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I found a simple way.For x+y+z=11, the probability is sum of 1/6*P(y+z=a), a is from 5 to 10For x+y+z=12, the probability is sum of 1/6*P(y+z=a), a is from 6 to 11The difference: 1/6*[P(y+z=5)-P(y+z=11)]>0so P(x+y+z=11) > P(x+y+z=12)
I found a simple way.
For x+y+z=11, the probability is sum of 1/6*P(y+z=a), a is from 5 to 10
For x+y+z=12, the probability is sum of 1/6*P(y+z=a), a is from 6 to 11
The difference: 1/6*[P(y+z=5)-P(y+z=11)]>0
so P(x+y+z=11) > P(x+y+z=12)
