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I have a question regarding a standard deck of 52 cards.

Say that you draw 5 cards. You are given that 4 of the cards are spades and then need to work out the probability that all 5 cards are spades. Naturally the probability would be the number of combinations of 5 spades over the sum of the number of combinations of 4 spades and the number of combinations of 5 spades.

I understand the above, but what I don't quite understand - but know is wrong - is the following. If 4 cards are spades then there should be 9 spades left in the deck. So the odds of drawing a fifth spade would be 9/48. 9/48 is a vastly different probability from what the above gives you. It would be very helpful if anyone had a fairly intuitive explanation as to why this second option doesn't work.

Edit: I meant a suite not a specific card.

RobPratt
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Steven Panagiotidis
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    Not following. Since there are only $4$ aces, you can't draw $5$. Or did you mean to say that you are drawing with replacement? – lulu Aug 20 '22 at 17:40
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    A standard deck of cards only have 4 aces, that of spade, hearts, diamond and clubs. – InanimateBeing Aug 20 '22 at 17:40
  • As there are 9 left after drawing 4, I think OP means a suit, not a value. – Peter Phipps Aug 20 '22 at 17:41
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    Did you mean to say that $5$ of the drawn cards were spades or something like that? – lulu Aug 20 '22 at 17:41
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    Guessing that you meant a specific suit, let's say spades, instead of "aces", note that there is a big difference between saying that the first four cards you drew were spades and saying that four out of five cards were spades. See the Boy/Girl probability paradox. – lulu Aug 20 '22 at 17:43
  • There is a difference between saying that of the five cards, the first four cards are spades (this would be equivalent to drawing one at a time and looking at them individually), as opposed to saying that of the five cards, at least four cards are spades. In the first case, you only have SSSSS, SSSSH, SSSSD, SSSSC; in the second case, you have SSSSS, HSSSS, SHSSS, SSHSS, SSSHS, SSSSH, DSSSS, SDSSS, etc. (Not that those are equiprobable in either case, but you can see that there are different sample spaces.) – Brian Tung Aug 20 '22 at 19:15

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This is drawing without replacement

Your $\frac 9{48}$ attempt, equal to $\frac 9{9+39}=0.1875$, is answering a different question, namely the probability that the fifth card drawn is a spade given that the first $4$ drawn were all spades.

But in the original question the (at least) $4$ spades can be any of the $5$, and if the other card is not a spade then it can be in any of the $5$ positions, which makes the answer $\frac 9{9+5\times 39} = \frac{9}{204} \approx 0.044$. There are several other ways of doing the calculation which produce the same result, including:

  • as combinations: $$\frac{{5\choose 5}{13 \choose 5}{39 \choose 0}}{{5\choose 5}{13 \choose 5}{39 \choose 0}+{5\choose 4}{13 \choose 4}{39 \choose 1}}$$

  • as probabilities: $$\tfrac{\frac{13}{52}\frac{12}{51}\frac{11}{50}\frac{10}{49}\frac{9}{48}}{\frac{13}{52}\frac{12}{51}\frac{11}{50}\frac{10}{49}\frac{9}{48}+\frac{13}{52}\frac{12}{51}\frac{11}{50}\frac{10}{49}\frac{39}{48} +\frac{13}{52}\frac{12}{51}\frac{11}{50}\frac{39}{49}\frac{10}{48} +\frac{13}{52}\frac{12}{51}\frac{39}{50}\frac{11}{49}\frac{10}{48} +\frac{13}{52}\frac{39}{51}\frac{12}{50}\frac{11}{49}\frac{10}{48} +\frac{39}{52}\frac{13}{51}\frac{12}{50}\frac{11}{49}\frac{10}{48}}$$

and you can see the extra factor of $5$ appearing in both

Henry
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