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My course says that if you draw two cards from a deck, define the events: S1 = ‘first card is a spade’ and S2 = ‘second card is a spade’ then P(S2)=13/52. Why 13/52? My thinking is this:

P(S2) is the probability that the second card is a spade given that we do not know whether the first card is a spade or not. If the first card is a spade then the chance that the second card is a spade is 12/51. If the first card is not a spade then the chance is 13/51. So we have a 13/52 chance that P(S2) is 12/51 and we have a 39/52 chance that P(S2) is 13/51. How can you say P(S2)=13/52?

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    $\begingroup$ Isn’t $\frac{13}{52}\frac{12}{51}+\frac{39}{52}\frac{13}{51}=\frac{13}{52}$ anyways? $\endgroup$
    – user700480
    Jun 2, 2022 at 19:37

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You already have a good start to apply Law of Total Probabilities to verify this result:

$$ \begin{align} P(S_2) &= P(S_2\mid S_1)P(S_1) + P(S_2 \mid S_1^c)P(S_1^c) \\ &= \frac {12} {51} \times \frac {13} {52} + \frac {13} {51} \times \frac {39} {52} \\ &= \frac {13} {52} \left(\frac {12} {51} + \frac {13} {51} \times 3 \right) \\ &= \frac {13} {52} \end{align}$$

And in fact the order of the card is not important if you do not have the result of other draw. The marginal probability of any of the card being spade is the same.

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