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So if we have a regular deck of $52$ cards that is shuffled, then what is the probability of drawing the first red ace on the $k$-th card? And which k would maximize the probability?

I wasn't entirely sure of computing the probability. Here are my thoughts:

Have $52$ cards, thinking about the $k$-th card. There are two red aces, so the probability would be $\frac{2}{52-(k+1)}$?

I am entirely lost on maximizing $k$. Would anyone be able to help me out? Or guide me on maximizing $k$? Is my probability correct?

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3 Answers

up vote 2 down vote accepted

In a well-shuffled deck, the positions of the two red aces are two randomly chosen numbers from $\{1,2,\dots, 52\}$. There are ${52\choose 2}$ such pairs of numbers.

If the first red ace is at position $k$, then the other red ace must belong to the set $\{k+1,\dots, 52\}$ and there are $52-k$ such possibilities.

Therefore the probability that the first red ace is in position $k$ is the ratio $${52-k\over {52\choose 2}},$$ which is maximized by taking $k$ as small as possible.

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So would 52 choose 2 be the possible placements of red aces in the deck? –  Wooooop Mar 4 '13 at 5:12
Yes that's right. –  Byron Schmuland Mar 4 '13 at 5:31
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There are two red aces in the deck, diamond ace and heart ace. The probability that the first card is not red ace is $\frac{52-2}{52}$. The probability that the second one is not the red ace, given than the first one was not is $\frac{51-2}{51}$. The probability the the first $k-1$ are not red ace and the $k$-th one is thus equals: $$ \left(\prod_{m=1}^{k-1} \frac{52-1-m}{52-m+1}\right) \cdot \frac{2}{52-(k-1)} = \left( \frac{(52-(k-1))(52-k)}{(52-1+1)(52-2+1)}\right) \cdot \frac{2}{52-(k-1)} = \frac{2(52-k)}{52 (52-1)} $$ Thus the probability is maximal for $k=1$.

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What does the (52-1-m)/(52-m+1) stand for? I understand that the second part would be the probability for the two red aces. Also, how did you find that the maximal is k=1? And thank you for your help!! –  Wooooop Mar 4 '13 at 3:51
It stands for the probability of not drawing a red ace at $m$-th step, given that no red ace had been drawn on prior steps. –  Sasha Mar 4 '13 at 5:05
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Another way of getting the same answer:

Consider a row of 52 empty slots, and put the ordered cards in randomly selected slots. The first two cards to be placed are the two red aces (why not?). The chances of the first red ace going in the k-th slot out of the 52 empty ones is $\frac{1}{52}$, and after that the chance of the second going in any slot after that, out of the 51 remaining, is $\frac{52-k}{51}$

Or, the first red ace could go in any of the "after" positions, probability $\frac{52-k}{52}$, and the second red ace could go in the k-th slot out of the 51 remaining, with probability $\frac{1}{51}$.

Multiplying out and adding the two results gives the same answer as above.

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