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(1/6)*(1/6)*(1/6)=1/216

Q: What is the probability of a regular 6-sided dice rolled three 6s in arow?

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50% chance

There are two possibilities for rolling three once, so the probability of this even is 2/36 or 1/18. Since each roll is an independent event, the total probability is: P(rolling three three times) = (1/18)3 = 1/5832

The answer depends on how many times in total the dice are rolled. As the total number of rolls increases, the probability rolling a 6 and 4 three times in a row increases towards 1.

1/6 x 1/6 x 1/6 = 1/216

Assuming the die is a standard die with a different number from the set {1, 2, 3, 4, 5, 6} on each side, then: Probability_of_success = number_of_ways_of_success/total_number_of_outcomes There is only 1 way to roll a 4 and there are 6 possible outcomes, therefore the probability of rolling a 4 is 1/6 The die has no knowledge of previous rolls; each roll is independent, thus: The probability of three fours is a row is the probability of a 4 times the probability of a 4 times the probability of a 4, which is: Probability = 1/6 × 1/6 × 1/6 = 1/216

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(1/6)^3 since each event is independent and each has a probability of 1/6.

50% chance

The probability of rolling a six is one in six. The probability of rolling three consecutive sixes is one in 216. (1/6 x 1/6 x 1/6 = 1/216)

The probability is 1 and you do not need Matlab to get that answer - only a little bit of thought.

The first roll doesn't matter for probability, it just sets the number to be rolled by the other two. So: P(rolling the same number three times) = P(rolling a particular number)2 = (1/6)2 = 1/36

There are two possibilities for rolling three once, so the probability of this even is 2/36 or 1/18. Since each roll is an independent event, the total probability is: P(rolling three three times) = (1/18)3 = 1/5832

The answer depends on how many times in total the dice are rolled. As the total number of rolls increases, the probability rolling a 6 and 4 three times in a row increases towards 1.

If the number cubes are standard dice cubes, the odds of rolling 3 ones is 1 in 216.

When a tetrahedral die is rolled, it will rest with three faces upwards. If the die is numberd from 1 to 4. therefore the sum of the upward facing numbers on 1 die is at least 6 and so for two dice, the minimum is 12. That being the case, the probability is 0.

2/12 or 1/6

If using regular dice, the probability is 0 since the minimum sum from four dice is 4.

It is (1/6)5 = 1/7776 = 0.00013 approx.