The other final totals can be similarly described, and 13 has the most possibilities, with 6 different final rolls reaching the total of 13.
The rolls preceding the final total are important to consider, but it turns out that a running total over 12 is nearly independent of the number of rolls it took to achieve that total. For smaller totals, such as over 6, the probabilities are not so easily defined, though a quick simulation will show that for any N (N>5), N+1 is the most likely final running total.
This solution appears to use too much hand-waving, so I welcome any comments or improvements!
To find the probability of each possible total (13-18) the probability of each
previous total must be calculated. The probability of a 1,2,3,4,5,6 is of
course 1 out of 6 on the first roll. The probability of receiving a sum of
2 on the 2nd roll is 1/36, the sum of 3 on the 2nd is 2/36, etc. The
probabilities must be carried to the maximum number of rolls which could be 13.
The odds of receiving any given sum on any given roll is shown in the table
below as the chance to receive that sum out of 6 raised to the nth roll.
1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th 11th 12th 13th
Roll Roll Roll Roll Roll Roll Roll Roll Roll Roll Roll Roll Roll
Prob Prob Prob Prob Prob Prob Prob Prob Prob Prob Prob Prob Prob
Out Out Out Out Out Out Out Out Out Out Out Out Out
SUM of of of of of of of of of of of of of
TOT 6 36 216 6^4 6^5 6^6 6^7 6^8 6^9 6^10 6^11 6^12 6^13
-------------------------------------------------------------------------------
1 1
2 1 1
3 1 2 1
4 1 3 3 1
5 1 4 6 4 1
6 1 5 10 10 5 1
7 6 15 20 15 6 1
8 5 21 35 35 21 7 1
9 4 25 56 70 56 28 8 1
10 3 27 80 126 126 84 36 9 1
11 2 27 104 205 252 210 120 45 10 1
12 1 25 125 305 456 462 330 165 55 11 1
-------------------------------------------------------------------------------
13 21 140 420 756 917 792 495 220 66 12 1
14 15 125 400 741 911 791 495 220 66 12 1
15 10 104 365 706 890 784 494 220 66 12 1
16 6 79 309 636 834 756 486 219 66 12 1
17 3 52 229 510 708 672 450 210 65 12 1
18 1 25 125 305 456 462 330 165 55 11 1
Note that the sequences are fibonacci of sorts, excluding probabilities of
sums greater than 12.
Then the probabilities must be added for each sum to achieve the probability of
that sum. Example: The probability of ending with 13 is
21 140 420 756 917 792 495 220 66 12 1 3647371105
--- + --- + --- + --- + --- + --- + --- + --- + --- + --- + --- = -----------
6^3 6^4 6^5 6^6 6^7 6^8 6^9 6^10 6^11 6^12 6^13 13060694016
Therefore the probability of a 13 rolled = 27.9263% (Most probable)
probability of a 14 rolled = 23.6996%
probability of a 15 rolled = 19.2313%
probability of a 16 rolled = 14.5585%
probability of a 17 rolled = 9.7371%
probability of a 18 rolled = 4.8472%
---------
Sum = 100.0000%