Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
1  
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. –  Arkamis 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

3 Answers 3

up vote 2 down vote accepted

use hypergometric http://en.wikipedia.org/wiki/Hypergeometric_distribution Probability exactly one of first n cards is king is:

$$4\frac{\binom{48}{n-1}}{\binom{52}{n}}$$

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

$$\frac{4\binom{48}{n-1}}{n\binom{52}{n}}$$

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

share|improve this answer
    
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 $n \le 48$.

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.