There are only 15 different types of racks to consider: 7 different letters (1111111), 6 different letters (211111), 5 different letters (31111 or 22111), etc, all the way down to a rack with all identical letters.

Assuming the bag contains at least 7 tiles for each letter (and we have 26 different tiles, that is, no blanks), it's easy to calculate how many different racks you can make for each type. For instance, type 111111 give you 26*25*24*23*22*21*20/7*6*5*4*3*2*1 == 657800 different racks. Six different letters on the board gives you the most combinations: 26*25*24*23*22*21*6/(5*4*3*2*1) == 8288280. Etc.

I leave it up to you to calculate the other numbers, but it quickly peters down. The number of different racks to consider is in the order of tens of millions, (less than 20 million I guess).

Enumerating all possible racks, given an sufficient number of tiles is doable. And given that list and the actual distribution, you can calculate the actual odds (some will have probability, as I doubt Wordfeud has 7 Zs in the bag).

Of course, you may want to be smart, and just derive the right number, using some combinatoric arguments, but for many people here in this forum, doing it brute-force is going to be easier.

In reply to Re: Wordfeud racks by Anonymous Monk
in thread Wordfeud racks by Anonymous Monk

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