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Proability with Multiple Trials of Independent Events

 
 
morphix
 
Reply Wed 27 Mar, 2013 05:52 am
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I am developing a game that uses dice to resolve conflcts involving two players. The two players each toss a pair of standard d6 dice in matches until only one of them obtains his or her target number, thus winning the challenge.
Assume the two players are seeking numbers with different probabilities, like a 6 for 5/36 and a 7 for 6/36

I know the formulas for independent events to figure out the probability that player A and/or B will obtain their bumber. And from that value you subtract the probability that both will get their number.

But what I want to know is how to find the probability that only one player will obtain his or her number over muliple matches/trials, like doing binomial calculations but for trials involving independent events. In other words, I not only want to know the probability that only one player will obtain his or her number in one match, but also over two or three successive matches, such as if neither or both get their number in the 1st match, in the 2nd match, etc. The probability that only one or the other player gets their number of course increases with each additional match in the series.

Does one just take the probability of only one player getting his or her number in one trial, and then use a binomial calculation on that value for whatever number of multiple trials, or does one perform a binomial calculation on the probabilies of obtaining each number for 2 trials, 3 trials etc., and plug each of those values into the independent event formula?

Hope I have made this question clear.
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markr
 
  1  
Reply Wed 27 Mar, 2013 02:10 pm
@morphix,
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.
markr
 
  1  
Reply Wed 27 Mar, 2013 02:18 pm
@morphix,
Side comment: This method of conflict resolution seems likely to slow down the play of the game.
morphix
 
  1  
Reply Wed 27 Mar, 2013 02:51 pm
@markr,
A singe toss each constitutes a match.
0 Replies
 
morphix
 
  1  
Reply Wed 27 Mar, 2013 03:02 pm
@markr,
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.
0 Replies
 
Kolyo
 
  1  
Reply Wed 27 Mar, 2013 03:50 pm
@markr,
markr wrote:

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


IIRC, it's really the whole point to the game. Wink
morphix
 
  1  
Reply Wed 27 Mar, 2013 03:53 pm
@markr,
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.
morphix
 
  1  
Reply Wed 27 Mar, 2013 03:58 pm
@Kolyo,
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.
0 Replies
 
Kolyo
 
  1  
Reply Wed 27 Mar, 2013 04:06 pm
@morphix,
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?


morphix
 
  1  
Reply Wed 27 Mar, 2013 04:09 pm
@Kolyo,
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.
Kolyo
 
  1  
Reply Wed 27 Mar, 2013 04:10 pm
@morphix,
Yeah, I fixed my post to read "challenge".
morphix
 
  1  
Reply Wed 27 Mar, 2013 04:20 pm
@Kolyo,
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.
0 Replies
 
markr
 
  1  
Reply Wed 27 Mar, 2013 04:41 pm
@morphix,
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
morphix
 
  1  
Reply Wed 27 Mar, 2013 04:52 pm
@markr,
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.
markr
 
  1  
Reply Wed 27 Mar, 2013 05:22 pm
@morphix,
No - if you want what I asked if you want, then you'd plug 0.5064 into the formula I posted.
morphix
 
  1  
Reply Wed 27 Mar, 2013 05:25 pm
@markr,
Thanks. I will plug it in and see what the result is.
markr
 
  1  
Reply Wed 27 Mar, 2013 05:34 pm
@morphix,
Note that I should have phrased it like this:
"the probability of terminating after exactly n rolls/tosses"
morphix
 
  1  
Reply Wed 27 Mar, 2013 06:06 pm
@markr,
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.


markr
 
  1  
Reply Wed 27 Mar, 2013 06:48 pm
@morphix,
I would use the formula I provided. These results look like they're turning a discrete problem into a continuous one.
morphix
 
  1  
Reply Wed 27 Mar, 2013 06:59 pm
@markr,
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?
 

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