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I'm supposed to compute a conditional probability and would like to make sure I've got this right.

Two cards are drawn from a 52 card deck (with four suits in it). The second card being a heart. What's the probability the first card is a heart?

The conditional probability to compute is then $$ P(\text{first=heart}|\text{second=heart}) \ = \ \frac{P(\text{first=heart AND second=heart})}{P(\text{second=heart})} $$ I believe the probability in the numerator should be $\frac{1}{4} \times \frac{12}{51} = \frac{1}{17}$ since you have a 1 in 4 chance to pull a heart to start, and then there are 12 hearts and 51 cards left after that.

The probability in the denominator is I believe $$ P(\text{first$\neq$heart AND second=heart}) + P(\text{first=heart AND second=heart}) = \frac{39}{52}\times \frac{13}{51} + \frac{1}{17} = \frac{13}{68} $$ At the end of the day I find that $$ P(\text{first=heart}|\text{second=heart}) = \frac{4}{17} $$ Is this correct? And is there a simpler/more intuitive way to calculate this?

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  • $\begingroup$ It should be $P(\text{first$\neq$heart AND second=heart}) + P(\text{first=heart AND second=heart}) = \frac{39}{52}\times \frac{13}{51} + \frac{1}{17} = \frac{1}{4}$. $\endgroup$
    – Klaas van Aarsen
    Commented Mar 3, 2020 at 22:36
  • $\begingroup$ Thanks, that was a typo. Fixed. $\endgroup$
    – QuantumEyedea
    Commented Mar 4, 2020 at 0:09

1 Answer 1

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Interchanging the two cards before looking at them will not change the probabilities. The conditional probability that the first is a heart, given that the second is a heart, is the same as the conditional probability that the second is a heart, given that the first is a heart. With the first heart removed from the deck, you have $51$ cards to choose from, of which $12$ are hearts. So $12/51 = 4/17$.

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