For each of the 4 faces on one die, each of the 4 faces on the other die can come up, so there are 4 x 4 = 16 possible outcomes.
Th dice have 6 sides and each side have one number from 1-6 so that there are 6 possible outcomes if you rolled the dice
6 sides on each dice. Each number on the sides is different, this represents the 1 out of all 6 amounts shown. So this is what you'll do for each dice... 1/6 + 1/6 = 2/6 = 1/3
There are 36 possible outcomes: 6 for each die.
50% chance
The question is underspecified since the answer depends on the numbers on the dice. If all the numbers on both the dice are the same, there is clearly only one outcome. If the dice have 4 different numbers, then there can be 16 different outcomes. If the numbers on each die are 1,2,3 and 4 (or any four numbers in arithmetic sequence) there will be 7 outcomes.
Th dice have 6 sides and each side have one number from 1-6 so that there are 6 possible outcomes if you rolled the dice
6 sides on each dice. Each number on the sides is different, this represents the 1 out of all 6 amounts shown. So this is what you'll do for each dice... 1/6 + 1/6 = 2/6 = 1/3
There are 36 possible outcomes: 6 for each die.
50% chance
The question is underspecified since the answer depends on the numbers on the dice. If all the numbers on both the dice are the same, there is clearly only one outcome. If the dice have 4 different numbers, then there can be 16 different outcomes. If the numbers on each die are 1,2,3 and 4 (or any four numbers in arithmetic sequence) there will be 7 outcomes.
You must notice that it is easier to find the probability that none of the dice are six. This is the complement of the answer. Pc= (55)/(65) = .4019 *the first die has 5 ways to not equal six, and so on for the remaining dice. [55] outcomes have no six. *the total number of outcomes gives each die 6 ways to fall. [65] outcomes overall. *Probability is the favorable outcome divided by the total number of outcomes. *Probability of an event is 1 minus the complement of the event. P = 1 - .4019 P = .5981 <<<<<<FINAL ANSWER
6 outcomes each roll, 3 rolls. 6*6*6 = 216.
6 outcomes for each die and so 6x6x6 outcomes for all three
I'm assuming your question is the same as this: "If 2 dice are rolled, what is the probability of not getting 1 on either die?" To answer this question, we need to look at what IS possible. If I'm 2 rolling normal, fair dice, then I have equal probability of getting each of the numbers 1-6 on either die. If I'm trying to NOT get 1, then I want to get any of the numbers 2-6 on both dice. This gives me 10 desired outcomes (5 numbers * 2 dice) out of 12 possible outcomes (6 numbers * 2 dice), so the probability is 10/12, which simplifies to 5/6.
The odds of rolling a word with the keyword "dice" on each side of a standard six-sided dice is 1 in 46,656.
If any numbers come up more often than any others over the long run, then the dice are loaded and the game is not honest. * * * * * That is true for the roll of a die, but not dice. With 2 dice, the probabilities of the numbers are as follows 2 or 12 : 1/36 each 3 or 11 : 2/36 or 1/18 each 4 or 10 : 3/36 or 1/12 each 5 or 9 : 4/36 or 1/9 each 6 or 8 : 5/36 each 7 : 6/36 or 1/6 So 7 is the most rolled number in the long run when 2 dice are rolled.
The game is rather simple and played with each player having 3 chips, and 3 dice that has "L", "C" and "R" on three sides and a dot on the others. If one rolls an "L" then a chip is given to the person on the left, if a "R" is rolled a chip must be passed to the right and if a "C" is rolled a chip goes to the center. The winner is the last player left with chips.