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Suppose we keep dealing cards from an ordinary 52-card deck until the first jack appears. What is the probability that at least 10 cards go by before the first jack?


The outcome space is ${52\choose10}$ Since you choosing $10$ spots from $52$ cards. Now how do I determine the probability that at least 10 card by by before the first jack?

There are 4 jacks in a 52 card deck. Pretty sure I have to figure out how many ways can I select 10 cards that are not jacks at the top. Unsure how do I go about this visually.

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  • $\begingroup$ So essentially you get no Jacks the first 9 goes. Then a Jack. So something like $\frac {52-4}{52} \frac {51-4}{51} \cdot \cdot \cdot \frac {4}{43}$. So we multiply all the times we don't get it in 9 goes and then all the ways we could get it in the 10th go. $\endgroup$ – Tony Hellmuth May 28 '18 at 0:38
  • $\begingroup$ $$P(1st \ Jack \ on \ n^{th} \ card)= \prod_{i=52}^{52-n+1} \frac {i-4}{i} \cdot \frac 4{52-n}$$ $\endgroup$ – Tony Hellmuth May 28 '18 at 0:44
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Hint:

  • There are ${52 \choose 10}$ equally probably sets of ten cards chosen from a full pack

  • There are ${48 \choose 10}$ equally probably sets of ten cards chosen from a full pack excluding the four Jacks

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The probability that the first card is not a Jack is 48/52. If this happens, the probability that the second card is not a Jack either is 47/51. The probability that none of the first ten are Jacks is $$\frac{48.47.46\ldots 39}{52.51.50\ldots 43}$$ $$=\frac{48!42!}{38!52!}=0.4134453782$$

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