Depends on what you are rolling - a cubic die or a more exotic shape.
Total number of possible rolls with 2 dice = 36.Total number of rolls that are doubles = 6.Probability of rolling doubles= 6/36 = 1/6 = (16 and 2/3) percent .
I'm going to assume you mean rolling the same number twice in a row in 25 rolls. The first won't cause a double, so you just need to consider the odds of rolling the same number as the last for the last 24 rolls. The easier approach is to realize that the probability of rolling at least one double is 1 minus the probability of rolling no doubles. One roll has this probability of not rolling the same as the last: P(different number from last) = 5/6 Since they are independent events: P(no doubles in 25 rolls) = (5/6)24 Now the final probability, of at least one double, is 1 - (5/6)24
The probability of not rolling it ever is 0.For n rolls it is (5/6)n sofor 10 rolls it is 0.1615for 20 rolls it is 2.608*10-2for 100 rolls it is 1.207*10-8 and so on.
If you keep rolling the die, then the probability of rolling a 6 and then a 1 on consecutive rolls is 1.The probability is 1/36 for the first two throws.
The answer depends on how many rolls. If there were n rolls, then the probability is n*(1/6)*(5/6)n-1/[1 - (5/6)n]
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% .
if a die is rolled.What is the probibil ty that an even number divisible by 3 appears
Total number of possible rolls with 2 dice = 36.Total number of rolls that are doubles = 6.Probability of rolling doubles= 6/36 = 1/6 = (16 and 2/3) percent .
With a fair die, it is 1/216 in three rolls, but the probability increases to 1 (a certainty) as the number of rolls is increased.
The probability of rolling a specific number on a fair six-sided dice is 1/6, as there are 6 equally likely outcomes. When rolling the dice 300 times, the probability of rolling that specific number on each roll remains 1/6, assuming the dice is fair and each roll is independent. Therefore, the probability of rolling that specific number at least once in 300 rolls can be calculated using the complement rule, which is 1 minus the probability of not rolling the specific number in all 300 rolls.
I'm going to assume you mean rolling the same number twice in a row in 25 rolls. The first won't cause a double, so you just need to consider the odds of rolling the same number as the last for the last 24 rolls. The easier approach is to realize that the probability of rolling at least one double is 1 minus the probability of rolling no doubles. One roll has this probability of not rolling the same as the last: P(different number from last) = 5/6 Since they are independent events: P(no doubles in 25 rolls) = (5/6)24 Now the final probability, of at least one double, is 1 - (5/6)24
if you multiply the number of rolls you did you will get the answer
The probability of not rolling it ever is 0.For n rolls it is (5/6)n sofor 10 rolls it is 0.1615for 20 rolls it is 2.608*10-2for 100 rolls it is 1.207*10-8 and so on.
If you keep rolling the die, then the probability of rolling a 6 and then a 1 on consecutive rolls is 1.The probability is 1/36 for the first two throws.
There are twenty six ways out of thirty six possible rolls of two dice to hit a number less than nine OR a seventy two percent probability.
If the die is rolled often enough, the probability is 1. With only two rolls of a fair die, the probability is 1/6.
The answer depends on how many rolls. If there were n rolls, then the probability is n*(1/6)*(5/6)n-1/[1 - (5/6)n]