Thread: Probability Question:

Say I rolling X number of 6 sided-dice. I'm attempting to get >=Y number of results of 5 or greater.


For instance, I'm trying to roll three dice and get two or more events. What's a reliable algorithm to determine the likelihood of that occurring? I'm trying to find a way to extrapolate that question for a game I'm making- so I can later apply it to four or five dice.

Do you know about the Binomial Distribution?

When I was doing some initial research, I found something on wikipedia about it. Couldn't make much sense of it though. I did find a binomial distribution calculator that may be able to sort this problem for me. It asks me for number of events, number of successes and probability.

For it to be useful to me, I have to assume that the probability of success part is per one die rolled... which must mean that an event is a single die. Number of successes... is that the number of required successes? Because... yeah, that'd be nice.
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Hmm. Tried it out with 3 dice, 33% for a success and req for 2 successes. It gave me 21.9%. My long hand came up with 25.9% I must not be doing this right. Binomial Distribution Probability Calculator Online

EDIT:
I think I figured out why that is. It seems the Binomial Distribution only tests for an exact number of successes. Over three dice, I'm looking for the percentage of 2+ successes. I wonder if there's anything in that vein?

Edit 2:
Ok- I think I found what I'm looking for. This calculator tracks the cumulitive distributionin multiple ways. If anyone needs it in the future, here it is. http://stattrek.com/online-calculator/binomial.aspx

Originally Posted by ClaimingLight

For it to be useful to me, I have to assume that the probability of success part is per one die rolled...

The probability of success should be the probability that a single dice roll returns a value of 5 or higher, for a six sided die this should be . Given X die are rolled, then the probability of Y successes is given by the following equation.



Where



is the number of combinations of 2 objects, drawn from a total of three objects. So for 3 dice the probability of 2 successes should be , the discrepancy between this answer and that of the calculator most likely being due to round off error.