Answer Patterns

There are 117 ways to split the 120 questions into chunks of four consecutive answers One way is to group them so: {{1234}, {5678}, …, {117 118 119 120}} The number of elements in that set is 30. There are other ways of grouping four consecutive answers, they are: {{2345} {6789} … {116 117 118 119}} 29 there {{3456} {789 10} … {115 116 117 118}} 29 there {{4567} {89 10 11} … {114 115 116 117}} 29 there, and we have exhausted the possible ways you can have four consecutive questions with the same answer. the total of those is 117 if you do the same with chunks of 5 answers, it’s 116 (24, 23 x 4)

ahh nicely done

The way to find the probability of a string of same answer choices is actually very easy, no need for permutations or “chunks”. Question: Starting at any given point, what is the probability of getting a run of N same choice? Answer: The probability of getting a run of N “a” answers is (1/3)^N Same for “b” and “c” answers. Now just add the above mutually exclusive probabilities. Hence the probability of getting a run of N same answers = 3 * (1/3)^N = (1/3)^N-1 A bit harder to compute what the probability of getting a run of N in a sample of M, where N > M, as there are different points at which the run could begin. NC

This smells a lot like a negative binomial, but it was easier for me to solve it with first principles. Anyone else have a cleaner solution?

nevcfa1 Wrote: ------------------------------------------------------- > The way to find the probability of a string of > same answer choices is actually very easy, no need > for permutations or “chunks”. > > Question: Starting at any given point, what is the > probability of getting a run of N same choice? > > Answer: The probability of getting a run of N “a” > answers is (1/3)^N > > Same for “b” and “c” answers. > > Now just add the above mutually exclusive > probabilities. > > Hence the probability of getting a run of N same > answers = 3 * (1/3)^N = (1/3)^N-1 > > A bit harder to compute what the probability of > getting a run of N in a sample of M, where N > M, > as there are different points at w hich the run > could begin. > > NC I meant N < M NC

hawgdriver Wrote: ------------------------------------------------------- > There are 117 ways to split the 120 questions into > chunks of four consecutive answers > > One way is to group them so: {{1234}, {5678}, …, > {117 118 119 120}} > > The number of elements in that set is 30. > > There are other ways of grouping four consecutive > answers, they are: > > {{2345} {6789} … {116 117 118 119}} > > 29 there > > {{3456} {789 10} … {115 116 117 118}} > > 29 there > > {{4567} {89 10 11} … {114 115 116 117}} > Sorry for the typo and I can’t edit my original post. But the final element of each of those sets is wrong. Should be: {{2345} {6789} … {114 115 116 117}} {{3456} {789 10} … {115 116 117 118}} {{4567} {89 10 11} … {116 117 118 119}} For those following along at home…

Yes my pattern was A BBB A CC A CC A BBB AAAAAA CC B AA CC CCCCCCCCCCCCCCC AA BB BBBB CC AA CC CC AA Then I just repeated until i got to 120 Are you kidding?

watch your six, devildog

devildog Wrote: ------------------------------------------------------- > Yes > my pattern was A BBB A CC A CC A BBB AAAAAA CC B > AA CC CCCCCCCCCCCCCCC AA BB BBBB CC AA CC CC AA > Then I just repeated until i got to 120 > > > Are you kidding? Seems like you got a lot wrong because the correct sequence was BACCCBAABAACAAACCCCCC

cfagoal2 Wrote: ------------------------------------------------------- > watch your six, devildog why do you say that?

modern warfare lol

There was a time I was swift, silent, and deadly, but I am fast becoming slow, noisy and harmless

> For the AM portion, did anyone of you have bunch > of c’s (4 or 5 in a row) in the begginning of the > second to last column…I think it was in the > derivatives section…I forgot what my form number > is, but can you guys confirm? >I think I had 5 in a row but went back and changed one of them. Thanks SJZ

I had many consecutive c’s in AM and b’s in PM (prob 5-6 of them)