Combinatorics in Words With Friends

[Lorizael]

I make the most boring threads. I'm sorry.

Anyway, my letters in Words With Friends right now are EDEQSEU. I happened to shuffle the letters and the 3 Es all came up next to each other. So I was like, hey, that's probably pretty unusual. But the question is, how unusual is it?

My guess is that you treat EEE as a single element and DQSU as 4 elements, which means there are 120 permutations (5!) that would produce EEE, right? And for all 7 letters, there are 5040 (7!) permutations. Which gives the odds of EEE coming up as ~2%.

So then I shuffled the letters 100 times and EEE came up 17 times. This tells me that (a) my math is wrong somehow, (b) I didn't run enough trials to conform to what's expected, or (c) Words With Friends shuffle isn't random enough.

I consider (a) to be the most likely answer. Apolyton?

[The Mad Monk]

(c). Randomizers are often pretty lame.

[Lorizael]

Thanks for the vote of confidence. I'm still pretty sure I'm wrong, though.

[KrazyHorse]

There are 6 different EEEs

[Dinner]

I'm coming up with six EEEs too. We have E1, E2, E3 so...

123
132
213
231
321
312

Now if we assume the final word has seven letters in it then we have multiple places we could put that EEE such as.

EEEXXXX
XEEEXXX
XXEEEXX
XXXEEEX
XXXXEEE

But there are still only six combinations of EEE.

[Lorizael]

You mean there's

E1 E2 E3
E1 E3 E2
E2 E1 E3
E2 E3 E1
E3 E1 E2
E3 E2 E1

Right. Which means 720 ways to produce EEE. Which is a ~14% chance of having it come up. Which means 17/100 is pretty damn close. Thanks.

x-post

[KrazyHorse]

oerdin:

Good lord, that was a useless response. The point of my post was not to tell lori something he already knew, but to remind him to account for it in his calculation (which he did not, but now is).

[Dinner]

[Tuberski]

Weird, I use the letters to spell words.

ACK!

[Provost Harrison]

Actually there are five possible combinations containing all 3 E's in a row (Oerdin gave them yet still quoted 6 for some strange reason). And there are 5040 possible combinations. 6 of those are legitimate which gives 1 in 1008. Slightly under 0.1%.

[Dauphin]

Actually there are five possible combinations containing all 3 E's in a row (Oerdin gave them yet still quoted 6 for some strange reason). And there are 5040 possible combinations. 6 of those are legitimate which gives 1 in 1008. Slightly under 0.1%.

No.

[Provost Harrison]

Well go on then, smart arse. Don't just do a KH.

[Provost Harrison]

Actually there is something I have overlooked there...you are quite right. Let me rethink...because all of the E's are interchangeable...there are therefore 30 combinations which gives 5040/30 = 168.

[Dauphin]

No again.

[Provost Harrison]

Are you sure?