1. ## Dice probability

Can anyone help me get started on this problem? Any help would be appreciated:

A die has one pink side, two red sides, and three green sides.

If the die is rolled twice, what's the probability that the two shapes on top are the same?

How many times do you need to roll the die so that the probability of seeing a pink side at least twice is greater than .5? Red? Green?

Thanks

2. ## Re: Dice probability

it's probability that the two shapes are red+ that they are green + that they are pink. so (1/6)²+(1/3)²+(1/2)²

you mean 0.5 ? the probability of seeing a pink side when you roll the die n times at least twice is 1-(probability of seeing 1 or 0 pink side when you roll the die n times).
probability of seeing 1 pink side when you roll the die n times is 6*(5/6)^(n-1) (Binomial Distribution, or just basic combinatorics).
probability of seeing 0 pink side when you roll the die n times is (5/6)^n (you have 5/6 chances to not have a pink side at each roll)
so you study the function to see when it becomes greater than 0.5

3. ## Re: Dice probability

Originally Posted by colerelm1
How many times do you need to roll the die so that the probability of seeing a pink side at least twice is greater than .5? Red? Green?
Why wouldn't this just be:

Pink: (1/6) + (1/6) + (1/6) + (1/6) = 4/6 > 0.5 -----> 4 times
Red: (2/6) + (2/6) = 4/6 > 0.5 ------> 2 times
Green: (3/6) + (3/6) > 0.5 -----> 2 times

4. ## Re: Dice probability

If you have 1/6 chances to see a pink side at each roll, the probability of seeing a pink side in two rolls is not 1/6+1/6.

btw, there is an error in my message, replace this "probability of seeing 1 pink side when you roll the die n times is 6*(5/6)^(n-1) (Binomial Distribution, or just basic combinatorics)". by this
probability of seeing 1 pink side when you roll the die n times is (5/6)^(n-1) (Binomial Distribution, or just basic combinatorics).