6/5, 6/4, 6/3, 6/2, 6/1, 5/4, 5/3, 5/2, 5/1, 4/3, 4/2, 4/1, 3/2, 3/1, 2/1.
This represents 15 of 36 possibilities, the odds are 7/5 against.
The first dice can show any number. However the second dice has a 1 in 6 chance of being the same as the first. Hence the probability of getting two numbers the same is 1/6.
Assuming that a dice roll is purely random, there is a 1 in 6 probability of landing on any number. Since the second roll depends on the probability of the first, they factor together: First Roll: 1/6 Second Roll: (1/6)(1/6) = 1/36 If you were rolling both dice at once, however, the math would be completely different.
This depends on if you want at least two of the dice to be the same number, or exactly two of the dice to be the same number.For the first scenario: Roll the first die, and get a number. Roll the second die, and there is 1/6 chance that it'll be the same as the first one. Now if it's not the same (5/6 chance) then the third die has 1/6 chance of being the same as the first, and 1/6 chance of being the same as the second. So we have:1/6 + 5/6*(1/6 + 1/6) = [simplified] 4/9 or about 44.44%chance that at least two are the same.For the second scenario: With three dice, there are 216 possible outcomes (6 x 6 x 6). So we know that there is a 4/9 chance that 2 or more will be the same: (4/9)*216 = 96 outcomes. Now 6 of these outcomes will have all three dice the same, so subtract 6 from 96 = 90. There is a 90/216 = 5/12 or 41.67% chance that exactly two dice are the same.
Well this is very self explanatory? The first top and bottom number on a dice are 6 and 1. 6 + 1 = 7 If you do this for the next two it will be 5 and 2. 5 + 2= 7 If you notice you have gone down by 1 on the high number of the dice (6 to 5) and gone up by 1 on the low number of the dice (1 to 2).
We calculate the numerator of the desired probability: There are 6 ways any number can show up on the first die AND There is only 1 way that same number can show up on the second die So since "AND" means "MULTIPLY", there are 6x1 or 6 ways that both dice can come up the same. So the numerator of the probability is 6. We calculate the denominator of the desired probability: There are 6 ways any number can show up on the first die AND There are 6 ways any number can show up on the second die So since "AND" means "MULTIPLY", there are 6x6 or 36 ways the two dice can all come up any number. So the denominator of the probability is 36.
None, because you cannot have the first or second dice: it is the first die or second die. The probability is 1/6 * 1/2 = 1/12
The first dice can show any number. However the second dice has a 1 in 6 chance of being the same as the first. Hence the probability of getting two numbers the same is 1/6.
11 = 6+5 is the only solution, so there are two combinations first dice : 6, second dice : 5 first dice : 5, second dice : 6
A. P(A ans B) = 1/6 &' D. P(B) = 1/6
It is 1/12.
There are only three even numbers on a "legal" six sided dice (2,4 & 6), which ever dice is designated as "first" whether rolled independently or in unison will have a 1 in 2 chance of being even. Put another way: legal dice, designate first, 3 even, 3 odd, 50/50. Bonus Answer: Applies to second dice also.
We'll assume standard six-sided dice. The first die rolls... well, whatever it rolls. It doesn't really matter what the specific number is. The second die can come up any of six different ways. One of those ways is with the same number as the first, the other five are different. So the chances of rolling two dice and getting the same number on both is 1/6.
I'm going to assume you mean standard dice, all three distinguishable, and exactly two of the same number. First break the problem into smaller parts. First, determine the odds of getting the same number on the first two dice, and a different number on the third. Then just multiply this result by 3 and 216 to account for the first and last dice being the same, or the last two dice, then compute the total number of possibilities. P(two dice being the same and last being different) = P(second die being the same as the first) * P(third die being different than the other two) = (1/6)(5/6) = 5/36 3*216*5/36 = 90 possibilities out of 216
(probably but very unlikely) ----------------------------------- A better answer would be as follows. Throw the first dice. It does not matter which number turns up. Let's suppose it was a 2; Now there is only one 2 out of six different numbers when you throw the second dice. So the probability of scoring another 2 is 1 out of 6 = 1/6 Now for each of the results from the two dice which we have read there is only one 2 from the 6 possible numbers, which again means a probability of 1 out of six = 1/6 So the final probability is found by multiplying 1 (certainty) for the first dice by 1/6 for the second dice by 1/6 for the third dice, which = 1/36
The probability of throwing just one one with two dice can be calculated by considering the different ways it can occur. There are two ways to get one one: rolling a one on the first die and any number on the second die, or rolling any number on the first die and a one on the second die. There are a total of 36 possible outcomes when rolling two dice, so the probability is 2/36, which simplifies to 1/18.
A die had 12 edges. The number of edges that dice have will depend on the number of dice!
Assuming that a dice roll is purely random, there is a 1 in 6 probability of landing on any number. Since the second roll depends on the probability of the first, they factor together: First Roll: 1/6 Second Roll: (1/6)(1/6) = 1/36 If you were rolling both dice at once, however, the math would be completely different.