regor60 wrote:This reads as each of the remaining three dice are thrown individually, which isn't the way I've played the game in the past.
If the three dice are thrown as a group, I wonder what the equivalent answer is ? I came up with 5/18
Whether the remaining dice are rolled one at a time or all at once, the probability is the same.
To illustrate:
If two dice are rolled, what is the probability of getting exactly one 3?
One at a time:
P(1st roll is 3 and 2nd roll is not 3) = 1/6 * 5/6 = 5/36.
P(1st roll is not 3 and 2nd roll is 3) = 5/6 * 1/6 = 5/36.
Since either case is a favorable outcome, we ADD the fractions:
5/36 + 5/36 = 10/36 = 5/18.
Both at once:
Total number of possible rolls = 6*6 = 36.
Ways that the pair formed by the die on the left and the die on the right will yield exactly one 3:
3-1, 3-2, 3-4, 3-5, 3-6, 1-3, 2-3, 4-3, 5-3, 6-3 --> 10 ways.
Of 36 possible rolls, 10 have exactly one 3.
Thus, P(exactly one 3) = 10/36 = 5/18.
Whether the two dice are rolled one at a time or both at once, the result is the same:
P(exactly one 3) = 5/18.
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