It is 5/18.
one fourth
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
It is 0.9459
There is a 1/2 chance of rolling an odd number on either die, so 1/2 X 1/2 = 1/4. 1/4 is the probability that both die will have odd numbers, so 3/4 is the probability that at least one is even.
Depends on what you are rolling - a cubic die or a more exotic shape.
one fourth
With one roll of three dice, the probability is 7/8.
The probability of rolling a sum of 8 on one roll of a pair of dice is 5/36.The probability of not rolling a sum of 8 on one roll of a pair of dice is 31/36.The probability of rolling a sum of 8 twice on two rolls of a pair of dice is(5/36)(5/36) = (5/36)2 .The probability of rolling first a sum of 8 and then rolling a sum that is not 8 on thesecond roll is (5/36)(31/36).The probability of rolling a sum that is not 8 on the first roll and rolling a sum of 8in the second roll is (31/36)(5/36).So The probability of rolling a sum of 8 at least one of two rolls of a pair of dice is(5/36)2 + (5/36)(31/38) + (31/36)(5/36) = 0.258487654... ≈ 25.8%.
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.
00
The probability of rolling at least one 2 when rolling a die 12 times is about 0.8878. Simply raise the probability of not rolling a 2 (5 in 6, or about 0.8333) to the 12th power, getting about 0.1122, and subtract from 1.
The probability of rolling at least one '1' with six dice is 66.5% [1-(5/6)^6]*100%
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
It is 0.9459
The probability of rolling a 4 in a die is 1 in 6, or about 0.1667. The probability, then, of rolling a 4 in at least one of two dice rolls is twice that, or 2 in 6, or 0.3333. The probability of rolling a sum of 4 in two dice is 3 in 36, or 1 in 18, or about 0.05556.
The probability of NOT rolling a 3 with one die is 5/6 so the probability of NOT rolling a 3 with a roll of two dice is 25/36. The probability of rolling at least one 3 is 1–25/36=11/36, a bit less than 1/3.
There is a 1/2 chance of rolling an odd number on either die, so 1/2 X 1/2 = 1/4. 1/4 is the probability that both die will have odd numbers, so 3/4 is the probability that at least one is even.