I am currently studying Statistical Inference class on Coursera. In one of the assignments, the following question comes up.

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| Suppose you rolled the fair die twice. What is the probability of rolling the same number two times in a row?1: 2/62: 1/363: 04: 1/6Selection: 2| You"re close...I can feel it! Try it again.| Since we don"t care what the outcome of the first roll is, its probability is 1. The second roll of the dice has to match the outcome of the first, so that has a probability of 1/6. The probability of both events occurring is 1 * 1/6.I do not understand this bit. I understand that the two die rolls are independent events and their probabilities can be multiplied, so the outcome should be 1/36.

Can you please explain, why I am wrong?

The probability of rolling a specific number twice in a row is indeed 1/36, because you have a 1/6 chance of getting that number on each of two rolls (1/6 x 1/6).

The probability of rolling any number twice in a row is 1/6, because there are six ways to roll a specific number twice in a row (6 x 1/36). Another way to think about it is that you don"t care what the first number is, you just need the second number to match it (with probability 1/6 ).

To make it perfectly clear, consider the sample space for rolling a die twice.

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

There are 36 equally likely possible outcomes, 6 of which define the event "rolling the same number two times in a row". Then, the probability of this event occurring is $frac636$, which is equal to $frac16$.

Conceptually, this is just asking "what are the chances a second die matches the result of the first". Suppose I rolled a die, secretly, and asked you to match the outcome with your own roll.

No matter which number I rolled, there is a 1/6 chance that your die matches my roll, as there is a 1/6 chance any die roll comes up any specific number.

If you roll a 1 then on the second roll (for a fair 6 sided die) the probability that the second roll is a 1 is 1/6 (assuming independence. This would be true for any other possible first roll.

i would look at it as a combination problem . where you are asked what possible combinations are thre that have same numbers on first and second roll. combinations are 6 (11,22,33,44,55,66) from a total possibilities 6*6=36so probability is 6/36

Hope this helps :

Probability for the first roll to turn up as 1 : 1/6Probability for the second roll also to turn up as 1 : 1/6

Thus , probability that the first two rolls turn up as 1 is (1/6*1/6) = 1/36

Now the probability that the first two rolls turn up as 2 is (1/6*1/6) = 1/36.... Same applies for 3,4,5,6

So the probability for any number to turn up consecutively twice is (1/36+1/36+1/36+1/36+1/36+1/36) = (6/36) = 1/6

Since I didn"t see this exact way of framing it above:

For your first roll there are 6 possible answers, and 6 acceptable answers (as any number 1-6 is acceptable).

6/6

For the second roll there are 6 possible answers, but now only 1 will match the first roll.

1/6

6/6 * 1/6 = 1/6

I guess you are confused because it did not mention an exact number. If the question said something like rolling a die twice and getting a 1 twice then you are correct but since it did not say so, the same is true for getting a 1,2 3 or any other number. Thus since there are 6 numbers that can turn up that way, it will be 6/36 which will give you 1/6.

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