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How do I calculate these odds of succes for a game system?

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> The chance of rolling at least one 5 or 6 is: (...)

These are all correct, well done (this is called the binomial distribution).

So the chance to succeed is equal to the chance that the number of 5s and 6s is greater than the number of 1s. Unfortunately there is no simple formula to calculate this as there is for binomial distribution. You have to go case by case.

Let's say X is the number of 5s and 6s you get in a given roll and Y is the number of 1s. Let's say you're rolling 4 dice. Your chance to win is P(X>Y).

    P(X>Y) = P(X>Y | Y=0)*P(Y=0) + P(X>Y | Y=1)*P(Y=1) + ... + P(X>Y | Y=4)*P*Y=4)

and

    P(X>N) = P(X=4)+P(X=3)+...+P(X=N+1)

so you see that the formula gets very complex very quickly. You would probably use a computer to figure something like that out rather than do it by hand.

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