Timeline for Calculate sum of probabilities in multinomial model
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 3, 2015 at 13:31 | vote | accept | chrisoutwright | ||
| Sep 2, 2015 at 22:14 | answer | added | ciao | timeline score: 4 | |
| Sep 2, 2015 at 18:58 | comment | added | chrisoutwright | It would be helpful if you could state how you would break it into different problems. For me this is essentially one probability problem: Find the complement of the calculation done over all these sums. But I can't find anything on this topic right now. | |
| Sep 2, 2015 at 17:28 | comment | added | ciao | You've over-thought this, I think. Think about breaking if into two different (well known) problems... | |
| Sep 2, 2015 at 14:28 | comment | added | chrisoutwright | I posted the code with multinomial in it as well now. Sadly it is not faster! | |
| Sep 2, 2015 at 14:27 | history | edited | chrisoutwright | CC BY-SA 3.0 | added 536 characters in body |
| Sep 2, 2015 at 14:01 | comment | added | chrisoutwright | @guess-who-it-is: Yes but as I ran into some issues I used the definition, this shouldn't pose any issue. Maybe I should pose this question to people familiar with this Probability Calculus as for getting theoretical assistance. Nevertheless the page linked above uses a different calculation I strongly assume, so maybe someone knows how to do this. This shouldn't be a world-moving topic. | |
| Sep 2, 2015 at 13:54 | comment | added | J. M.'s missing motivation | Did you know that Multinomial[] is built-in? | |
| Sep 2, 2015 at 13:51 | history | edited | chrisoutwright | CC BY-SA 3.0 | added 62 characters in body |
| Sep 2, 2015 at 13:48 | comment | added | chrisoutwright | The code is nothing else than all the applicable summation of the Probability mass function of the Multinomial distribution, when looking at the possible series in question (e.g. sample n=10, two distinct special figures and all other 8 are duplicates of them). This is one series but there are more where I get these two figures (2 special out of 15) actually! So therefore I have to sum. The mathematics shouldn't be the difficult part in it. So I can't ramify over every detail when the topic is actually mainstream. | |
| Sep 2, 2015 at 13:40 | history | edited | chrisoutwright | CC BY-SA 3.0 | added 4 characters in body |
| Sep 2, 2015 at 13:35 | comment | added | chrisoutwright | You would like to get at least one specific special figure out of 15, these 15 only appear on average once in every 7th egg, but you don't know which of course. So therefore $$\frac{104}{105}$$ chance of not getting this special figure in the first draw. n is the sample size, so the number of eggs you buy for your observation. The Problem I have is the solving time, so I would appreciate help in this. | |
| Sep 2, 2015 at 13:30 | comment | added | yohbs | First, please define everything properly (e.g. what is $n$?). Second: I don't understand the problem. Why isn't the answer for $n=1$ simply $$1-\left(\frac{6}{7}\right)^n\ ?$$ | |
| Sep 2, 2015 at 13:28 | history | edited | chrisoutwright | CC BY-SA 3.0 | added 256 characters in body |
| Sep 2, 2015 at 13:28 | comment | added | demm | There is an English version of the site that you refer to: verklagekasper.de/ueei-en.html | |
| Sep 2, 2015 at 13:25 | history | edited | chrisoutwright | CC BY-SA 3.0 | added 256 characters in body |
| Sep 2, 2015 at 13:04 | history | asked | chrisoutwright | CC BY-SA 3.0 |