# Probability finding a card in $n$ draws? [closed]

I remember seeing this as a trick before. I tried to recreate it but it's proving to be difficult.

I have a deck of $13$ cards, consisting of ace through king. I have a card number in my head (ace, $2$-$10$, jack, queen, king). What is the probability of finding it in $n$ draws? After each draw I put the card back and shuffle the deck.

## closed as off-topic by JMoravitz, Jean-Claude Arbaut, Daniel W. Farlow, Shailesh, LeucippusApr 15 '17 at 0:07

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• First, you don't need to have infinitely many cards. Consider simply a deck of $13$ cards (All the numbers) and once you have drawn a card, you put it back in the deck and reshuffle. – Zubzub Apr 14 '17 at 16:25
• As it is a deck with infinitely many cards (i.e. an infinite shoe), we may treat this as though it were drawing with replacement. The probability of seeing the first ace after exactly $n$ draws will correspond to finding a non-ace in the first $n-1$ draws followed by an ace for the $n$'th draw. See the geometric distribution on wikipedia. – JMoravitz Apr 14 '17 at 16:25
• @Zubzub Fixed it – Duncan Whyte Apr 14 '17 at 16:48
• You've been back to visit the page since my comment (as evidenced by your edit) but you did not seem to acknowledge it. Did you, or did you not understand my comment and still have a question? It seemed pretty clear to me... – JMoravitz Apr 14 '17 at 16:57
• Sorry, I' m quite new at probability. I was checking the wiki in your link. – Duncan Whyte Apr 14 '17 at 17:09

As it is essentially drawing with replacement, we notice that any particular draw will be the card we are looking for with probability $\frac{1}{13}$ and will not be with probability $\frac{12}{13}$.

As each draw is independent of one another, the probability of a specific sequence of draws will be the product of the probabilities of each draw individually. E.g. Ace-Ace-Nonace in that order will occur with probability $\frac{1}{13}\cdot\frac{1}{13}\cdot\frac{12}{13}$.

There are two possible interpretations of the question I can think of.

What is the probability that the first ace is drawn on the $n$'th draw?

This follows the geometric distribution with $p=\frac{1}{13}$. This corresponds to getting $n-1$ non-ace cards in a row followed by an ace. This occurs then with probability $\frac{12}{13}\cdot\frac{12}{13}\cdots\frac{12}{13}\cdot\frac{1}{13}=(\frac{12}{13})^{n-1}\cdot\frac{1}{13}$

What is the probability that within the first $n$ draws, at least one ace is drawn? (but it doesn't necessarily need to be the $n$'th draw where the ace occurs)

Getting at least one ace within the first $n$ draws means that the first ace we see occurs within the first $n$ draws, so we may simply sum over the geometric distribution up to $n$. $\sum\limits_{k=1}^n(\frac{12}{13})^{k-1}\frac{1}{13}$ which may be simplified using what you know about geometric series.

An alternate method to calculate this is by recognizing that getting at least one ace within the first $n$ draws is the opposite event of getting no aces within the first $n$ draws. Getting no aces corresponds to not getting an ace for the first draw followed by not getting an ace for the second draw followed by...etc...

Getting no aces in $n$ draws has probability then $\frac{12}{13}\cdot\frac{12}{13}\cdots\frac{12}{13}=(\frac{12}{13})^n$ so getting at least one ace in $n$ draws has probability $1-(\frac{12}{13})^n$. Had you simplified your geometric sum above, it would have arrived at this answer as well.

A word of warning, I mentioned that $Pr(A\cap B)=Pr(A)\cdot Pr(B)$, that is the probability event $A$ and event $B$ simultaneously occur is the probability of the product of their respective probabilities. This is true only for independent events, and is in general not true for arbitrary events.