Probability of getting a specific sequence of length $4$ in $10$ coin tosses This is more of a thinking question maybe, hope that's ok.
Suppose I toss a coin $10$ times. What is the probability that within these 10 tosses I get the sequence THHT.
My attempt:
If I have 10 coin tosses THHT can happen in only 1 way, so similar to seating arrangements I treated it as one outcome.
Then I have THHT _ _ _ _ _ _ six more spots to fill, each of which have 2 possible outcomes, heads or tails.
THHT can take any of the 7 spots so we have $ 7*2^6 $ ways this can happen.
Then because there is six spots, we could have the sequence THHT twice.
THHT THHT _ _ there is twelve ways this can happen times 4 because of the double T's.
Now I have $$\frac{7*2^6-48}{2^{10}}$$ $2^{10}$ are all the possible coin toss combinations.
This gives me $$\frac{400}{1024}$$
The correct answer is $$\frac{393}{1024}$$.
This is a question to a homework that I did a few months ago.
I am quite enthusiastic about these types of problems (counting, probability, combinatorics) and that's why I did it differently than how we learned in the lectures (inclusion, exclusion).
My answer was wrong and I couldn't find out why during the semester, I miscounted something.
I was told (by TA) my answer is wrong because I didn't do inclusion-exclusion.
I know how to do it inclusion exclusion way, after all it's just following a few repetitive steps.
But I enjoy finding the correct way to count $$\frac{\text{possibilities this can happen}}{ \text{all possibilities}}$$
However after spending all morning tinkering, I can't find out why.
I think my solution could have been correct as it was very close, but I counted 7 variations extra.
Because 7 is the number of "seats" I can have THHT take, I feel like this isn't a coincidence.
Thank you very much for any insight you can share in what I missed, or if indeed, me being close to the answer was pure coincidence (the irony) and this can only be solved with inclusion and exclusion mechanics.
 A: You started with THHT_ _ _ _ _ _ and similar in $7 \times 2^6$ different ways
But you have double counted:

*

*THHTHHT_ _ _ and _ THHTHHT_ _ and  _ _ THHTHHT_ and _ _ _ THHTHHT in $4\times 2^3$ ways


*THHT_ _ THHT and _ _ THHTTHHT and  _ THHTTHHT_ and _ THHT_THHT and THHT_THHT_ and THHTTHHT_ _  in $6\times 2^2$ ways
If you subtract these, then you oversubtract THHTHHTHHT.  That string appears three times in the initial count, twice in the first double count and once in the second double count, and you only want it counted once overall
So I think the total count is
$$7 \times 2^6 - 4\times 2^3-6\times 2^2 + 1=393$$
A: You are doing a partial inclusion--exclusion. After all you claim to be over-counting and then having to subtract off strings with two occurrences of THHT. Doesn't this look like you are exactly on the way to inclusion -exclusion except that you haven't quite set up everything properly?
Let $A_i$ for $i=1,\dots,7$ denote the strings in which $THHT$ occurs starting from $i^{th}$ position i.e. positions $i$, $i+1$, $i+2$ and $i+3$ are $THHT$. You start by counting $$ |A_1| + \dots + |A_7|.$$ The you try to subtract off $$\sum_{i<j} |A_i \cap A_j|.$$ But some errors start to creep in. And then you miss the fact that $$|A_1 \cap A_4 \cap A_7| = 1,$$ too.
I'm not sure your TA has explained things very well (apparently not well enough since you still had questions about your attempt), but basically you really do need to do inclusion--exclusion more 'properly'.
