Scrabble Probability:

luca-deltodesco · 2009-02-10 22:34:19

Challenge a friend has given me; he has been given it as a homework assignment, i'm just doing it for fun:

Calculate the probablity of drawing your first 7 tiles from a full bag, and getting a 7 letter word (He has been given a full list of valid scrabble words, i'm just doing the maths bit for fun)

So far I've found the probablity given that you totally ignore the blank tiles: What i've gotten is this:

Where T is the total number of tiles in the bag
O(z[sub]j[/sub]) denotes how many of each tile is in a full bag
and 'n[sub]j[/sub]' is the number of each tile in the word
'i' iterates over all valid 7 letter words

I'm struggling on trying to include blank tiles in the calculation though! (I'm not asking for help, since this is someone else's homework assignment)

Last edited by luca-deltodesco (2009-02-10 22:39:49)

JaneFairfax · 2009-02-10 23:32:20

I did that once. From the bag I drew out the letters spelling out the word EVIDENT. Not only that, the letters actually came out in that order (from left to right)!!

It was really scary. I couldnt sleep for a few nights after that because I was so scared.

mathsyperson · 2009-02-11 02:51:28

I might well be reading this formula wrong, but if it's doing what I think it is then it has a small flaw.

Consider instead the probability of being able to make a 3-letter word by drawing three tiles.
Also, we use a bag from the end of a game, that only contains three tiles. These are P, T and O.

For each valid three letter word, we check the probability of drawing tiles that will arrange to make it.
The probability of being able to spell POT is 1.
The probability of being able to spell TOP is 1.
The probability of being able to spell OPT is 1.
The probability of being able to spell any other word is 0.

Summing these, we get that the total probability is 3.

Also, are you considering that the tiles are drawn without replacement?
Given that a bag contains three S's, what would the formula say the probability was of spelling POSSESS?

luca-deltodesco · 2009-02-11 04:26:34

Scrabble has no replacement, and an easy modification to make to solve the problem you've pointed out, is to say:

'i' iterates over all 7 letter words with unique letter combinations:

Which would make POT TOP and OPT in your example treat as the same word since the use the same combination of letters.

or alternatively:

'i' iterates over all combinations of 7 letters that form valid words. Allowing repeated letters obviously.

Last edited by luca-deltodesco (2009-02-11 04:27:49)

JaneFairfax · 2009-02-11 05:14:47

I agree with Luca. When you have a selection of tiles on your rack, it doesnt matter how many different valid words you can make out of them only thing that matters is is that you can make at least one valid word

On the other hand, if Mathsy wants to make life difficult for himself, then this is what he should have done:

For each valid three letter word, we check the probability of drawing tiles that will arrange to make it.
The probability of being able to spell POT is
.
The probability of being able to spell TOP is
.
The probability of being able to spell OPT is
.
The probability of being able to spell any other word is 0.
Summing these, we get that the total probability is
.

Then Mathsy is clearly calculating the probability of being able to spot a valid word from the selection of letters. This is clearly a different problem altogether from the one Luca is considering!!

TheDude · 2009-02-11 06:41:36

I can't follow exactly what you did, but it looks like you took each word, calculated the probability of getting that word with your first 7 tiles, and then added all of those probabilities together.

Another way you might go about it is to consider sets. Let T be the set of all possible tiles you can draw, and let A be the set of all of the subsets of T that contain exactly 7 elements. Then your answer is simply the number of unique combinations of letters that form a valid 7 letter word divided by the cardinality of A.

The benefit to this approach is that you can, without too much hardship, calculate the probability with blank tiles as well. The only real tricky part is to make sure that if you have 2 different words that share 5 or 6 letters in common you don't count them twice when considering blank tiles.

Math Is Fun Forum

#1 2009-02-10 22:34:19

Scrabble Probability:

#2 2009-02-10 23:32:20

Re: Scrabble Probability:

#3 2009-02-11 02:51:28

Re: Scrabble Probability:

#4 2009-02-11 04:26:34

Re: Scrabble Probability:

#5 2009-02-11 05:14:47

Re: Scrabble Probability:

#6 2009-02-11 06:41:36

Re: Scrabble Probability:

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