Patrick B. answered • 05/11/19
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(2/6)^N = (1/3)^N = 1^n / 3^n = 1/ 3^n
Stephen M.
Here’s some more information: I directly computed the probabilities for rolling runs of two numbers by exhaustively calculating the possibilities, for various rolls of dice. For N = 1, p = 1.000 (obviously). For N = 2, p = 0.833. For N = 3, p = 0.417. For N = 4, p = 0.162. For N = 5, p = 0.058. And for N = 6, p = 0.020. I’d like a general formula.
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05/12/19
Stephen M.
Oops! Obviously, for N = 1, the probability isn’t 1.000, since it can’t happen!
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05/12/19
Stephen M.
No, that’s definitely incorrect as it includes cases in which only one number shows up repeatedly. Although I asked the question for runs of two numbers, I’m really interested in the general case of runs of M numbers, (1<=M<=6), and in fact the more general case of dice with any number of sides S, where S > 1.05/11/19