March 27, 2014 7:27 PM Subscribe

I'm an answer-checker who is rusty at probability. Can you solve this problem?
What is the probability that a random arrangement of the letters in the word THRUSTS will have the two T's next to each other?

Also, if you could quickly explain your answer, that'd be awesome.
posted by 23skidoo to Education (11 answers total) 2 users marked this as a favorite

Also, if you could quickly explain your answer, that'd be awesome.

Crap, forgot to mention that you need to multiply that last result by 2, because the double-T's can be in any order. So I think 1/7 * 2 = 2/7 is the answer.

posted by Androgenes at 7:48 PM on March 27 [3 favorites]

posted by Androgenes at 7:48 PM on March 27 [3 favorites]

There are 42 possibilities for pairs of locations that the T's can end up in the permutation: (1,2), (1,3), (1,4), ... , (2,1), (2,3), (2,4), ..., (7,6). Out of these, twelve end up with the two T's together: (1,2), (2,1), (2,3), (3,2), .... So the probability is 12/42 = 2/7.

On preview: I agree with Androgenes's corrected answer.

posted by Johnny Assay at 7:49 PM on March 27

On preview: I agree with Androgenes's corrected answer.

posted by Johnny Assay at 7:49 PM on March 27

Yeah, I also think the answer is 2/7.

You can arrange all of the letters in 7! ways.

If TT occurs in the first two positions, there are 5! ways to arrange the other letters in positions 2-6; if TT occurs in the 2nd and 3rd position, 5!; etc, through TT in the 6,7 position.

So, the probability of the two T's next to each other is (2 * 6 * 5!) / 7! = 2/7

posted by JumpW at 7:49 PM on March 27

You can arrange all of the letters in 7! ways.

If TT occurs in the first two positions, there are 5! ways to arrange the other letters in positions 2-6; if TT occurs in the 2nd and 3rd position, 5!; etc, through TT in the 6,7 position.

So, the probability of the two T's next to each other is (2 * 6 * 5!) / 7! = 2/7

posted by JumpW at 7:49 PM on March 27

Does the double S's affect this at all?

posted by 23skidoo at 7:49 PM on March 27

No. For this sort of random permutation/probability question, they could be different letters and the answer wouldn't change.

posted by supercres at 7:55 PM on March 27

posted by supercres at 7:55 PM on March 27

No, you're rearranging individual letters randomly as if they were Scrabble tiles, so each letter is in fact unique for the purposes of computing the probability (i.e. just imagine that you have S1 and S2 as well as T1 and T2; for example, 7! counts both THTS1S2UR and THTS2S1UR.)

posted by Androgenes at 7:57 PM on March 27

posted by Androgenes at 7:57 PM on March 27

To me, it's slightly easier to look at the equivalent question: out of the possible permutations of the letters ABCDEFG, in how many of them are A and B next to each other? Then, do what Androgenes did, with 5! possible arrangements of CDEFG, inserting A and B, but note the pair can appear as AB or BA.

posted by supercres at 7:59 PM on March 27 [1 favorite]

posted by supercres at 7:59 PM on March 27 [1 favorite]

I agree w/ the previous posters about this problem not caring about the double letters because of the randomness.

If, for funsies, you wanted to change the problem to only deal with*distinguishable * permutations, then it would go like this:

1) Total distinguishable arrangements of THRUSTS: 7!/(2!2!). Each 2! in the denominator represents the double Ts and Ss respectively.

2) Total distinguishable arrangements w/ TT. Treat this as a SIX letter word, _ _ _ _ _ _, where one "letter is" TT. This should then be 6!/2! (the 2! for the double S's)

3) If you were looking at this list of distinguishable permutations, the chance of a randomly selected item having a TT would be: (6!x2)/7!

posted by Wulfhere at 8:03 PM on March 27

If, for funsies, you wanted to change the problem to only deal with

1) Total distinguishable arrangements of THRUSTS: 7!/(2!2!). Each 2! in the denominator represents the double Ts and Ss respectively.

2) Total distinguishable arrangements w/ TT. Treat this as a SIX letter word, _ _ _ _ _ _, where one "letter is" TT. This should then be 6!/2! (the 2! for the double S's)

3) If you were looking at this list of distinguishable permutations, the chance of a randomly selected item having a TT would be: (6!x2)/7!

posted by Wulfhere at 8:03 PM on March 27

Excellent. The double S's was throwing me off. Thanks for the help all.

posted by 23skidoo at 8:14 PM on March 27

posted by 23skidoo at 8:14 PM on March 27

If you want to avoid multiplication...

You have 6 letters in some order, one of which is a T. You add another T. What are the odds that it is next to the first T?

There are 7 spots to put in the second T (before the first letter, between the first and second letters, etc) and 2 of those are next to the first T.

posted by leopard at 9:05 PM on March 27 [8 favorites]

You have 6 letters in some order, one of which is a T. You add another T. What are the odds that it is next to the first T?

There are 7 spots to put in the second T (before the first letter, between the first and second letters, etc) and 2 of those are next to the first T.

posted by leopard at 9:05 PM on March 27 [8 favorites]

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First find the total number of arrangements. Seven letters; pick one of seven, you have six left to choose from, pick one of those six, now you have five, etc. = 7*6*5*... = 7!)

Next thing you want to know is all the possible arrangements of HRUSS (THRUSTS minus T's), which is 5!

The double-T's may be present at any point in the word; for example, you want to count not only TTRUHSS and TTSHRSU but also STTURHS, URTTSSH and HSSURTT. You can 'insert' TT into the five-letter word at any of six different points.

So, multiply the HRUSS permutation by that number and you get all the different possible arrangements with a double-T: 5!*6 = 6!

Divide the total number of arrangements by that number and you have the probability: 6!/7! = 1/7.

posted by Androgenes at 7:46 PM on March 27 [3 favorites]