It's not that the dice have memory, but it is like a series of coin tosses. The probability of heads or tails for each toss is always the same, but for a specific number of tosses in a series the probability that you will get x number of heads or tails in a row does change.
So, if you repeatedly tossed two coins and were seeking say two heads at the same time, for whatever number of tosses it becomes increasingly likely for that to happen, which is the standard binomial calculation. From that calculation you can definitely come up with the probability for getting two heads in x number of tosses, and it becomes increasingly likely to occur within the sequential set with every added toss, even though each toss has exactly the same probability.
The question I am asking can be built around two players each tossing some coins in matches where each gets one toss. Let's say player A tosses four coins in each match and needs to get three heads, and player B tosses three coins per match and needs to get two tails.
The probabilities are different for each player, so this resembles my example with dice and players trying to get a 6 or a 7 respectively. One can run the formula for independent events by plugging in the two different probabilities and arrive at the proability that neither will succeed, that both will succeed, or that only one will succed in any given match. In my scenario, it is when just one succeeds that the winner is determined and the chalenge is resolved.
My question is: what is the proabibility that either player A or B, but not both, will obtain their number in two matches, three atches, four matches, etc.