Two cards are randomly drawn from a shuffled deck of 52. Find the probability that both are spades, given that at least one of the cards is a spade. Would you get a different answer if you were given that one of the cards is the ace of spades?


$a = \{\text{both cards are spades}\}$

$b= \{\text{at least one spade}\}$

$$P(a \mid b)= \frac{P(a \cap b)}{P(b)} = \frac{\binom{13}{2}}{?}$$

  • $\begingroup$ If you are given that one of the cards is a spade. This would be the same even if you were told this card was the ace of spades because the event of interest is whether or not the card is a spade. $\endgroup$ – Derek Sep 8 '17 at 2:51
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 8 '17 at 9:20

attempt: a={both cards spade}, b={at least one spade}, P(a|b)= P(a∩b)/P(b) (13 choose 2)/?

$b=\{1\text{ spade and }1\text{ other}\}\cup\{\text{both spades}\}$

$$\mathsf P(a\mid b) =\dfrac{\lvert a\cap b\rvert}{\lvert b\rvert} =\dfrac{\binom {13}{2}}{\binom{13}{1}\binom{39}{1}+\binom{13}{2}} =\dfrac{2}{15}$$

$c=\{1\text{ aces of spade and }1\text{ non-spade}\}\cup\{\text{both spades and one is ace}\}$

$$\mathsf P(a\mid c)=\dfrac{\lvert a\cap c\rvert}{\lvert c\rvert} =\dfrac{{\binom 11\binom {12}{1}}}{\phantom{\binom{1}{1}\binom{39}{1}}+\phantom{\binom 11\binom{12}{1}} }=\dfrac{\phantom{4}}{\phantom{17}}$$

That is all.


Note that if you have two spades, then you automatically have at least one spade. In other words, if $a$ holds, then $b$ also holds. This means that $P(a \cap b) = P(a)$

Now, $P(a)=\frac{13}{52}\cdot\frac{12}{51}$

Also, $P(b)=1-P(b^C)=1-\frac{39}{52}\cdot\frac{38}{51}$

If we let $c$ be the event that one of the cards is the ace of spades, then both cards being spades means that one of them is the ace of spades, so let's define $d$ as the event of having the ace of spades plus one other spade.

Then: $P(d|c)=\frac{P(d \cap c)}{P(c}=\frac{P(d)}{P(c)}$



  • $\begingroup$ $P(d)=2 \cdot \frac{1}{52} \cdot \frac{12}{51}$ right? $\endgroup$ – Derek Sep 8 '17 at 3:02
  • $\begingroup$ @Derek of course! I'll change that, thanks! $\endgroup$ – Bram28 Sep 8 '17 at 3:11

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