4 ms·

"there are more than 16 billion ways to draw seven tiles" Wrong. Even assuming the order matter (which is false, ABCDEFG is the same draw as GFEDCBA) and even

by mrb 12y ago

"there are more than 16 billion ways to draw seven tiles" Wrong. Even assuming the order matter (which is false, ABCDEFG is the same draw as GFEDCBA) and even assuming there are 7 tiles of each letter and 7 blanks in a bag (which is false), that would be 27^7 = 10.5 billion ways. Because these assumptions are false, the real number of combinations is lower.

kqr2 12y ago
For calculating the number of combinations, it looks like they are simply using 100 choose 7, or C(100, 7) which yields 16,007,560,800. http://what-are-the-chances.blogspot.com/2008/02/7-letter-words-and-8-card-suits.html http://what-are-the-chances.blogspot.com/2008/02/7-letter-wo...
- psychometry 12y ago
  That doesn't take into account repeated letters. You actually want the number of 7-combinations of the multiset { A * 5, B * 3, C * 4, ... } or whatever the letter frequencies are. That's for the first draw from the bag.
DanBC 12y ago
I was noodling around to find the correct answer to this question, when I found this. It has nothing to do with why Nigel is the best Scrabble player. "It is very hard to draw a rack that does not contain a valid word" - http://stats.stackexchange.com/a/6075 http://stats.stackexchange.com/a/6075 This statsSE answer includes an R program.
Terr_ 12y ago
I started working on a python script to monte-carlo the problem, but it's pretty complex. Even if we assume that you always start with a full bag of 100 tiles, every tile drawn changes the odds of the next draw in a different way. Right now I've got a series of "unique draws" that grow slightly slower than the number of tries, but I'm not yet sure what single number it would converge to at infinity tries.
- Someone 12y ago
  If you want to estimate it that way, google "estimating wildlife populations", "population estimation methods", or something like it. It will give you, a.o, http://en.m.wikipedia.org/wiki/Mark_and_recapture http://en.m.wikipedia.org/wiki/Mark_and_recapture. I think it isn't that hard to exactly compute the number on modern hardware, though.

by mrb 12y ago

kqr2 12y ago

For calculating the number of combinations, it looks like they are simply using 100 choose 7, or C(100, 7) which yields 16,007,560,800. http://what-are-the-chances.blogspot.com/2008/02/7-letter-words-and-8-card-suits.html http://what-are-the-chances.blogspot.com/2008/02/7-letter-wo...

psychometry 12y ago
That doesn't take into account repeated letters. You actually want the number of 7-combinations of the multiset { A * 5, B * 3, C * 4, ... } or whatever the letter frequencies are. That's for the first draw from the bag.

DanBC 12y ago

I was noodling around to find the correct answer to this question, when I found this. It has nothing to do with why Nigel is the best Scrabble player. "It is very hard to draw a rack that does not contain a valid word" - http://stats.stackexchange.com/a/6075 http://stats.stackexchange.com/a/6075 This statsSE answer includes an R program.

Terr_ 12y ago

I started working on a python script to monte-carlo the problem, but it's pretty complex. Even if we assume that you always start with a full bag of 100 tiles, every tile drawn changes the odds of the next draw in a different way. Right now I've got a series of "unique draws" that grow slightly slower than the number of tries, but I'm not yet sure what single number it would converge to at infinity tries.

Someone 12y ago
If you want to estimate it that way, google "estimating wildlife populations", "population estimation methods", or something like it. It will give you, a.o, http://en.m.wikipedia.org/wiki/Mark_and_recapture http://en.m.wikipedia.org/wiki/Mark_and_recapture. I think it isn't that hard to exactly compute the number on modern hardware, though.