Timeline for Dice roll expectation question [closed]
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 9, 2020 at 17:52 | comment | added | Jonah765 | @MattF. seems as though a moderator closed it, not much I can do. It was my error posting in the wrong place originally. If you decide to look further into it perhaps let me know | |
| Jul 9, 2020 at 17:51 | vote | accept | Jonah765 | ||
| Jul 9, 2020 at 17:13 | comment | added | user44143 | It is oddly close to 22. If the question is reopened and the current answer is unaccepted, I might play around more. | |
| Jul 9, 2020 at 16:24 | comment | added | Jonah765 | @MattF. interesting, thanks. 100 rolls is quite arbitrary but I wonder of there is any significance to this number. | |
| Jul 9, 2020 at 5:11 | review | Reopen votes | |||
| Jul 9, 2020 at 11:52 | |||||
| Jul 9, 2020 at 4:47 | comment | added | user44143 | Mathematica gives 21.997 from a recursion for the ways to get a $k-$fold mode of $max$ after $tot$ $n-$faced dice-rolls: $$\texttt{ f[tot_,n_,max_,k_]:=f[tot,n,max,k]=If[max<=tot/n,0,}$$ $$\texttt{Binomial[n,k]Product[Binomial[tot-j max,max],{j,0,k-1}]Sum[f[tot-k max,n-k,m,j],{m,1,max-1},{j,1,n-k}]];}$$ $$\texttt{ f[tot_,n_,max_,n_]:=If[max==tot/n,Product[Binomial[j max,max],{j,1,n}],0];}$$ $$\texttt{ f[tot_,1,max_,k_]:=If[And[max==tot,k==1],1,0]}$$ $$\texttt{Sum[max f[100, 6, max, j], {max, 0, 100}, {j, 1, 6}]/6^100 // N}$$ | |
| Jul 9, 2020 at 2:39 | history | closed |
YCor Gerald Edgar R.P. Yemon Choi Yoav Kallus |
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| Jul 8, 2020 at 21:44 | vote | accept | Jonah765 | ||
| Jul 9, 2020 at 17:50 | |||||
| Jul 8, 2020 at 21:23 | answer | added | Iosif Pinelis | timeline score: 2 | |
| Jul 8, 2020 at 21:19 | review | Close votes | |||
| Jul 9, 2020 at 2:39 | |||||
| Jul 8, 2020 at 20:26 | review | First posts | |||
| Jul 8, 2020 at 21:02 | |||||
| Jul 8, 2020 at 20:26 | history | asked | Jonah765 | CC BY-SA 4.0 |