This depends on if you want at least two of the dice to be the same number, or exactly two of the dice to be the same number.
For the first scenario: Roll the first die, and get a number. Roll the second die, and there is 1/6 chance that it'll be the same as the first one. Now if it's not the same (5/6 chance) then the third die has 1/6 chance of being the same as the first, and 1/6 chance of being the same as the second. So we have:
1/6 + 5/6*(1/6 + 1/6) = [simplified] 4/9 or about 44.44%chance that at least two are the same.
For the second scenario: With three dice, there are 216 possible outcomes (6 x 6 x 6). So we know that there is a 4/9 chance that 2 or more will be the same: (4/9)*216 = 96 outcomes. Now 6 of these outcomes will have all three dice the same, so subtract 6 from 96 = 90. There is a 90/216 = 5/12 or 41.67% chance that exactly two dice are the same.
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
The probability is 1 and you do not need Matlab to get that answer - only a little bit of thought.
If we are thinking of getting a '6', here are the odds. Wth one dice, its 1 in 6. So,with two dice its 1 in 216 with three dice its 1 in 7776 with four dice its 1 in 279936 with five dice its a huge 1 in 10077696
The probability of flipping three heads when flipping three coins is 1 in 8, or 0.125. It does not matter if the coins are flipped sequentially or simultaneously, because they are independent events.
the chances of rolling doubles once is 1 in 6; 3 times in a row it is 1 in 216. It does not make any difference after how many times you rolled the dice before.
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
If the number cubes are standard dice cubes, the odds of rolling 3 ones is 1 in 216.
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)^3 since each event is independent and each has a probability of 1/6.
50% chance
Possible outcomes of one roll = 6.Probability of an even number on one roll = 3/6 = 0.5 .Probability of an even number on the second roll = 0.5 .Probability of an even number on the third roll = 0.5 .Probability of an even number on all three rolls = (0.5 x 0.5 x 0.5) = 0.125 = 1/8The probability of at least one odd number is the probability of not gettingan even number on all 3 rolls. That's (1 - 1/8) = 7/8 or 0.875 or 87.5% .
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.
Three out of twelve
The probability is 1/36
If you mean rolling at least one three, the 1/3. If you mean exactly one three, then 5/18. and If you mean a sum of 3 for the two throws, the answer is 1/18
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