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What is the probability that the first card is spades given that the second and third cards are spades?

Attempt: Use Bayes' rule to calculate $P(E_1|E_2E_3)$

  • $E_1$ = first card is spade
  • $E_2$ = second card is spade
  • $E_3$ = third card is spade

$$ P(E_1) = \frac{13}{52} $$ $$ P(E_2) = \frac{13}{52} $$ $$ P(E_3) = \frac{13}{52} $$

$$ P(E_1|E_2E_3) = \frac {P(E_2E_3|E_1)P(E_1) }{ P(E_2E_3)} $$ So this $$= \frac{(\frac{12}{51} \cdot \frac{12}{51})(\frac{13}{52})} {\frac{13}{52} \cdot \frac{13}{52}} = 0.2214$$

Is this correct?

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so the probability is 0.12? – lord12 Sep 19 '11 at 3:36
up vote 4 down vote accepted

Assume that we are dealing from a thoroughly shuffled standard deck of cards.

All sequences of cards are equally likely. Thus the required probability is the probability that the third card is a spade, given that the first two are spades. This probability is (clearly) equal to $11/50$.

Comment: We can obtain the same result by using Bayes' Rule, and working quite a bit harder. And in any complicated calculation, there is always the possibility of error. For example, the idea in the OP's calculation is correct. However, $$P(E_2E_3|E_1)=\frac{12}{51}\cdot \frac{11}{50}.$$ Also, the denominator should be $$P(E_2E_3)=\frac{13}{52}\cdot \frac{12}{51}\cdot\frac{11}{50}+\frac{39}{52}\cdot \frac{13}{51}\cdot\frac{12}{50}.$$ With these modifications, the OP's argument goes through just fine. After some cancellation, we obtain $11/50$.

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