Mathematical odds of picking letters What are the odds of picking two pre-selected non-sequential letters in order out of a $26$ letter alphabet, i.e. $D$ then $S$, not $S$ then $D$? One chance only.
if I understand the question correctly, what's the probability that you select a two letters from the English alphabet, such that the second one comes after the first. Use the law of total probability:
$$
P(X) = \sum_{k=1}^{26}P(X|Y=k)P(Y=k) = \frac{25}{26} \cdot P(X=A) + \frac{24}{26} \cdot P(X=B) +\ldots
$$

of course $P(X=j) = \frac{1}{26}$. Can you handle from here? 

EDIT: you may want ot use the identity $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
First are the selection of the two letters the same event? Or separate, and are you done selecting letters, or will you continue to select letters. If you continue to select letters that needs to be specified or the answer would assume you selected wrong (1/26)(1/26), or 1/26^2 but that wouldn't be correct for your scenario is suggesting either two separate events (like above) however selecting from a different amount of letters than 26. What exactly do you mean sequential? Picking D before the next event of selecting letters? That's wouldn't be 1/26 as mentioned by others because selecting any that come before S and is correct is what I believe you are asking? If that's what you mean by sequential (picking D then picking another letter in the same event (so d is taken out of selections) However does the person know, or aware that the letter is sequential and after D?) If they know the letter is after D then 1/22 for the second guess if they are in the same event (or turn picking letters). Also do they know it is sequential, that makes a difference on the first guess, because you could not guess Z, so the first guess witch was D would have to be 1/25 and then knowing it was after the letter D and guessing 1/22 so with that knowledge (knowing the second letter is after in the alphabet or sequential). 

1/25*1/22

That only works if they know that the second letter is sequential. If they are picking as two separate events and after combining them and deeming them sequential it is just 1/26^2. This is just the rate of guessing the correct letter and then doing it again.

If your looking for all sequential possibilities and guessing the correct pair, that is a lot, especially because they are not even letters that are adjacent to each other or one after the other, they seem to have very random placement. That's would be something like 1/26^26*26 * 1/IDK Someone else can figure that out. I probably started it wrong.

(Edit) Oops I meant to just make this a comment. So I'll add some thought above.The letters "D" and "S" make no difference, it could be any letter, or even numbers for that matter as long as your using 26.  @David