Assuming equally distributed probability of the "other" throws:
6: 2.1504e-06
7: 1.516032e-05
8: 6.105599999999999e-05
9: 0.0001866496
10: 0.0004811571199999999
11: 0.0011080704000000004
12: 0.002317765120000012
13: 0.00445283840000005
14: 0.007932728320000082
15: 0.013234946560000219
16: 0.02078419456000026
17: 0.03073397760000035
18: 0.042942725119999754
19: 0.05675394815999913
20: 0.07107977727999641
21: 0.08419953407999499
22: 0.09412483327999363
23: 0.09939161855999414
24: 0.09888333311999539
25: 0.09254429439999612
26: 0.08094589695999811
27: 0.06610059775999934
28: 0.050275064320000035
29: 0.03517282560000011
30: 0.022496632320000085
31: 0.012905395200000015
32: 0.006652177920000004
33: 0.0029200895999999987
34: 0.0010063871999999997
35: 0.0002651443199999999
36: 2.9030399999999992e-05
Or, if you prefer the exact answer:
6: 21/9765625
7: 2961/195312500
8: 477/7812500
9: 7291/39062500
10: 23494/48828125
11: 10821/9765625
12: 905377/390625000
13: 173939/39062500
14: 1549361/195312500
15: 5169901/390625000
16: 4059413/195312500
17: 600273/19531250
18: 8387251/195312500
19: 22169511/390625000
20: 13882769/195312500
21: 32890443/390625000
22: 36767513/390625000
23: 38824851/390625000
24: 19313151/195312500
25: 7230023/78125000
26: 31619491/390625000
27: 12910273/195312500
28: 19638697/390625000
29: 2747877/78125000
30: 8787747/390625000
31: 504117/39062500
32: 2598507/390625000
33: 57033/19531250
34: 9828/9765625
35: 25893/97656250
36: 567/19531250
Here's a brief explanation of how I did this:
First, here's a list of the probabilities associated with the different dice throws:
Dice 1 probabilities
Probability of a 1 is 1/10
Probability of a 2 is 9/50
Probability of a 3 is 9/50
Probability of a 4 is 9/50
Probability of a 5 is 9/50
Probability of a 6 is 9/50
Dice 2 probabilities
Probability of a 1 is 4/25
Probability of a 2 is 1/5
Probability of a 3 is 4/25
Probability of a 4 is 4/25
Probability of a 5 is 4/25
Probability of a 6 is 4/25
Dice 3 probabilities
Probability of a 1 is 7/50
Probability of a 2 is 7/50
Probability of a 3 is 3/10
Probability of a 4 is 7/50
Probability of a 5 is 7/50
Probability of a 6 is 7/50
Dice 4 probabilities
Probability of a 1 is 3/25
Probability of a 2 is 3/25
Probability of a 3 is 3/25
Probability of a 4 is 2/5
Probability of a 5 is 3/25
Probability of a 6 is 3/25
Dice 5 probabilities
Probability of a 1 is 1/10
Probability of a 2 is 1/10
Probability of a 3 is 1/10
Probability of a 4 is 1/10
Probability of a 5 is 1/2
Probability of a 6 is 1/10
Dice 6 probabilities
Probability of a 1 is 2/25
Probability of a 2 is 2/25
Probability of a 3 is 2/25
Probability of a 4 is 2/25
Probability of a 5 is 2/25
Probability of a 6 is 3/5
Then, there are only $6^6 = 46656$ possible throws. This is a small number for a computer to iterate through, so I do so. For each throw, I calculate what the probability of it occurring is, and what the sum of the dice is. I accumulate that sum to the probability of getting that number.
Here are some examples of ways to throw 21:
Die 1 has a 1 (probability 1/10 )
Die 2 has a 1 (probability 4/25 )
Die 3 has a 1 (probability 7/50 )
Die 4 has a 6 (probability 3/25 )
Die 5 has a 6 (probability 1/10 )
Die 6 has a 6 (probability 3/5 )
Probability of this throw: 63/3906250
Die 1 has a 1 (probability 1/10 )
Die 2 has a 1 (probability 4/25 )
Die 3 has a 2 (probability 7/50 )
Die 4 has a 5 (probability 3/25 )
Die 5 has a 6 (probability 1/10 )
Die 6 has a 6 (probability 3/5 )
Probability of this throw: 63/3906250
Die 1 has a 1 (probability 1/10 )
Die 2 has a 1 (probability 4/25 )
Die 3 has a 2 (probability 7/50 )
Die 4 has a 6 (probability 3/25 )
Die 5 has a 5 (probability 1/2 )
Die 6 has a 6 (probability 3/5 )
Probability of this throw: 63/781250
Die 1 has a 1 (probability 1/10 )
Die 2 has a 1 (probability 4/25 )
Die 3 has a 2 (probability 7/50 )
Die 4 has a 6 (probability 3/25 )
Die 5 has a 6 (probability 1/10 )
Die 6 has a 5 (probability 2/25 )
Probability of this throw: 21/9765625
Die 1 has a 1 (probability 1/10 )
Die 2 has a 1 (probability 4/25 )
Die 3 has a 3 (probability 3/10 )
Die 4 has a 4 (probability 2/5 )
Die 5 has a 6 (probability 1/10 )
Die 6 has a 6 (probability 3/5 )
Probability of this throw: 9/78125
Die 1 has a 1 (probability 1/10 )
Die 2 has a 1 (probability 4/25 )
Die 3 has a 3 (probability 3/10 )
Die 4 has a 5 (probability 3/25 )
Die 5 has a 5 (probability 1/2 )
Die 6 has a 6 (probability 3/5 )
Probability of this throw: 27/156250
Obviously this goes on for some time, but in the end you get the answer. Below is the python script:
from collections import defaultdict
import itertools
import fractions
#Generate the probabilities for the dice
dice=defaultdict(list)
for die in range(1,7):
for throw in range(7):
if throw==0:
dice[die].append(0)
elif die==throw:
dice[die].append(fractions.Fraction(die,10))
else:
dice[die].append(fractions.Fraction(10-die,50))
#Generate the probabilities for the different throws
results = defaultdict(fractions.Fraction)
for throw in itertools.product(range(1,7),repeat=6):
prob=fractions.Fraction(1,1)
for die,t in zip(range(1,7),throw):
prob*=dice[die][t]
results[sum(throw)]+=prob
#Print results
for r in results:
print("%2d: %s"%(r,results[r]))