Timeline for From 52 cards, we draw N times one card with the return. Calculate the probability that each card in the deck will be drawn at least once.
Current License: CC BY-SA 4.0
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| when toggle format | what | by | license | comment | |
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| Jun 23 at 21:08 | vote | accept | thefool | ||
| Jun 16 at 18:44 | comment | added | thefool | @Henry does my answer look correct? | |
| Jun 16 at 18:44 | answer | added | thefool | timeline score: 1 | |
| Jun 16 at 18:16 | comment | added | thefool | Ok, now I understand. Thank you for help. I will update my original question or write an answer containing the use of inclusion-exclusion to bring it home. Thanks! | |
| Jun 16 at 18:01 | comment | added | Henry | @thefool You do not need Stirling numbers of second kind to find the values $p(52,52)\approx 4.7\times 10^{-22}$ or $p(225,52)\approx 0.50579$ or $p(1000,52)\approx 0.9999998$; you can use a spreadsheet to get all these values simultaneously. But an exact compact expression will require Stirling numbers of the second kind or something equivalent. | |
| Jun 16 at 17:48 | comment | added | thefool | @Henry, so the answer is: $p(N, 52) = p(N-1, 52) + \frac{1}{52} p(N-1, 51)$ and it is impossible to solve that recoursion (express it as a compact function of N) without Stirling numbers of second kind? | |
| Jun 16 at 17:43 | comment | added | Henry | @thefool: you want $P(N,52)$ for all the cards - in the linked recursion answer $P(N,365)$ for all the $365$ days to be somebody's birthday. "Unfolding" may not be the best approach; starting with $m=0$ and then increasing towards $52$ may be more efficient. If you want to avoid sums (i.e. reject recursion or inclusion-exclusion) then exact results and compact expressions require Stirling numbers of the second kind or something equivalent. | |
| Jun 16 at 17:39 | comment | added | thefool | Should I therefore unfold the recoursion for $m = 52$ and 'go down with m' until $m = 0$? That will be my function of N, right? | |
| Jun 16 at 17:30 | comment | added | thefool | @Henry I see the recursion.If I replace 365 days with 52 cards I get: $p(N,m) = \frac{m}{52}p(N-1,m)+\frac{53-m}{52}p(N-1,m-1)$. But that is a recoursion on $N$ and $m$. In my case, I need compact function of $N$ (not including sums). | |
| Jun 16 at 5:20 | comment | added | user2661923 | See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula. | |
| Jun 16 at 3:24 | history | edited | RobPratt |
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| Jun 16 at 1:57 | comment | added | Henry | My answer to the linked question gives a recursion if you replace $365$ days with $52$ cards. For an example of inclusion-exclusion see math.stackexchange.com/a/4813267/6460 using $m=52$ | |
| Jun 16 at 1:27 | comment | added | thefool | @Henry thank you for help! Could you show me the inclusion-exclusion and recursion ideas? Provided link and the answer mention Stirling numbers of second kind and I would love to avoid those. | |
| Jun 16 at 0:21 | comment | added | Henry | The answer is $\dfrac{52!\, S_2(N,52)}{52^N}$ where $S_2(N,52)$ is a Stirling number of the second kind. You can also use inclusion-exclusion or recursion to find it. math.stackexchange.com/questions/4506194/… is one of many similar questions on this site. | |
| Jun 16 at 0:08 | history | asked | thefool | CC BY-SA 4.0 |