Choosing Scrabble Letters

Date: 01/07/2002 at 15:41:43
From: Matthew Turgeon
Subject: Choosing Scrabble letters--All at once, or one at a time?

In a Scrabble game with two players, each must randomly grab seven 
letters from a bag. Does it matter whether each player chooses all 
seven letters at once, or they alternate back and forth, taking one at 
a time?

Please help me settle this bet with my wife. She argued that the act 
of choosing second after seven letters are gone means the event is 
dependent on the first. I agreed, but countered that it does not give 
the first person an advantage because the letters chosen are not 
known, and that it is as likely that desirable letters are left as it 
is that undesirable letters remain.

I tried calculating the probabilities, but became overwhelmed.

Thanks,
Matt

Date: 01/07/2002 at 17:43:39
From: Doctor Douglas
Subject: Re: Choosing Scrabble letters--All at once, or one at a time?

Hi, Matt.

Thanks for submitting your question to the Math Forum.

You are both right. The act of choosing after seven letters are gone 
IS dependent on the first seven letters. So your wife is correct.  And 
you are right in that this does not confer an advantage (or 
disadvantage) on drawing first, at least as far as the game of 
Scrabble is concerned. Both ways of drawing tiles are equivalent in 
that both methods lead to the same probabilities of the various 
letters in the Scrabble racks of both players. But for a different 
game (such as "First-person-to-draw-a-vowel-wins" - I know that's not 
a real game, but it illustrates the point), the two methods of drawing 
tiles are clearly not equivalent; for this game I would want to draw 
my seven letters before my opponent got a chance to draw even once.

To illustrate this problem mathematically, let's consider not the 100
tiles of Scrabble and the racks of seven letters each, but instead a 
"deck" of 4 tiles (A,B,C,D), and let each player draw two tiles using
either of the two methods above:

Using the twenty-four possible arrangements of the "deck" that are 
obviously equally likely, we can enumerate all of the possibilities.
I won't list all twenty-four orderings of the drawn letters, but just
the subset consisting of those orderings in which the first letter
drawn is A.  You can verify that the other 18 orderings also lead to
the same result.

                   alternately            two then two
  ordering     player 1   player 2      player 1  player 2
  ---------    -------------------     --------------------
   A B C D      A C        B D           A B        C D
   A B D C      A D        B C           A B        D C
   A C B D      A B        C D           A C        B D
   A C D B      A D        C B (*)       A C        D B
   A D B C      A B        D C           A D        B C
   A D C B      A C        D B           A D        C B (*)

Now, we see that each of the "hands" appears once in each column.
For example, the situation when player 1 gets AD and player 2 gets CB 
is shown by the asterisks. Since each outcome occurs exactly once
(with the same probability) in both drawing methods, these two methods
are equivalent.  

You can extend this argument by enumerating all twenty-four orderings 
to show that for every chance that player 1 gets a particular hand 
(such as that containing A and B, in either order), there is an 
equivalent set of possibilities in which player 2 gets that type of 
hand.

I hope this helps. Please write back if you need more of an 
explanation.

-- Doctor Douglas, The Math Forum
   http://mathforum.org/dr.math/