[Math] Calculating Probability of Die Rolls with a single reroll

dicestatistics

I am trying to figure out the odds of success for this problem, but keep running into issues around the ability to reroll one die.

Scenario: I am making two rolls of special dice. I am allowed to reroll one die in the first set, and in total I need 3 "successes"

Set 1, where I am allowed to reroll one die, has the following dice:
A Die with success on 1 of 6 faces
A Die with success on 2 of 6 faces
A Die with success on 3 of 6 faces

Set 2, where no reroll is allowed, has the following dice:
A Die with sucess on 1 of 6 faces
Two Dice with successes on 2 of 6 faces

I need a total of 3 successes among all of the dice rolled (with the reroll replacing the result of the rolled die).

Calculating the odds of each group succeeding is easy (disregarding reroll 6/216 or 2.8% and 4/216 or 1.85% respectively). I am at a total loss for how to figure in that reroll for the larger problem.

EDIT FOR CLARITY: Three success must happen among all six dice, the sets are just defining where a reroll can happen. You can choose which die in the first set to reroll (but obviously if it rolled a success and you reroll, the success is ignored and the new result is used).

Best Answer

Simulation may indeed be faster, though a systematic calculation of the probabilities is possible.

For the first set of dice the possible outcomes are as follows, where for example S F F(S) 2 6/216 means success on the first die and failure on the second and failure on the third with the third rerolled for a success with an outcome of $2$ successes with probability $\frac16 \times \frac46 \times \frac36 \times \frac36 = \frac{6}{216}$ 

Pattern     Successes Probability
S   S   S       3      6/216
S   S   F(S)    3      3/216
S   S   F(F)    2      3/216
S   F(S)S       3      4/216
S   F(F)S       2      8/216
S   F   F(S)    2      6/216
S   F   F(F)    1      6/216
F(S)S   S       3      5/216
F(F)S   S       2     25/216
F   S   F(S)    2     15/216
F   S   F(F)    1     15/216
F   F(S)S       2     20/216
F   F(F)S       1     40/216
F   F   F(S)    1     30/216
F   F   F(F)    0     30/216

to give probabilities of the number of successes on the first set of dice 

Successes   Probability
    0       30/216
    1       91/216
    2       77/216
    3       18/216

Similarly for the second set of dice 

Pattern     Successes   Probability
S   S   S       3        1/54
S   S   F       2        2/54
S   F   S       2        2/54
S   F   F       1        4/54
F   S   S       2        5/54
F   S   F       1       10/54
F   F   S       1       10/54
F   F   F       0       20/54

to give

Successes   Probability
    0       20/54
    1       24/54
    2        9/54
    3        1/54

Combine the first and second sets of dice to give

Successes   Probability
    0        600/11664
    1       2540/11664
    2       3994/11664
    3       3057/11664
    4       1216/11664
    5        239/11664
    6         18/11664

And so the probability of at least $3$ successes in total is then $\frac{4350}{11664} = \frac{755}{1944} \approx 0.38837$, similar to the result of your simulation

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