac{13!}{12!} = 12! \times 13 \]

\[ P(52,13) = 635,013,559,600 \]

\[ P(12,12) \times P(13,1) = 12! \times 13 = 479,001,600 \]

\[ Probability = \frac{479,001,600}{635,013,559,600} \approx 0.000753 \]