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Do I consider the probability before drawing both cards or after?

Question more clearly: A single card is removed at random from a deck of $52$ cards. From the remainder we draw $2$ cards at random and find that they are both spades. What is the probability that the first card removed was also a spade?

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  • $\begingroup$ +1 because this question even confused me, and will confuse others who are seeing it for the $1$st time . Nice one $\endgroup$
    – Anonymous
    Commented Oct 6, 2020 at 15:44

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The trick here is to realize that the situation described is equivalent to removing two spades from the deck and then drawing a card at random from what remains. If you can calculate the probability of a spade occurring as the third card in that simpler setting, you're good to go.

Many probability problems can be similarly simplified by realizing that the specific order in which certain things are said to happen is irrelevant. That's not to say the order in which things happen is never relevant, just that the meanings of the letter strings (aka words) "first," "second," "third," etc., are sometimes interchangeable.

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    $\begingroup$ This answer is the correct answer. The probability of an event happening given other conditions is very relevant. $\endgroup$
    – Clayton
    Commented Oct 6, 2020 at 15:41
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To emphasize, @Barry's answer is correct and is the easiest way to think of the answer.

Since that confuses people for whatever reason, a way to convince people of this is instead to approach directly via definitions.

Recall that $\Pr(A\mid B) = \dfrac{\Pr(A\cap B)}{\Pr(B)}$ by definition. That is to say, the probability of an event $A$ occurring given that an event $B$ also occurs (whether that is past, present, or future... irrelevant) is the ratio of the probability of both $A$ and $B$ occurring over the probability that $B$ occurs regardless.

Here, letting $A$ be the event that the first card is a spade, $B$ the event that both the second and third cards are spades, we have:

$$\Pr(A\mid B) = \dfrac{\Pr(A\cap B)}{\Pr(B)}=\dfrac{\frac{13\cdot 12\cdot 11}{52\cdot 51\cdot 50}}{~~~~~\frac{13\cdot 12}{52\cdot 51}~~~~~} = \dfrac{11}{50}$$

In case $\Pr(B)$ confuses you, see this related question and/or again approach directly via definition. If you insist on doing this the long way, then recognize $\Pr(B) = \Pr(B\mid A)\Pr(A)+\Pr(B\mid A^c)\Pr(A^c)$ by law of total probability and see that it simplifies to what I claimed above.

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You are calculating the probability of the first card being a spade, therefor the next two additional draws mean nothing. The chance of drawing a spade from a deck of 52 would be 25%. It's as simple as that.

I hope this helps.

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    $\begingroup$ What if instead of drawing two additional cards and finding them both to be spades the problem had said we draw thirteen additional cards and find all thirteen to be spades? Would that still mean nothing? $\endgroup$
    – Barry Cipra
    Commented Oct 6, 2020 at 15:14
  • $\begingroup$ @BarryCipra it is actually $25$% , when you are taking the $1st$ card the probability of the later cards has no meaning. If it would have been your case, I would say the answer is $0$% , since the OP already mentions in his question that the $1$st card is not a spade (with a bit of logic). If it would have been $12$ spade cards I would still say it is $25$% . I can see you are trying to confuse the one who answered :) $\endgroup$
    – Anonymous
    Commented Oct 6, 2020 at 15:23
  • $\begingroup$ Yes, because the question is asking what the probability of the first card being a spade is. They remove one card initially, unaware of what it is. The chance of it being a spade is 25% They then remove 2 cards from the deck, these 2 cards are spades. The chance that the initially removed card is also a spade is still 25% as removing cards from the deck doesn't change the fact that the chance of the initial draw from deck of 52 being a spade is 25%. $\endgroup$
    – Brice Cawley
    Commented Oct 6, 2020 at 15:34
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    $\begingroup$ @Anonymous, I am not trying to confuse anyone, I'm trying to point out the flaw in the answerer's logic. $\endgroup$
    – Barry Cipra
    Commented Oct 6, 2020 at 15:34
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    $\begingroup$ So they key word here is (also) insinuating that all 3 draws would have to be spades and what is the probability of that occurring? (What is the probability that the first card removed was also a spade?) $\endgroup$
    – Brice Cawley
    Commented Oct 6, 2020 at 15:57

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