If you assign each face a value starting from 1 and ascending with each new glyph that appears the average of p_k will converge on (n+1)/2 assuming a fair die
-
-
-
Alternately, and faster, if each face is unique you could count how many rolls is required before a repeat glyph
End of conversation
New conversation -
-
-
This Tweet is unavailable.
-
A normal die seems unlikely. I’d assume uniform
- Show replies
-
-
-
i expect you'll like this... https://en.wikipedia.org/wiki/German_tank_problem …
-
is this similar? https://en.wikipedia.org/wiki/Urn_problem …
- Show replies
New conversation -
-
-
Not unless you're a decent deadlifter.
Thanks. Twitter will use this to make your timeline better. UndoUndo
-
-
-
max(p_k) for sufficiently large k. Else, estimate assuming a binomial distribution with P(X=1) for max(p_k) i.e p=1/n. Mean would be k/n. If (k/max(p_k)) estimates the number of observations max(p_k) then n = ~max(p_k)
Thanks. Twitter will use this to make your timeline better. UndoUndo
-
-
-
If I understand your question correctly, yes. Use Beta updating to estimate the possible proportion of each side after each roll. I have a short tutorial on this.
-
Ack! I meant Bayesian updating (although Beta updating can be used for a similar question). See http://www.incitedecisiontech.com/tutorials/Bayes_discrete_sequential_2.nb.html … (note - this is still a WIP)
End of conversation
New conversation -
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.