It is 1/3
land on a 3? as in a dice? well if so then it is 1/6 chance because there are 6 numbers on a dice and 3 is one of them so 1/6
you will land on 1 or 6 5/6 times
If a standard number cube or die is rolled, the probability that a 4 does not land face up is five out of six, or (six minus one) out of six.
The probability of getting heads on three tosses of a coin is 0.125. Each head has a probability of 0.5. Since the events are sequentially unrelated, simply raise 0.5 to the power of the number of tosses (3) and get 0.125, or 1 in 8.
Each cube can land in 6 different ways.Two cubes can land in (6 x 6) = 36 different ways.There are 3 different ways of rolling a ten:4 - 65 - 56 - 4The probability of rolling a ten is (3 / 36) = 1/12 = 8 and 1/3 percent
If using regular dice, the probability is 0 since the minimum sum from four dice is 4.
1 out of 6
You cannot roll "a dice" because it is one die, many dice. If you roll an ordinary, 6 faced die, the probability that it will land on 1 is 1/6.
You roll it many times. The probability that it lands on a six is the number of times that it lands on a six divided by the number of times the die has been rolled.
If it lands on a six 140 times then the estimated probability of a six is 140/400 = 0.35
Answer 1:The odds are very easy to calculate. Simply divide the number of "valid" rolls against all possible rolls. For ease, you can write down all possible combination for the 2 dice.1-1; 1-2; 1-3; 1-4....and so on, remember 1-4 and 4-1 are different rollsThere are 36 unique possible combination, and 6 of them are doubles, so that's 6/36 chances (and since 6 goes into 36, 6 times, this reduces to 1/6) or about 17%Answer 2:Another way to look at this problem, generically, is to assume we have an 'n' face dice. In most cases, dice have 6 faces (1, 2, 3, 4, 5, 6). But why not create a solution that works for any number of sides? Well, if we are trying to calculate the probability of rolling two dice (dice-1 and dice-2) of 'n' sides at the same time and having them turn up as doubles, only one of the dice really matters. Here's why. Dice-1 is guaranteed to land on a number 1-n. This will happen every time (on a fair dice, disregarding freak incidents). What we are trying to calculate is the probability that dice-2 will land on the SAME number as dice-1. Dice-2 can only land on one of 'n' values: 1, 2, 3, 4, 5, 6, ... , 'n'. For you non math folks, this just means it must land on a number from 1 to 'n' where 'n' is the number of sides on your dice. Out of all of the sides that dice-2 can PHYSICALLY land on, one of the sides MUST necessarily have the same as the value that dice-1 landed on. That is to say, if dice-1 landed on the value 3, there must be some chance that dice-2 will also land on the value 3. The probability of this occurring on a fair die is 1 divided by the total number of possible outcomes, which would be 'n'. So, really, there is a 1/n chance that dice-2 will land on the same number as dice-1. Thus, our probability for rolling doubles is simple 1/n. For our 6 sided dice example, our dice-1 lands on some value between 1 and 6 and there is a 1/6 chance dice-2 will match it.
land on a 3? as in a dice? well if so then it is 1/6 chance because there are 6 numbers on a dice and 3 is one of them so 1/6
The probability of getting 11 is 2/36 = 1/18. This is because of the 36 ways the dice might land, two of them sum to 11, namely 5 + 6 and 6 + 5.
you will land on 1 or 6 5/6 times
The probability of landing on any 1 given number is 1:6. Specific numbers don't matter. It is all odds. Just multiply by two so that we get a 2:6 probability that you will land on a 1 or 2. this reduces to 1:3 or 33.3%
1/8
There are 2 possible answers. If you meant is either one of the dice land on a 1, then the answer is 1/3. If you meant if they both ADD UP to 1, then it is impossible and the answer is 0. =========================================== The first answer accurately states the theoretical probabilities. But the question clearly specifies "the experimental probability". That number is the result of an experiment which, as far as we know, has not yet been perfrormed. Also, in the parlance of frequent rollers, "rolling a (x)" means a roll of two dice after which the sum of the spots showing on the top surfaces of both is (x).