Rolling Two 3s: Probability & Combinations

[new_at_math]

Say you roll a dice twice.You want to calculate the probability of getting both dice to land on 3. Using the formula for combinations: the total number of combinations is 21. so is the probability of getting a 3 on both dice 1 /21 ?

 

[pwsnafu]

The probability of rolling a three on a six sided die is $\frac16$. Doing it twice is $\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$

 

[new_at_math]

explain

I don't see the logic
how is it permuation formula(combination formula ).
What is wrong with my method?

 

[economicsnerd]

Whatever formula you have in mind, but there are 36 possible outcomes: $\{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
If each of these seems equally likely to you, then the answer is $\frac1{36}$.

 

[economicsnerd]

I'm guessing your method was to plug in "6 choose 2", which is the formula that tells you how many ways to pick a pair of people from a collection of 6 people. That doesn't describe the situation you named.

 

[tiny-tim]

hi new_at_math!

Say you roll a dice twice.You want to calculate the probability of getting both dice to land on 3. Using the formula for combinations: the total number of combinations is 21.

no, you're completely misunderstanding what combinations are for

21 is the number of different results you can get from two seven-sided dice if you're not allowed doubles …

12 13 14 15 16 17 23 24 25 26 27 34 35 36 37 45 46 47 56 57 67 …

start again: write out the possible combinations for a 3 (you did mean 3-total?)

 

[new_at_math]

I get it now it was a permutation with repetition;my bad.

 

[economicsnerd]

Say you roll a dice twice.You want to calculate the probability of getting both dice to land on 3.

I don't believe any formula with the word "permutation" or the word "combination" is an effective way to approach this problem.

 

[pwsnafu]

I get it now it was a permutation with repetition;my bad.

No. Permutation is a rearrangement of a collection of objects.
Your example is a Bernoulli trial.

 

[HallsofIvy]

Say you roll a dice twice.

You roll a die twice. (Or you roll two dice.) "Dice" is the plural of "die".

You want to calculate the probability of getting both dice to land on 3. Using the formula for combinations: the total number of combinations is 21. so is the probability of getting a 3 on both dice 1 /21 ?

Another way of looking at this is that if "A" and "B" are independent events, then the probability of "A and B" is the probability of A times the probability of B.

There are 6 faces on a die, one of which is a "3". As long as the faces are all equally likely to come up, the probability of a "3" is 1/6. The second roll of the die is independent of the first so the probability that both will come up "3" is (1/6)(1/6)= 1/36.