Scrabble Statistics

Should be a standard math problem (and it's been a while since I've done math like this, so please bear with me. I'm sure @engineer can show a much more elegant solution). In any event, for every tile, there's a 1% chance of it turning up.

We treat the duplicates as interchangeably but, IIRC, probability does not.

So let's call the O tiles:

1. Alpha O
   
2. Beta O
   
3. Gamma O
   
4. Delta O
   
5. Epsilon O
   
6. Zeta O
   
7. Eta O
   
8. Theta O

There's a 1% chance that you'll get a Theta O, and there's a 1% chance that you'll get a Delta O. There are lots of combinations where you can get four O tiles. It could be, for example, Delta O + Eta O + Epsilon O + Zeta O. Or it could be Beta O + Zeta O + Gamma O + Theta O. Or anything else.

Your chance of getting Zeta O is 1%. Your chance of getting Delta O is also 1%.

Pulling 4 O tiles can be in combinations such as:

• Alpha, Beta, Gamma, Delta
  
• Alpha, Beta, Gamma, Epsilon etc

For each set of three O tiles, there are five possibilities for the fourth tile.

Also IIRC, the pulling of, say, the Gamma O does not change the probability of pulling the Beta O. They're both 1/100, AKA 1%.

I think it throws people off when you consider duplicates. But consider any combination of 7 letters. Let's say it was the letters to spell the world pickles (you can tell what I had on the side at dinner tonight). Those letters all also have a 1% chance of being pulled.

Anyway, I suspect engineer will show me the error of my ways but my belief is that it's just 7%.

There are 8C4 times 92C3 ways of selecting 4 letters O in 7 random letters at the beginning of the game out of the 100C7 possibilities. Given that subsequent letter collection is not a random variable in scrabble the answer to the rest of your question is indeterminate.

I may have been remiss in my original post. I failed to provide the distribution of tiles in the population. There are 9 "A's", 2 "B's".... So in a draw of 7 without replacement, what is the probability of getting 4 "O's"? (There are 8 "O's" in the population)


A - 9
B -2
C - 2
D - 4
E - 12
F - 2
G - 3
H - 2
I - 9
J - 1
K -1
L - 4
M - 2
N - 6
O -8
P - 2
Q - 1
R - 6
S - 4
T - 6
U - 4
V - 2
W - 2
X - 1
Y - 2
Z - 1
? - 2

Fobvious is correct. All you care about is "O" and not "O" (the letters printed on the other tiles are irrelevant.

To elaborate a bit on what fobius. You will be selecting four tiles of which 8 are "o" (that is the 8C4). And 3 tiles of which 92 are not-"o" (that is the 92C3). You multiply the two to ensure that both happened.

Then you divide by the number of possible combinations (which is 100C7).

A good calculator will have combinations as a built in function. If not, you can do out the math by using this page.

https://en.m.wikipedia.org/wiki/Combination

The answer I get is 0.00054915299

That is about 1 in every 2000 times.