1. ## Probability computation questions

I want to compute a)The probability of getting at least one 3 in the rollings of 6 dice b) The probability of getting at least two '3's in the rollings of 12 dice. c)The probability of getting at least three '3's in the rollings of 18 dice.

Solution:- I computed the probability for a)=$\frac{3906}{46656}$

b)$\frac{\frac{1(5^{11}-1)}{4}}{6^{12}}$

c)$\frac{\frac{1(5^{16}-1)}{4}}{6^{18}}$

2. ## Re: Probability computation questions

this is just a binomial distribution with $n=6,~12,~18,~p=\dfrac 1 6$

$P[n,k] = \dbinom{n}{k}p^k (1-p)^{n-k}$

a) $P[\text{at least one 3 in 6 rolls}] = 1 - P[\text{no 3s}] = 1 - P[6,0] = \dfrac{31031}{46656}$

b) $P[\text{at least 2 3s in 12 rolls}] =1 - P[12,0]-P[12,1] = \dfrac{1346704211}{2176782336}$

c) $1-P[18,0]-P[18,1]-P[18,2] = \dfrac{15166600495229}{25389989167104}$

3. ## Re: Probability computation questions

In general, the probability of getting at least $n$ 3's in $6n$ rolls is:

$$\dfrac{5^{5n}}{6^{6n}}\dbinom{6n}{n}{_2F_1}\left (1,-5n;n+1;-\dfrac{1}{5}\right)$$

And the limit as $n$ approaches infinity that at least a sixth of the rolls will be 3 turns out to be 0.5.

4. ## Re: Probability computation questions

Originally Posted by SlipEternal
In general, the probability of getting at least $n$ 3's in $6n$ rolls is:

$$\dfrac{5^{5n}}{6^{6n}}\dbinom{6n}{n}{_2F_1}\left (1,-5n;n+1;-\dfrac{1}{5}\right)$$

And the limit as $n$ approaches infinity that at least a sixth of the rolls will be 3 turns out to be 0.5.
Hello,
I didn't understand what do you mean to say? What is ${_2F_1}\bigg(1,-5n;n+1;-\frac15\bigg)$. What is the meaning of at least a sixth of rolls will be 3 turns out to be 0.5. Please explain.

5. ## Re: Probability computation questions

Originally Posted by Vinod
Hello,
I didn't understand what do you mean to say? What is ${_2F_1}\bigg(1,-5n;n+1;-\frac15\bigg)$. What is the meaning of at least a sixth of rolls will be 3 turns out to be 0.5. Please explain.
That is one of the Hypergeometric Functions.

If you roll six dice and get at least one 3, then at least a sixth of the dice rolled a 3.
If you roll 12 dice and get at least two 3's, then at least a sixth of the dice rolled a 3.
If you roll 18 dice and get at least three 3's, then at least a sixth of the dice rolled a 3.

6. ## Re: Probability computation questions

Originally Posted by Vinod
Hello,
I didn't understand what do you mean to say? What is ${_2F_1}\bigg(1,-5n;n+1;-\frac15\bigg)$. What is the meaning of at least a sixth of rolls will be 3 turns out to be 0.5. Please explain.
I'm going to guess he's talking about the Hypergeometric Function