Combinatorics in Words With Friends

[Dauphin]

Think about interchanging all the non-Es....

[Dauphin]

BTW, the clue is in the OP.

[KrazyHorse]

Good god, provost. Lori now has the right answer. Treat the EEE as a single element, there are 5! = 120 permutations. Now multiply by 6 because each of those permutations represents 3! internal permutations of the EEE substring. That makes 720, so the probability is 720/5040 = 1/7

[Dauphin]

Spoil sport, I wanted to see him get it wrong...again.

[Provost Harrison]

You should be at the gym anyway. Interesting though.

[Dauphin]

I will be this arvo, but I've got an eye test I have to go to first.

[Provost Harrison]

Good god, provost. Lori now has the right answer. Treat the EEE as a single element, there are 5! = 120 permutations. Now multiply by 6 because each of those permutations represents 3! internal permutations of the EEE substring. That makes 720, so the probability is 720/5040 = 1/7

Well I have very little education in mathematics but with this I can see the error of my logic in terms of how I attempted to process it. I really don't have enough practice at this kind of stuff.

[Provost Harrison]

I will be this arvo, but I've got an eye test I have to go to first.

You want to lay off the you-know-what

[Dinner]

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%.

Very no.

[Provost Harrison]

Very no.

Like you'd know

[Lorizael]

Weird, I use the letters to spell words.

ACK!

[Lorizael]

Good god, provost. Lori now has the right answer. Treat the EEE as a single element, there are 5! = 120 permutations. Now multiply by 6 because each of those permutations represents 3! internal permutations of the EEE substring. That makes 720, so the probability is 720/5040 = 1/7

Hah. I didn't even realize that 720/5040 simplifies to 1/7. Is it just a coincidence that 1/7 is the answer when there are 7 letters, or is there some deeper relationship that would hold true for other combinations?

[Dauphin]

It's a quirk because 3! x 5! = 6!

[Lul Thyme]

Hah. I didn't even realize that 720/5040 simplifies to 1/7. Is it just a coincidence that 1/7 is the answer when there are 7 letters, or is there some deeper relationship that would hold true for other combinations?

Try a few more examples and see if you think there is a pattern. If so, try to prove it.

[Lorizael]

Try a few more examples and see if you think there is a pattern. If so, try to prove it.

Lori's Last Theorem: The odds of three like letters in a set appearing in a row is equal to 1/n, where n is the total number of letters in the set. I have discovered a truly marvelous proof of this, which this post is too small to contain.