# Thread: Dice Problem

1. ## Dice Problem

B throws 12 dice and wins if he scores at least two aces.
Find the probability:

So I was wondering if anyone could confirm or deny that
P(A) = 1 - (5/6)^12 - ((5/6)^11)(1/6)

Thanks

2. Originally Posted by tbyou87
B throws 12 dice and wins if he scores at least two aces.
Find the probability:

So I was wondering if anyone could confirm or deny that
P(A) = 1 - (5/6)^12 - ((5/6)^11)(1/6)

Thanks
It should be: $1 - \left( \frac56 \right)^{12} - {{12}\choose 1}\left( \frac56 \right)^{11} \left( \frac16\right)$

3. i'm confused

what does "aces" mean when we're talking about dice?

4. It means rolling a one.
But why do you need to multiply by 12 in the expression?
Thanks

5. Originally Posted by tbyou87
It means rolling a one.
But why do you need to multiply by 12 in the expression?
Thanks
Because we are thinking of this probably as an ordered pair:
X X X X X X X X X X X X
Now you want the probably of getting exactly one 1.
There are twelve's locations for that 1. And each have (5/6)^11 (1/6) probability.

6. Originally Posted by tbyou87
It means rolling a one.
But why do you need to multiply by 12 in the expression?
Thanks
i hope you noticed that he used a binomial distribution method here. it might not have been what he thought about, but it's what he ended up using. the expression he has is 1 - Bin(12,0) - Bin(12,1), which is the probability of NOT rolling 0 aces or one ace, that is, at least two