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 odds of rolling a single one on a 20-sided die is 1 in 20 (1/20). Since the rolls are independent events, the odds of rolling two ones on two 20-sided dice would be (1/20) * (1/20), which simplifies to 1 in 400 (1/400).
50%
On a single die it is 1 in 6.
The probability of rolling a 1 on a die is 1/6 if you roll it once.
one half.
Since there are six sides on a die, the odds of rolling a 4, or any digit for that matter is 1/6
0.5
if you rollid a die once the odds of getting less than four would be 3/6 or 50%.
The odds of rolling a single one on a 20-sided die is 1 in 20 (1/20). Since the rolls are independent events, the odds of rolling two ones on two 20-sided dice would be (1/20) * (1/20), which simplifies to 1 in 400 (1/400).
50%
On a single die it is 1 in 6.
The probability of rolling a 1 on a die is 1/6 if you roll it once.
The answer depends on what you are rolling. With an ordinary die, the answer is 1 (a certainty)..
5 to 1
one half.
One in Six
1 out of 6