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Scrabbling for Mastery

a patzer's journey

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Calculating the probability of drawing JANITOR twice
me again
boshvark
I know I have some mathematically inclined LJ friends, and I was wondering if you would be willing to check my math. I've received an email containing a story about someone in the UK who drew JANITOR to begin a game at age 19, and then at age 59 did the exact same thing again. It didn't specify whether there were any blanks involved. He was wondering what is the probability of this happening.

Here is what I'm planning to write in response. Does this look about right to you? All these numbers only pertain to the very first draw (once per game), not all the draws during the course of an entire game. Also, I think I should be able to figure out how to calculate the answers to the two unanswered questions near the end, but if you have any pointers I would appreciate it!

I'm not showing my work at the moment, but if necessary I will detail all the steps I took to get these numbers. Note that the "number of combinations" numbers that I quote were calculated by Zyzzyva; I didn't calculate those by hand. All numbers assume a 100-tile bag with 2 blanks. And of course, I'm using CSW07 since the emailer is from Britain.

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Zyzzyva has the ability to display the relative probability rankings of words compared to each other. For example, it can tell you that if you draw 7 tiles out of a full bag of Scrabble tiles, the most likely word you could form is OTARINE (pertaining to an otary, a big-eared seal).

Zyzzyva doesn't display absolute probabilities, but I'm happy to calculate these for you. Below are some questions and answers you may find useful. If you have more questions, please let me know.

- What is the probability of drawing the word JANITOR on a single try?

There are 16,007,560,800 distinct ways to draw 7 tiles from a full bag of 100 Scrabble tiles. There are 5,423,112 distinct ways to draw tiles that will allow you to play OTARINE, so the absolute probability is about 0.034%, or about 3 chances in 10,000. There are 817,236 ways to draw JANITOR, so its probability is about 0.0051%, or about 5 chances in 100,000. Therefore JANITOR is about 6.6 times less likely than the most probable word, OTARINE. Overall, the probability of JANITOR gives it a ranking of 2,391 out of 32,909 total 7-letter words. In other words, even though it contains the unlikely J, JANITOR is among the top 10% most probable words to draw on a single try.

- What is the probability of drawing the word JANITOR twice in only two tries?

The odds of drawing JANITOR in 2 out of 2 tries is 0.0051% squared, or about 0.00000026%, about 2 chances in a billion.

- What is the probability of drawing the word JANITOR twice over the course of many hundreds of games?

The email you received stated that the person has played Scrabble many hundreds of times over 40 years or so. For simplicity, let's assume that he has played 1000 games of Scrabble, or about 25 games per year. Over the course of 1000 games, the probability of drawing JANITOR at least one time is about 5%, or about 1 chance in 20. The probability of drawing JANITOR at least twice is about 0.25%, or about 1 chance in 400.

Here are a couple more questions for which I haven't calculated the answers yet.

- What is the probability of drawing any word (not particularly JANITOR) twice in a row?
- What is the probability of drawing any word (not particularly JANITOR) twice over the course of many hundreds of games?

These are actually key questions, since the coincidence may have been equally astonishing if the word had been any other word besides JANITOR. However, adding the possibility that it could have been *any* word significantly increases the likelihood of the coincidence. I haven't yet calculated exactly how the probability is affected, but I wouldn't be surprised to learn that the total likelihood becomes greater than 50%. In other words, over the course of 1000 games a truly astonishing occurrence might be that one *never* draws the same word twice!

This is similar to the well-known birthday problem, where the likelihood that two people in a room share a birthday becomes greater than 50%, counterintuitively, when there are only 23 people in the room.
http://en.wikipedia.org/wiki/Birthday_problem

Of course, not all valid words are as likely to be spotted as JANITOR, since JANITOR is a word in common use. For example, someone might have twice drawn the word TRIAXON (a sponge spicule with three axes), which has exactly the same probability as JANITOR. But it would probably go unnoticed, since TRIAXON is a very uncommon word.

Here are a couple other tidbits you might find interesting:

There are three 7-letter words that are absolutely impossible to play. These words are BEZZAZZ, PAZZAZZ, and PIZZAZZ. These words all mean the same thing; they are variations of the spelling of PIZZAZZ, meaning panache or flamboyance. They are unplayable because the standard set of Scrabble tiles only contains one Z and two blanks, so a word with four Zs cannot be formed.

The very least likely 7-letter word that is not impossible to play is ZYZZYVA. There are only 18 distinct ways to draw ZYZZYVA, since you would need to draw both blanks, both Ys, one of two Vs, and the only Z. That makes the probability of drawing ZYZZYVA about 0.0000001%, or about 1 chance in a billion. Note that this is even less likely than drawing JANITOR twice in a row!


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nota bene, NOTAIRE+# isn't good yet.

Whoooops! Thank you! I'll get rid of it and only refer to OTARINE.

Okay, I've edited the post and my email draft to remove NOTAIRE. Thanks for the correction.

Yep, this is all in the context of CSW, and adding hashes would simply confuse my email correspondent unnecessarily (I don't think he's a Scrabble player).

There are 16,007,560,800 distinct ways to draw 7 tiles from a full bag of 100 Scrabble tiles.

Well, there are 16,007,560,800 ways to choose 7 tiles out of 100. Many of those aren't distinct though. For example there are 12C7 = 792 ways to draw 7 Es but they're all the same rack.

There are 817,236 ways to draw JANITOR

I guess you're including blanks? I get 139,968 otherwise. I am too lazy to do the with-blanks calculation to see if I get the same number as you :-)

I think that's okay since I'm comparing distinct draws, not distinct rack contents. Zyzzyva's combinations values are based on distinct draws, and I am using the values for 2 blanks.

right, i was just nit-picking the usage of 'distinct' :-)

Ah, cool! Is there a better word I should use? Seriously, I have no idea; I'm just winging it here.

Well, it really is a nit-pick. It's true that we're counting distinct ways to draw the tiles, if you number them on their backs 1-100 and ignore the letters printed on the front. I guess I was concerned about possible ambiguity (especially if the correspondant is not mathematically inclined) with whether the alphagrams are distinct.

So I think I would just say "there are 16,007,560,800 ways" without the qualifier :-)

Oh, I don't think I was really clear either... I'm not too worried about the "number of combinations" values. Mostly I want to make sure my probability calculations are correct, given that the combo numbers are correct. Also any suggestion of an approach to the last problems (probability of drawing *any* word twice) would be appreciated. I think I can figure it out, but I'm also lazy and not as much of a math genius as I used to fancy myself. :-)

Your last sentence applies to me to :-)

Yeah, I think I've decided to punt on those last questions. I think I would be able to calculate the probability of drawing any word twice in a row given a 98-tile blankless bag. That's just the sum of all the probabilities of drawing each individual word twice in a row. Once the blanks get in there and start creating overlap, my head asplode. Ditto for any number of draws where N > 2. I need some math skills I just ain't got.

probability of drawing JANITOR (no blanks) on opening draw is 1/100 * 9/99 * 6/98 * 9/97 * 6/96 * 8/95 * 6/94 = 1.7E-9

for two times, just square that.


That's the odds of drawing a J, then an A, then an N, then an I, then a T, then an O, then an R... in sequence.

probability of drawing JANITOR (no blanks) on opening draw is 1/100 * 9/99 * 6/98 * 9/97 * 6/96 * 8/95 * 6/94 = 1.7E-9

Using this method the probability of drawing JANITOR out of a bag that only contained JANITOR would be 1/7 * 1/6 * 1/5 * 1/4 * 1/3 * 1/2 * 1.

forgot to multiply by the proper combination

1.7E-9 * (1*9*8*6*6*6) = (9*8*6*6*6*1)^2 / (100*..*94) = 3.0E-6

I played around with the probability of scrabble so much, that I created You Go Scrabble (http://www.yougoscrabble.com) because of it. For the word Janitor (http://www.yougoscrabble.com/word/janitor), I got the chance of having this word, is 0.000079% or better explained as a chance of 19 out of 241,439 of getting the 7 letter word, janitor, in your scrabble hand. To get it twice, dramatically alters my math. So now, because of this page, I'm going to have to add the probability of getting each word twice, lol! Thanks... excellent article, brilliant read :)! Also, the probability of drawing zyzzyva is 0! unless you're referring to drawing a z and then putting it back in the bag.

There is also a lot more than 3 7 letter words that are impossible to play, lol. Just check all the 7 letter words! Tons!!

I think you may be discounting the two blanks in the bag... you could play ZYZZYVA by using one A, one V, both Ys, the only Z, and both blanks designated as Zs. Considering both blanks, the three words I listed are the only 7-letter words that are impossible to play. Without considering the blanks, you're right, there are many, many more.

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