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Suppose I draw two cards, without replacement, one at a time, from a well-shuffled deck.

I want to know the probability that the second card is red.

Let A = "1st card is red"

Let B = "2nd card is red"

So I want to know P(B).

Since the probability that the second card is red depends on whether or not the first card was red, I can express this using the Law of Total Probability:

P(B) = P(B|A)P(A) + P(B | Abar)P(Abar) = (25/51)(26/52) + (26/51)(26/52)

which is approximately equal to 0.4902.

I understand that much, but then P(Bbar) should be approx. 1 - 0.4902 = 0.5098

But if I use the Law of Total Probability here, then I get:

P(Bbar) = P(Bbar | A)P(A) + P(Bbar | Abar)P(Abar)

= (26/51)(26/52) + (25/51)(26/52) = 0.5

Why are these numbers so different? Shouldn't I have gotten exactly 0.5 for P(B)?

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  • $\begingroup$ Your expression for $P(B)$ is identical to your expression for $P(A).$ $\endgroup$
    – Bridgeburners
    Commented Sep 30, 2019 at 14:42
  • $\begingroup$ how did you get .4902? (25/51)*(1/2)+(26/51)*(1/2) = (1/2)*(25/51+26/51) = (51/51)*1/2 = 0.5 $\endgroup$
    – Ryan
    Commented Sep 30, 2019 at 14:45
  • $\begingroup$ Thank you @Ryan! I plugged it in three or four times, got 0.4902. I plugged it in again and got 0.5. Whew! $\endgroup$
    – StatsSorceress
    Commented Sep 30, 2019 at 15:10

1 Answer 1

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You are asking the question about the second card before you look at the first card. Whether any given card in a shuffled deck is red, without any other knowledge about any of the cards, is 50%.

Also, you are giving two different answers for the same calculation. (25/51)(26/52) + (26/51)(26/52) = (26/51)(26/52) + (25/51)(26/52) ... and that equals 0.50.

In any event, the probability that a red card is the first card on a deck with 51 cards, having 25 red cards and 26 black cards is 25/51 = 0.4901961

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