"A" is singular, "dice" is plural. One "dice" is a "die".
Tossing one six sided die means the odds of the number 1 displaying is 1 out of 6.
Tossing two six sided dice means the odds of the number 1 displaying is zero, as at a minimum it will be 1 and 1, or 2.
you have one twelfth chance
1/3
With fair regular dice, the answer is 4/36 = 1/9
The probability of both dice showing the same number is 1/6 and the probability of different numbers is 5/6.
Two dice can come up in any one of 36 ways.You can roll a five with:1 + 44 + 12 + 33 + 2That's 4 ways out of the 36 possibilities.The probability of a 5 is (4/36) = (1/9) = 0.1111... ≈ 0.111 = 11.1%The probability of rolling a 5 on the second toss is also ≈ 0.111 = 11.1%Honest dice have no memory; they don't know how the first toss came up.The probability of getting a 5 on both the first and second rolls is (1/9)2 = (1/81) = 0.0123456789... ≈ 0.012 ≈ 1.2% (rounded)
1/3
you have one twelfth chance
With fair regular dice, the answer is 4/36 = 1/9
The probability of the event is 25/36.
The probability of both dice showing the same number is 1/6 and the probability of different numbers is 5/6.
It's has a 1/6 chance of being a 6.
It is 7/12.
Two dice can come up in any one of 36 ways.You can roll a five with:1 + 44 + 12 + 33 + 2That's 4 ways out of the 36 possibilities.The probability of a 5 is (4/36) = (1/9) = 0.1111... ≈ 0.111 = 11.1%The probability of rolling a 5 on the second toss is also ≈ 0.111 = 11.1%Honest dice have no memory; they don't know how the first toss came up.The probability of getting a 5 on both the first and second rolls is (1/9)2 = (1/81) = 0.0123456789... ≈ 0.012 ≈ 1.2% (rounded)
One in six
If you toss the coins once only, it is 1/4.
The probability that when you toss a dye two times (or toss two dice) both times thedye shows the same number (or the two dice roll doubles) is calculated as follows.First dye toss roll any outcome (1, 2, 3, 4, 5 or 6). The probability the second tosscomes out the same as first toss is 1/6. So, for two dice to roll doubles:P(doubles) = 1/6Probability of complement event (not rolling doubles) is;P(not doubles) 5/6So the probability of first two dice trial 'not doubles' and following trial of the two dice'doubles is;P([N doubles]1 and [doubles]2) = 5/6 ∙ 1/6 = 5/36
The probability of getting 11 with one throw of 2 dice is 1/6*1/6*2 = 1/18 So the probability of not getting 11 with 1 throw of the dice is 17/18. Tossing the dice 54 times, the probability of not getting 11 54 times is (17/18)54 = 0.0456... So the probability of at least 1 roll of 11 is 1 - 0.0456 = 0.954