What are the odds of hitting exactly 100 rolling a fair die I roll a fair die and sequentially sum the numbers the die shows. What are the odds the summation will hit exactly 100?

More generally, what are the odds of hitting an exact target number t while summing the results of numbers sequentially drawn uniformly from a set S.

I have experimentally tested this and the result is (warning - spoilers ahead):


   1 over the expectation of the drawn numbers. In the case of the fair die 1/((1+2+3+4+5+6)/6) = 6/21 = 2/7


However I do not have a strong intuition, let alone a formal proof, why this is the case.

I'll be happy to get your thoughts!
The probability of hitting $n$ on the nose is a function of $n$, call it $f(n)$. It satisfies the recurrence
$$f(n)=\frac{f(n-1)+f(n-2)+f(n-3)+f(n-4)+f(n-5)+f(n-6)}6$$
with boundary values $f(0)=1$ and $f(n)=0$ for $n\lt0$.
I didn't try to solve the recurrence, but my numerical calculations agree with yours, it looks like $f(n)\to2/7$.
I would say since the possible outcomes are $1,2,3,4,5,6$ on the die that the chances of getting exactly $100$ are the same as that of getting any single desired digit on the die once you are in the $94$ to $99$ range. Eventually you will get into that range such that you need either a $1,2,3,4,5$ or $6$ to make a $100$ sum.  Then you must get that number to achieve the $100$ sum.  So I would say $1$ in $6$ chance at that point.  For example, suppose the partial sum is $94$, then you would need a $6$.  If you instead roll a $2$ and get $96$ then you would need a $4$.  If instead you roll a $3$ then you need a $1$.  So anytime you get into the $94$ to $99$ range, you have a $1$ in $6$ chance of hitting exactly $100$ on a single final roll.  If you instead consider the probability from the beginning when the partial sum is $0$ or when the partial sum is in the range $94$ to $99$ and considering multiple "final" rolls, then the answer will be different. 

So the answer depends on when you start counting the probability but in my case, I simplified and only care about the one final roll that would give you a $100$ sum.

One can also argue/claim that the probability is $0$ if the number of die rolls for the $100$ sum case is less than $17$ rolls.  The original question didn't specify how many rolls are allowed or will take place.

Another answer is to interpret the question to mean only 1 die roll and sum (as many times as needed) only the numbers showing.  This could be interpreted many ways such as only the top number (in which case $1, 2, 4$, and $5$ would "always" sum to exactly $100$), or all sides except the "hidden" bottom side (assuming it wasn't rolled on a see thru table or equivalent), or even summing all $6$ sides if they are all visible.

Also the question doesn't state only base $10$ of the sum $100$ so if I go past that, I could change to base $11$ and keep rolling (and go for $121$ in base $10$ which is $100$ in base $11$).  If I miss that I can go for $144$ which is $100$ in base $12$....

How about if you interpret the question as strictly $1$ die roll and all $6$ faces are visible thus adding up to $21$ and then expressing that sum in base $21^{1/2}$ which would be written as $100$?  Then it is $100$% chance of getting it.

I don't know the answer to the more general question yet but I will think about it some.