mersenneforum.org Counting primes in Scrabble racks
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 2008-02-20, 17:08 #1 retina Undefined     "The unspeakable one" Jun 2006 My evil lair 22·5·307 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 Count the number of unique racks you can draw, from a full bag, for draws of 1 tile through 50 tiles. Factorise, and post the largest prime found among all draw sizes. 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) And post the largest number among a...z (there is only one unique answer). BTW: The point value of each tile means nothing for this puzzle, you can safely ignore it.  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 2008-02-20 at 17:21
 2008-03-05, 17:59 #2 Brian-E     "Brian" Jul 2007 The Netherlands CC516 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)
 2008-03-05, 21:39 #3 Brian-E     "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?
2008-03-06, 01:38   #4
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

17FC16 Posts

Quote:
 Originally Posted by Brian-E 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)
Correct for 2 and 3 tiles
Quote:
 Originally Posted by Brian-E 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?
The method for this and the recent darts puzzle are closely related. Anything more than 1ms CPU time to get all 50 values and you have some room for improvement.

Last fiddled with by retina on 2008-03-06 at 01:39

 2008-03-06, 20:12 #5 Brian-E     "Brian" Jul 2007 The Netherlands 326910 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)
2008-03-07, 02:57   #6
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

22×5×307 Posts

Quote:
 Originally Posted by Brian-E 4 tiles: 18104 (2^3*31*73) 5 tiles: 148024 (2^3*18503)
Doesn't look right to me.

2008-03-07, 09:57   #7
Brian-E

"Brian"
Jul 2007
The Netherlands

7·467 Posts

Quote:
 Originally Posted by retina Doesn't look right to me.
My way of tackling this problem is much too laborious and error-prone. I found an error in my working of both 4- and 5-tiles. Here are corrected versions, but the chance that they are now error-free 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.

2008-03-07, 10:15   #8
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

614010 Posts

Quote:
 Originally Posted by Brian-E 4 tiles: Total: 25254 (2*3^2*23*61) 5 tiles: Total: 148150 (2*5^2*2963)
This looks correct now.

Quote:
 Originally Posted by Brian-E Generalising this method and programming it doesn't really appeal to me. I am wondering is there is a more elegant way of counting.
So far, you have done it the hard way, and I am impressed that you took it all the way to 5 tiles.

 2008-03-07, 10:15 #9 ckdo     Dec 2007 Cleves, Germany 2×5×53 Posts I'll take a shot at answering the original question: 0xC75407108F5D?
2008-03-07, 10:20   #10
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

22×5×307 Posts

Quote:
 Originally Posted by ckdo I'll take a shot at answering the original question: 0xC75407108F5D?
Correct.

2008-03-07, 10:27   #11
ckdo

Dec 2007
Cleves, Germany

2·5·53 Posts

Quote:
 Originally Posted by retina Correct.
Boy am I glad. Getting the 50 values certainly is a millisecond task, but factoring them takes me somewhat over a second. I'll try to work on this.

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