A die is thrown but the nuber is not looked at. If thrown a second time, what is the probabilty?

Question

That the number is
a) the same
b)greater than the frist time?

please explain how you get the answer

Answer 1

a) 1/6
b) depends on the number

Answer 2

a) 1/6. This part is easy: whether you look at the first die or not, the second has a 1-in-6 chance of being any particular number, including whatever number is required to match the first die.

b) This is more complicated. To solve it strictly, you'd have to consider all of the possible rolls of the first die, and the second, as independent events, and add up the combinations:

1/6 (first die is 1) * 5/6 (second die is 2 to 6) +
1/6 (first die is 2) * 4/6 (second die is 3 to 6) +
1/6 (first die is 3) * 3/6 (second die is 4 to 6) +
1/6 (first die is 4) * 2/6 (second die is 5 or 6) +
1/6 (first die is 5) * 1/6 (second die is 6) +
1/6 (first die is 6) * 0 (no values greater than 6)

that equals:
5/36 + 4/36 + 3/36 + 2/36 + 1/36 + 0 =
(5+4+3+2+1)/36 =
15/36 =
5/12

Note that you could also solve the first problem the same way as the second, but it takes longer:

1/6 (first die is 1) * 1/6 (second die is 1 also) +
1/6 (first die is 2) * 1/6 (second die is 2 also) +
1/6 (first die is 3) * 1/6 (second die is 3 also) +
1/6 (first die is 4) * 1/6 (second die is 4 also) +
1/6 (first die is 5) * 1/6 (second die is 5 also) +
1/6 (first die is 6) * 1/6 (second die is 6 also) =

1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 =
(1+1+1+1+1+1)/36 =
6/36 =
1/6

Answer 3

a) 1/6

I never took combinatorics or probability... so I am forced to guess at technique for solving this.

b) I would say that it is true that you cannot know for sure without knowing the first number. But, on average regardless of the first number, I would say that the equation would be:
1/6 * 5/6 + 1/6 * 4/6 + 1/6 * 3/6 + 1/6 * 2/6 + 1/6 * 1/6 + 1/6 * 0/6
= 5/36 + 4/36 + 3/36 + 2/36 + 1/36 = 15/36

That is the sum of the probabilities for each first die possibility followed by each second die greater value possibiltiy

Answer 4

The probability that you get one of any six numbers on a die is 1/6.

The probability that you get the number again is the same, 1/6.

The probability that you get 1 is 1/6, and the probability that you get a number greater than that the second time is 5/6. (2, 3, 4, 5, 6)

The probability that you get 2 is 1/6, and the probability that you get a number greater than that the second time is 4/6. (3, 4, 5, 6)

The probability that you get 3 is 1/6, and the probability that you get a number greater than that the second time is 3/6. (4, 5, 6)

The probability that you get 4 is 1/6, and the probability that you get a number greater than that the second time is 2/6. (5, 6)

The probability that you get 5 is 1/6, and the probability that you get a number greater than that the second time is 1/6. (6)

The probability that you get 6 is 1/6, and the probability that you get a number greater than that the second time is 0. You can only get 6 and below.

Answer 5

I'd look at it like this: there are 36 possible outcomes when you roll the die twice.

a) There are six outcomes where the die has the same number for the second roll as for the first : (1,1), (2,2), ... (6,6), so the probability for this one is 6/36 or 1/6.

b) If the first roll is a 1, there are 5 outcomes where the second roll is higher: (1,2), (1,3), (1,4), (1,5), (1,6).

You could show in the same way that for a first roll of 2, there are four such outcomes,
for a roll of 3 there are three such outcomes,
for a roll of 4 there are two,
for a roll of 5 there is one, and for a roll of 6 there are none.

Add up all the possibilities: 5+4+3+2+1=15, so the probability of getting a higher number on the second roll is 15/36.

This is pretty easy to see if you draw a table (rows and columns) of all the possible rolls.

Answer 6

Actually, this is easier than other posters have indicated. It is clear that the probability of getting a duplicate number on the second roll is 1/6, and it is also clear from symmetry that the probability of getting a larger number the second time is the same as getting a smaller number the second time. So the probability of a larger (or for that matter a smaller) result on the second roll is half of (1-1/6), or 1/2*5/6 = 5/12

Answer 7

The dice has six numbers on it - so if you throw it there is a one in six chance (16.666%) that the number that falls is a specified number - one you pick or one that has been thrown before.
Your second question's answer depends on the first number you threw. So take for instance if the first number you threw was a three, there is 4,5 and six on the dice greater than 3 so if you throw again you have three chances in 6 of throwing a greater number the second time, i.e. 50% chance
if you threw a 2, 4 numbers greater, a 4 in 6 chance (66.666%)... etc.

Answer 8

a) 1/6

b) Probability first die is 1 --- 1/6
Probability second die > 1 ---- 5/6

Probability first die is 2 --- 1/6
Probability second die > 2 ---- 4/6

Probability first die is 3 --- 1/6
Probability second die > 3 ---- 3/6

Probability first die is 4 --- 1/6
Probability second die > 4 ---- 2/6

Probability first die is 5 --- 1/6
Probability second die > 5 ---- 1/6

Probability first die is 6 --- 1/6
Probability second die > 6 ---- 0/6

Probability that second die is greater than the first
= (1/6)*(5/6) + (1/6)*(4/6) + (1/6)*(3/6) + (1/6)*(2/6) + (1/6)*(1/6) + (1/6)*(0/6)

= (1/6) * (5+4+3+2+1+0)/6
= (1/6) * (15/6)
= 15/36
= 5/12

Answer 9

for the first it's 1/6. For the second, it's half of the cases of the 5/6 remaining. So it's 5/12 if strictly greater, 7/12 otherwise.

Answer 10

a) 1/6. There has been one number rolled, and therefore there is only one chance that the number could be the same on the second roll.
b) cannot be determined with the information given.