20080220, 17:08  #1 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6075_{10} Posts 
Counting primes in Scrabble racks
Take a standard English set of Scrabble tiles:
Code:
0 points: Blank ×2 1 point: E ×12, A ×9, I ×9, O ×8, N ×6, R ×6, T ×6, L ×4, S ×4, U ×4 2 points: D ×4, G ×3 3 points: B ×2, C ×2, M ×2, P ×2 4 points: F ×2, H ×2, V ×2, W ×2, Y ×2 5 points: K ×1 8 points: J ×1, X ×1 10 points: Q ×1, Z ×1 e.g. Code:
1 tile draw  27 unique racks = (3*3*3) 2 tile draw  ?? unique racks = (a*b*...*c) ... 50 tile draw  ?? unique racks = (x*y*...*z) BTW: The point value of each tile means nothing for this puzzle, you can safely ignore it. [edit] Just to make sure this is clear. Each new draw is started with a full set of 100 tiles. Once you have started drawing, the tiles are not replaced into the bag, they put on the rack. So, for 1 tile draws: Start with 100 tiles, draw an "A". That counts as 1 possible draw set. Then start again with 100 tiles, draw a "B". Now we have 2 possible sets. ... Until lastly we draw the blank, to make 27 unique sets of draws. Permutations are not counted as unique, "ABC"="CBA"="BCA"="ACB"="BAC"="CAB" is only counted as one set. [/edit] Last fiddled with by retina on 20080220 at 17:21 
20080305, 17:59  #2 
"Brian"
Jul 2007
The Netherlands
7·467 Posts 
My calculations for drawing 2 tiles and 3 tiles are below. The method I use is too laborious for much further progress to be feasible. Plus I'm not sure if I'm not making a mistake.
2 tiles: 22 ways if both tiles are the same (because 5 of the letters cannot be used) 27*26/2! if they are different Total: 373 (prime) 3 tiles: 12 ways if all three tiles are the same 2*5*22 ways if two of the tiles are the same 27*26*25/3! if all three are different Total: 3157 (=7*11*41) 
20080305, 21:39  #3 
"Brian"
Jul 2007
The Netherlands
7×467 Posts 
My solution for the 3 tiles was wrong, correction:
3 tiles: 12 ways if all three tiles are the same 22*26 if two of them are the same (these two cannot be any of the 5 single letters) 27*26*25/3! if all three are different Total: 3509 (=11^2*29) Going further seems to require lots of effort in dealing with all the different possibilities. I would undoubtably make more mistakes if I tried. Does anyone have a better method? 
20080306, 01:38  #4  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
3^{5}×5^{2} Posts 
Quote:
Last fiddled with by retina on 20080306 at 01:39 

20080306, 20:12  #5 
"Brian"
Jul 2007
The Netherlands
7·467 Posts 
Extended to 4 and 5 tiles but I'm not succeeding in generalising the method so that it can be programmed.
2 tiles: 373 (prime) 3 tiles: 3509 (11^2*29) 4 tiles: 18104 (2^3*31*73) 5 tiles: 148024 (2^3*18503) 
20080307, 02:57  #6 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
3^{5}×5^{2} Posts 

20080307, 09:57  #7 
"Brian"
Jul 2007
The Netherlands
7·467 Posts 
My way of tackling this problem is much too laborious and errorprone. I found an error in my working of both 4 and 5tiles. Here are corrected versions, but the chance that they are now errorfree is still not good.
4 tiles: All 4 the same letter: 11 ways 3 the same, 1 of another: 12*26 ways "2+2": 22*21/2! ways "2+1+1": 22*26*25/2! ways (forgot this possibility before) "1+1+1+1": 27*26*25*24/4! ways Total: 25254 (2*3^2*23*61) 5 tiles: "5 all the same": 7 ways "4+1": 11*26 ways "3+2": 12*21 ways (mistakenly divided this by 2 before, but the ordering is important here) "3+1+1": 12*26*25/2! ways "2+2+1": 22*21*25/2! ways "2+1+1+1": 22*26*25*24/3! ways "1+1+1+1+1": 27*26*25*24*23/5! ways Total: 148150 (2*5^2*2963) Generalising this method and programming it doesn't really appeal to me. I am wondering is there is a more elegant way of counting. 
20080307, 10:15  #8  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
3^{5}·5^{2} Posts 
Quote:
So far, you have done it the hard way, and I am impressed that you took it all the way to 5 tiles. 

20080307, 10:15  #9 
Dec 2007
Cleves, Germany
2×5×53 Posts 
I'll take a shot at answering the original question: 0xC75407108F5D?

20080307, 10:20  #10 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
3^{5}·5^{2} Posts 

20080307, 10:27  #11 
Dec 2007
Cleves, Germany
212_{16} Posts 

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