# Doubts on a probability problem

Problem:

From the deck of the 52 cards,cards are drawn randomly without replacement.What is the probability of drawing a king of hearts at the third attempt? If it was drawn at the 15th attempt, what is the probability ?

I have solved the first by finding P(getting the card at third attempt) = P(not getting at first) $\times$ P(not getting at second) $\times$ P(getting at third attempt) = $\frac{51}{52} \times \frac{50}{51} \times \frac{1}{50} = \frac{1}{52}$.

This method seems fine when the times to be drawn is less but for 15 or say for $x$ ... I have solved for $15$ by just writing just two lines in C and got $\frac{1}{52}$, i do agree that 15 is not much big and it may be possible to use this during exam but does there any tricky method exist?

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No calculation is required and certainly no C. As you should observe when you start writing out the same product, the denominator of each next term will cancel the numerator of the previous, so the only terms left will be the denominator of the first and the numerator of the last terms. So the answer will always be $\frac{1}{52}$.

An easy way to see this is to stand the problem on its head: at any attempt, you are equally likely to draw any given card from the deck (why should the kind of hearts be any different from the 9 of clubs?). Since the are 52 cards, you get your probability.

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Thanks, +1 for explaining in the second paragraph :) – Quixotic Dec 9 '10 at 13:36
"at any attempt, you are equally likely to draw any given card from the deck" seems likely to me to be misinterpreted, as it's only true if you know nothing about earlier draws (which is, of course, exactly what's being talked about here). The analogy that I prefer is to think of the deck as being laid out all in a row, face down. The probability that turning over the first card yields the king of hearts is the same as the probability that turning over the 15th (or any other) card yields the king of hearts. – Isaac Dec 9 '10 at 16:42
@Isaac Of course I was talking about the probabilities before you start drawing. I don't understand what you mean by "misinterpreted". Clearly, once you have drawn the third card, the probabilities become 1 (for the card you drew) and 0 (for all others), so this can't be what I meant. Of course, you can think of the deck being laid out in a row, if you prefer. – Alex B. Dec 10 '10 at 1:11

A combinatorial approach is useful here: 51!/52! = 1/52.

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And why is this work ? – Quixotic Dec 9 '10 at 13:50
There are $52!$ possible outcomes, from which you are interested only in $51!$. – Shai Covo Dec 9 '10 at 13:58