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I am reviewing my probability lecture notes and I decided to try and do the exercises that were solved in the lecture.

I tried to solve the following problem, but came up with a different answer than the one given in the lecture, so I suspect I got it wrong.

We draw cards from a deck of cards (with $52$ cards), what is the probability that the first king was drawn at the $n-th$ draw ?

My attempt:

The total number of sequences of $n$ cards is $\binom{52}{n}\cdot n!$ .

There are $\binom{4}{1}$ ways to get the king in the $n-th$ draw.

There are $\binom{52-4}{n-1}\cdot(n-1)!$ sequences of $n-1$ cards with no king in them.

Hence the answer is $$ \frac{\binom{4}{1}\cdot\binom{48}{n-1}\cdot(n-1)!}{\binom{52}{n}\cdot n!} $$

Can someone please help me understand my mistake ?

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Are you sure your answer is different, as opposed to just looking different? –  Gerry Myerson Feb 4 '13 at 11:42
    
@GerryMyerson - Yes (unless I made another mistake...it is not the same for $n=3$) –  Belgi Feb 4 '13 at 11:48
    
For $n=3$, your formula gives $(4)(48)(47)/(52)(51)(50)$, and Damien's gives $(48/52)(47/51)(4/50)$, which is the same thing. If Damien is right, so are you (at least, for $n=3$). –  Gerry Myerson Feb 4 '13 at 12:30
    
@GerryMyerson - I will check this again and post a comment –  Belgi Feb 4 '13 at 12:53
    
Your answer is the same as Damien’s: $$\frac{\binom41\cdot\binom{48}{n-1}\cdot(n-1)!}{\binom{52}{n}\cdot n!}=\frac{4\cdot 48^{\underline{n-1}}}{52^{\underline n}}\;,$$ where $x^{\underline n}$ is the falling factorial. –  Brian M. Scott Feb 4 '13 at 22:57

2 Answers 2

up vote 2 down vote accepted

Your answer is the same as Damien’s:

$$\frac{\binom41\cdot\binom{48}{n-1}\cdot(n-1)!}{\binom{52}n\cdot n!}=\frac{4\cdot 48^{\underline{n-1}}}{52^{\underline n}}\;,$$

where $x^{\underline n}$ is the falling factorial. You just got there by a different route.

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To get the first knight at the n-th draw, it means that from draw 1 to draw $n-1$ you are not getting a knight and at the n-th draw it is a knight. There are 4 knights in $52$ cards, so the probability not to get a night at draw 1 is $48/52$.

Then for the second draw, there is one card less in the deck, so the probability to have no nights both draw 1 and draw 2 is $48/52 \times 47/51$ etc.

And the probability to get specifically a knight at draw $n$ is $4/(52 - n + 1)$.

Multiply everything and you'll get the answer.

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knight? king!${}$ –  Gerry Myerson Feb 4 '13 at 11:38
    
Yes, I'll go shoot myself :\ –  Damien L Feb 4 '13 at 11:39
    
Thank you for your answer, but this does not explain what was my mistake –  Belgi Feb 4 '13 at 11:41

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