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I'm looking for the probability that we first get a king on the nth card draw when drawing from a pack of 52 cards.

Here's what I have done -

Let $A_i$ be the event that we don't get a king on card $i$.

Let $K$ be the event of getting a king.

Want to find

P(King occurs on nth card)

$= P(A_1 \cap A_2 \cap ... \cap A_{n-1} \cap K)$

$= P(K | A_1 \cap A_2 \cap ... \cap A_{n-1})P(A_{n-1}| A_1 \cap A_2 \cap ... \cap A_{n-2})...P(A_2 | A_1)P(A_1)$

$$= \frac{48}{52}\frac{48}{51}\frac{48}{50}...\frac{48}{52-n+2}\frac{4}{52-n+1}$$

$$= \frac{(4)(48)^{52-n+2}}{^{52} P_n}$$

Is that correct?

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I assume you dont put the cards back. In this case you get $\frac{48}{52}\frac{47}{51}\frac{46}{50}\ldots$, since the non-King cards get reduced after every draw. – A.Schulz Nov 30 '12 at 14:00
you also must account for the possibility of drawing a king on any $n-i$ draw. Your first three cards could be kings, and there is still a king left in the deck. Alternatively, you should clarify the problem if you mean that the first king drawn is on the nth card. – Emily Nov 30 '12 at 14:31
@EdGorcenski Sorry that should be 'first get a king'. I've edit the original post. – sonicboom Nov 30 '12 at 14:36
up vote 2 down vote accepted

use hypergometric Probability exactly one of first n cards is king is:


So the probability the nth card is the king is this multiplied by probability it's the nth card of n, which is, I think


$n=1$ gives $$4*\frac{\binom{48}{0}}{\binom{52}{1}} = 1/13$$ which is correct.

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aagh, thought it looked too easy. That's the probability that exactly one of the first n cards drawn is a king. So it has to be adjusted for position. Divide the above by n? – hal c yon Nov 30 '12 at 17:43

Here's a quick solution. Let $E_n$ represent drawing a King on the $n$th draw. Then,

$$P(E_n) = \prod_{i=0}^{n-2} \left (\frac{48-i}{52-i} \right )^{n-i} \frac{4}{52-n+1} $$

for all $n \le 48$.

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Subbing in $1$ I get $\frac{4}{51}$ when it should be $\frac{4}{52}$? – sonicboom Nov 30 '12 at 16:27

Here is something similar worked out. You can easily extend it from here. There are also other similar questions right here on stackexchange.

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