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For this set of data use the formula nCr(p^r)[q^(n-r)]
p = P(a number) = 1/6
q = P(not that number) = 5/6
n = 3 trials
r = 3 successes
n - r = 0 failures
P(the same number in 3 trials) = nCr(p^r)[q^(n-r)]
Substitute what you know into the formula:
3C3[(1/6)^3][(5/6)^0] = 1 x 1/216 x 1 = 1/216 = 0.0046 approximately.

Or shortly we can say :
Since each number has the same probability, 1/6, to appear in this experiment, then the probability of the three dices to show the same number is 1/6 x 1/6 x 1/6 = 1/ 216

The chances are 1 in 36. It doesn't matter what the first one is. There is a 1 in 6 chance of the second one being the same, and same with the third one. 1/6 x 1/6 = 1/36. I have no idea what the rubbish above is all about, but its complete &^%*

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Q: You have 3 fair dice each numbered 1 to 6 what is the chance all three dice show the same number?
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Continue Learning about Algebra

If a dice has the numbers 1 2 2 3 4 and 5 how many times would the number 2 come up?

A dice has six sides, so it has a chance of one-in-six of landing on one particular side. Therefore, one with two 2's would have a chance of one in three of coming up.


What are the odds of rolling a small straight 12345 with five dice?

With 5 dice, if the first die must equal 1, the second must equal 2, and so on, then each die has a one in six chance of being the correct number. So the chance that all five of them will be correct is (1/6)5 = 1/7776 But, that's assuming that the order of the numbers must be 12345. The dice could also fall as 12453, and it would be considered a small straight. So how many ways can you get 1, 2, 3, 4 and 5? The 1 could be in any one of 5 places. Then the 2 could be in any one of the remaining 4, and so on. There are 5 x 4 x 3... ways of arranging 12345. This is also written as 5! (five factorial). 5!=120 So now, we have 120 different ways of getting a straight. Each straight, as shown at the top, has a one in 7776 chance. Therefore, (1/7776) x 120 possibilities = probability of a small straight. (1/7776) x 120 = 120/7776 = 5/324 So there is a 5/324 probability, or about one out of 65, or 1.5% chance of getting the numbers 1, 2, 3, 4, and 5 on a single roll of 5 dice.


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25 percent


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It would have 72 total sides, or faces. 6 per die, 12 dice, 12 x 6 = 72.


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The chance is 10/216 = 5/108


What is the probability of tossing two number cubes with faces numbered 1 though 6 and getting a number greater than 2 on one cube and a number less than 2 on the other?

If you were throwing one numbered cube (dice) with each face numbered 1 to 6, then the chance of getting a number greater than two is 4/6 (or 2/3) since there are 4 numbers greater than 2, and six numbers in total.Similarly, if you were throwing one dice, the chance of getting a number less than 2 is 1/6.To determine the probability of getting both of these outcomes on two dice (greater than 2 on one dice and less than 2 on the other) we simply multiply these two values together.2/3 x 1/6 = 2/18 = 1/9.So there is a one in nine chance of this outcome occuring.


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