I have some probability I'm not sure how to calculate for a system. The system rolls three pairs of numbers where each pair (A, B) is two independent variables. The actual values of A and B are chosen according to their own weighted table. The task is to determine the odds of a desired pair (A, B) out of three pairs where there are no 'half-successes.' A 'half-success' is the desired (A, B) occurring after any prior pair that has the desired A. See examples at the end for more details.
This initially looks like the typical binomial distribution at this point where p is the chance of a single result for A and B, right? But there is a difference in this system because an earlier half-success will invalidate a later success.
How would this be calculated? I think it should be the binomial distribution for n = 3 and x >= 1 to start. The success probabilities here include trials that would be considered duplicates and they need to be removed from the count, right?
Initial thoughts: if there was a successful pair in one of the three rolls but there were two pairs with the desired A, then there's a 50% chance it was a duplicate and not a true success. So it seems like half of the binomial success cases with two pairs of desired A need to be removed somehow. And then 2/3 of the success cases with three of the desired A values? But how would these be calculated?
Or is there another way to calculate this?
EDIT: Some more information. The system rolls a random number out of 100 and assigns A one of [0, 5] depending on the result. The system then rolls a different random number out of 10000 and assigns B one of [5, 90] depending on the random number. To simplify this question, let's assume the desired the value for A has a 5% success rate, and the desired value for B has a 1% success rate.
If there was a single generated pair, then clearly the chance of success is 5% * 1%. However, the system actually gives us three chances at this result, but getting a successful A and unsuccessful B in an earlier pair prevents a successful A and successful B in a later pair.
Some examples below. Suppose (5, 10) is the desired result.
Successful: (5, 10); (0, 0); (1, 4)
Unsuccessful: (5, 4); (2, 3); (5, 10) because (5, X) was already rolled
Successful: (1, 10); (4, 5); (5, 10)
Unsuccessful: (1, 2); (3, 4); (5, 4) because no (5, 10)
Unsuccessful: (1, 2); (3, 4); (4, 4) because no (5, 10)
Successful: (1, 3); (5, 10); (5, 2) because (5, 10) occurred before any half-successes.
Following the rule that you present, to calculate the probability of a pair $(a,b)$ you must use the law of total probability. Let $A_i$ and $B_i$ (i=1,2,3) denote the $i$-th output of A and B (here I consider the output before the transformation of the duplicates values in A).
Chosen the desired pair $(a,b)$, let $S_i$ denote that the $i$-th draw was successful and $U_i$, that the mentioned draw was unsuccessful. Let $S$ denote that some draw was successful.
In general terms, we have that: $$ P(S) = P(S_1) + P(S_2) \cdot P(U_1) + P(S_2) \cdot P(U_1) + P(S_3) \cdot P(U_2) \cdot P(U_1)$$
Now, for $a \neq 0$, we have:
$$P( S) = \left[P(A_1 = a) \cdot P(B_1 = b)\right] + \left[P(A_2 = a) \cdot P(B_2 = b)\right] \cdot P(A_1 \neq a) + \\ \left[P(A_3 = a) \cdot P(B_3 = b) \right]\cdot \left[P(A_2 \neq a) \cdot P(A_1 \neq a)\right]$$
Note that $P(A_1 = a) = P(A_2 = a) = P(A_3 = a)$ since the drawing are independents, (and the same for $P(B_i = b)$).
Since you have the table of probabilities, it is easy to calculate the probabilities presented in the equation above (note that $P(A_i \neq a) = 1 - P(A_i = a)$.
** EDIT** According to the comments in the OP, when a value $A = a$ appears again, the pairs is transformed into $(0,0)$.
In that way, for a pair $(0,b)$, we have the same calculation as for $(a,b)$ with $a \neq 0$.