TITLE: Most Probable Sum by rolling $2$ dice in $[100,111]$.
QUESTION [6 upvotes]: Given two dices, we roll them and add the result to a sum (initialised to 0) till sum is $\ge 100.$ The resultant sum can be any number in [100 111]. Which among them have the highest probability of being the resultant sum.

REPLY [5 votes]: The single die version of this question is analyzed here:
Probability of dice sum just greater than 100
In my paper Linear recurrences via probability,
 American Mathematical Monthly 122, 386-389 (April 2015), 
I consider the general version of this game. 
In particular, equation (5) gives the asymptotic hitting probabilities as the lower limit (100 in this problem) gets large.
For the 2 dice game, the asymptotics say that 
$$\mathbb{P}(\mbox{games ends at state }99+k)\approx{\mathbb{P}(\xi\geq k)\over\mathbb{E}(\xi)},
 \mbox{ for }k=1,2,\dots, 12,$$ where $\xi$ is the sum on a pair of dice. 
These probabilities are 
$$1/7, 1/7, 5/36, 11/84, 5/42, 13/126, 1/12, 5/84, 5/126, 1/42, 1/84, 1/252,$$
or in decimal form 
$$.1429, .1429, .1389, .1310, .1190, .1032, .08333, .05952, .03968, .02381, .01190, .003968.$$