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If you deal one card at a time from a shuffled deck, what is the probability that a red ace will first appear on the k th card? Example- the probability that a red ace will first appear on the first card dealt is 2/52. If k=10, what is the probability that a red ace will first appear when you deal the 10th card. Give the expression for arbitrary k.

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closed as off-topic by Parcly Taxel, Rohan, Ove Ahlman, Graham Kemp, Brian Borchers Dec 12 '17 at 0:02

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    $\begingroup$ The title question is different from the body question. The probability that the $52$nd card dealt is a red ace is $2/52.$ The probability that a red ace will first appear on the $52$nd card is $0.$ $\endgroup$ – bof Dec 11 '17 at 7:45
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There are $\binom{52}2$ possible locations for the two red aces, all equally likely. You are asking about the event that one of two red aces is in the $k^{\text{th}}$ position, and the other one is in one of the $52-k$ subsequent positions. The probability of that is $$\frac{52-k}{\binom{52}2}=\frac{52-k}{1326}.$$

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  • $\begingroup$ An elegant solution. I calculated the probability as $$\frac{\binom{50}{k - 1}}{\binom{52}{k - 1}} \cdot \frac{2}{52 - (k - 1)}$$ by finding the probability that a red ace was not selected among the first $k - 1$ cards and that a red ace was selected with the $k$th card. Of course, it reduces to the expression you found. $\endgroup$ – N. F. Taussig Dec 11 '17 at 11:29
  • $\begingroup$ bof's succinct solution can be put into elementary words. The Probability that the first red ace shows up on the kth card, is proportional to (52-k), the available places the second red ace can appear. i.e. Prob of k = Cx(52-k) where C is the constant of proportionality. The sum of probabilities from k=1 to k=51 adds up to one. 1= sum over k = 1 to k=51 of C x ( 52-k ) The sum of k from k=1 to k=51 is equal to (51x52)/2, so C is the inverse of 51x (52 -52/2) = 51x26 =1326 and the probability sought is (52-k)/1326. $\endgroup$ – steve newman Dec 11 '17 at 20:19
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For a formal proof I would use the hypergeometric distribution. The event you are interested in can be stated as

after k cards exactly one red ace was drawn, but no red ace was drawn among the first k-1 cards.

Using the formula

$$P(A \backslash B) = P(A) - P(A \cap B)$$

one can compute the probability of interest by calculating the probability for the events

after k cards exactly one red ace was drawn

and

after k-1 cards exactly one red ace was drawn and the k-th card was not a red ace.

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