From 52 card deck 5 cards are draw in succession. What is the probability of second or third card being diamonds? Looks simple but this kind of problems are my weak spot. All I can think of is to sum the probabilities of this events: (not diamond, diamond, not diamond) + (not diamond, diamond, diamond) + ... But this calculation doesn't look too pleasant... There must be a better way.
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asked Jun 10, 2014 at 14:10
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2 Answers
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A slightly better way is to use the probability of the negation, which is a common approach to certain "OR" problems, such as this one. That is, the probability that the second or third card is a diamond is one minus the probability that neither the second nor the third card is a diamond.
So, our answer will be $$ \begin{align} 1 - [&\mathbb{P}(\text{first non-diamond, then non-diamond, then non-diamond})\\ &+ \mathbb{P}(\text{first diamond, then non-diamond, then non-diamond})] \end{align} $$
answered Jun 10, 2014 at 14:16
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$\begingroup$ Note that you still have to sum multiple probabilities here, but there are only two possibilities (just for the first card) instead of 7 in your original approach. $\endgroup$– DuncanCommented Jun 10, 2014 at 14:19
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$\begingroup$ Alternatively, you can think of this as one event if you remember that we don't care what the first card is. So the events "first not diamond, second not diamond, third not diamond" and "first diamond, second not diamond, third not diamond" are really part of the same event, which is "first is something, second not diamond, third not diamond". In the same way, your original approach can be thought of as the sum over $3$ events rather than $6$. $\endgroup$ Commented Jun 10, 2014 at 14:23
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$\begingroup$ I thought the problem was about drawing with replacement... which I suppose is not accurate. Thanks. $\endgroup$ Commented Jun 10, 2014 at 15:05
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$\begingroup$ I agree that "in succession" probably means without replacement. $\endgroup$ Commented Jun 10, 2014 at 15:12
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Imagine you draw the five cards, but instead of looking at any of them you just put them face-down on the table. Now, turn over card number two and three and look at their suit. This is the same as taking the top card, putting it on the bottom (which could be seen as part of the shuffle process), and leaving the fourth and fifth card on top of the deck.
What I'm trying to get to here is that you can ignore the three other cards. It's not easy to accept, and I therefore show you the calculations on an easier example, like this: If I draw two cards, what is the probability that the second card is a diamonds?
If you do it the hard way, you have to take into account the first card, and you get $$ \frac{13}{52}\cdot \frac{12}{51} + \frac{39}{52} \cdot \frac{13}{51} $$ but this turns out to be $\frac{1}{4}$, just as if you'd completely ignored the first card. You can do the same in your example.
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$\begingroup$ So, if I understand correctly, then the answer simply is 1 - (39*38)/(52*51) ? $\endgroup$ Commented Jun 10, 2014 at 14:27
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$\begingroup$ This is one of those instances where "common sense" fails. $\endgroup$– tpb261Commented Jun 10, 2014 at 14:40
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$\begingroup$ @tpb261 Or it succeeds: The problem statement does not care at all what the first, fourth or fifth cards are, and thus, neither do we. But yes, it is counterintuitive on some level. $\endgroup$– ArthurCommented Jun 10, 2014 at 14:42