Proability with Multiple Trials of Independent Events

What is a match, and what is a trial? In the opening paragraph, you state that they toss until only one obtains his/her number. Is a match/trial a single toss each, or is it a sequence of tosses until a winner is declared?

"The probability that only one or the other player gets their number of course increases with each additional match in the series."
Why? The dice have no memory.

Side comment: This method of conflict resolution seems likely to slow down the play of the game.

A singe toss each constitutes a match.

Well, I am not completely describing what is actually happening in the game because it would take time, and the example I gave of a challenger and defender respectively trying for a 6 and a 7 involves the same probability question. In fact, the way I have thses challenges structured it takes less than two matches on average to determine a winner. My interest here is to know mathematically what's going on. If needed I can give more specific details.

IIRC, it's really the whole point to the game.

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.

No, and that would be a silly game. These challenge matches and uneven probabilities for the players involved are one mechanism in a larger game, and I have not given all that background here. I am just trying to understand the problem in terms of calculating the increasing probability that matches will be resolved with each added match in a series.

Terminology explained: match = a round of play in which players roll simultaneously

As soon as a player hits his "target" number in such a round and his opponent fails to do so the challenge ends.

Is that right?

Yes, but it is not the game that ends, just that challenge is resolved and the outcome has consequences for the two players within the larger game.

Yeah, I fixed my post to read "challenge".

Thanks. Wish I had given a better explanation from the start:)

What I could have said is that two players engage in challenges where each tosses a pair of dice in repeated matches until ony one of them obtains his or her target number, and that the probabilities for each number sought in a match are not the same, so player A and B never share the same probability to get their number, it is always lopsided, like 7 to 5 or 3 to 2 odds.

So, are you just looking for the probabilities that a challenge will be terminated after 1 roll/toss, after 2 rolls/tosses, ..., after n rolls/tosses?

If so, since you already know how to compute the probability of exactly one of them achieving his/her goal in one roll/toss, call it P, the probability of terminating after n rolls/tosses is:
[(1-P)^(n-1)]*P

That is exactly what I want to confirm. So let's say each match involves two unrelated events, with the probability of event A ocurring at 48% and event B at 66%. When you run the formula you get a probability of 50.64% that one or the other, but not both, will occur. Do I just take that 0.5064 and project it to whatever number of matches/trials in a binomial distribution calculation?

I have found it extremely useful, accurate and quick to use online calculators for this, such as http://stattrek.com/online-calculator/binomial.aspx for binomial and http://www.calculator.net/probability-calculator.html for two event.

No - if you want what I asked if you want, then you'd plug 0.5064 into the formula I posted.

Thanks. I will plug it in and see what the result is.

Note that I should have phrased it like this:
"the probability of terminating after exactly n rolls/tosses"

Yes, I ran your formula and it provides the same answer as delivered from the online binomial calculator I referenced for "Cumulative Probability: P(X > 1)" which is good.

Here is its entire output for the probability of .5064

Probability of success on a single trial .5064
Number of trials 2
Number of successes (x) 1
Binomial Probability: P(X = 1) 0.49991808
Cumulative Probability: P(X < 1) 0.24364096
Cumulative Probability: P(X < or = 1) 0.74355904
Cumulative Probability: P(X > 1) 0.25644096
Cumulative Probability: P(X > or = 1) 0.75635904


I believe the relevant result sufficient for resolving a match is the last one of "Cumulative Probability: P(X > or = 1) 0.75635904"
which is I guess yet another question for me.

I would use the formula I provided. These results look like they're turning a discrete problem into a continuous one.

Well, I know that the probability must increase as we increase the number of trials, so we must find a greater probability than the .5064 we started with for a single trial, which leaves only "P < or = 1 " or "P > or = 1". I think the "< or =" won't do because "less than" means the event does not occur. Thoughts?