Calculating the probability of drawing JANITOR twice

[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

Comments

[ar_raqis]

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

[boshvark]

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

[boshvark]

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

[badqoph]

Also, TRIAXON#.

[boshvark]

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).

[evwhore]

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 :-)

[boshvark]

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.

[evwhore]

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

[boshvark]

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

[anonymous]

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.

[evwhore]

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.

[ext_1693992]

I played around with the probability of scrabble so much, that I created You Go Scrabble because of it. For the 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!!

[boshvark]

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.