Timeline for What is the amount of draws necessary to see all red cards from a standard deck of 52 cards if you draw 5 cards from the deck?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 5, 2019 at 12:59 | vote | accept | user3223162 | ||
| Feb 5, 2019 at 12:59 | comment | added | user3223162 | You are correct, my tests contained an error and after correction confirm your results. I have ran the formula alongside with some simulations and the results correspond. For now, I am marking this as solved, I will try and confirm my findings in some existing literature on the subject. Thanks! | |
| Feb 4, 2019 at 20:20 | comment | added | awkward | @user3223162 There must be something amiss, then, because my simulation agrees with the analytical result I posted. I have added the simulation to the solution, above. | |
| Feb 4, 2019 at 20:19 | history | edited | awkward | CC BY-SA 4.0 |
added Monte Carlo results
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| Feb 4, 2019 at 15:46 | comment | added | user3223162 | In that case, that equation does not solve my issue. Furthermore, simulations I ran disagree with your attempt. Still, I appreciate the effort | |
| Feb 4, 2019 at 12:56 | comment | added | awkward | @user3223162 I haven't read the paper you refer to, but in the question you link to, the assumption is that the urn contains one each of $n$ distinct balls, so the number of distinct balls is the same as the number of balls in the urn. | |
| Feb 4, 2019 at 11:11 | comment | added | user3223162 | Thank you for your answer! After some digging, in a paper The Coupon Subset Collection Problem by Adler, Ilan and Ross, Sheldon M I have found equation 7 that solves the issue, and link this question as well. What keeps me confused is that the expectation depends only on the number of distinct balls in the set and the size of the drawn subset, but not on the size of the entire set. Am I misunderstanding something? | |
| Feb 1, 2019 at 13:28 | history | edited | awkward | CC BY-SA 4.0 |
additional explanation
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| Jan 31, 2019 at 16:37 | history | answered | awkward | CC BY-SA 4.0 |