Unit 5 15 Answer

Unit 5 15 Answer

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  Howto & Style

[Dailos Guerra]

Don't understand the reason why he used 11 words in spite the dictionary has 12. Does anyone know?

[Andrew Wilkie]

I got 25 based on the arc-count method we learnt in Unit 3 i.e. 1 for the root node (as it has no incoming arcs) and 2^n-arcs (i.e. n incoming arcs) for each child node. So his explanation and result still doesn't make any sense to me.

[santropedro]

Of course we know. If 12 numbers must sum a total of 1, you only need to specify 11 of them, the 12ve is 1-all the others.

[AnnanFay]

P(HAM) = 1 - P(SPAM)

[AnnanFay]

I know this video is old. However, maybe another explanation will help someone...

You are given ONE word and told that it is SPAM (or HAM), this word is from the 12 word Dictionary. When given a word and told it comes from a SPAM message there is a 1/3 chance it will be "SECRET" and 1/9 probability of it being "IS" (using the example messages). These 12 P values must add up to 1 because the word you are given must come from the dictionary. (The question seems to assume this)

[Иван Заярный]

where is prior p(HAM) ?

[popacrovac]

He counts only 11 words times 2 parameter plus one for prior, because the 12th pair of parameter can be derived from the previous 11 (1 minus sum of these). However, doesn't that imply that the upcoming messages can contain only these 12 different words, that is - the whole language lexicon consist of 12 words only?

[TheMolind]

probability of 12'th word = 1 - sum of probabilities of 11 previous words. So we can just calculate it.

[PhantomAct]

12 words => 12 children => 24 parameters + 1 for the root = 25! But it's wrong.

[uniquename57]

I agree with 13Septem13, it's not consistent with the previous conventions

[trombonemunroe]

No, because in each dictionary the occurrence probabilities of only 11 of the 12 words are unknowns; since the 12th probability is known, it doesn't need to be supplied as a parameter. Hence, 1 + 11 + 11 = 23.