# Pair of dice question, P(at least 4 rolls)

• Oct 4th 2009, 06:16 PM
Flipz4226
Pair of dice question, P(at least 4 rolls)
Hi, I need some help with this...

Two dice are rolled until a 6 appears on at least one of the dice. What is the probability that it takes at least four rolls(of two dice) before the first 6 is obtained?

Is this a binomial process: X~B(4, 2/12)

P(X >= 4) = 1 - [f(0)+f(1)+f(2)+f(3)] ?????

Thanks
• Oct 4th 2009, 07:12 PM
novice
Quote:

Originally Posted by Flipz4226
Hi, I need some help with this...

Two dice are rolled until a 6 appears on at least one of the dice. What is the probability that it takes at least four rolls(of two dice) before the first 6 is obtained?

Is this a binomial process: X~B(4, 2/12)

P(X >= 4) = 1 - [f(0)+f(1)+f(2)+f(3)] ?????

Thanks

It is related to binomial process, but it's closer to the multiplication rule.

I assume you know how to find the probability, p, of get a 6 at each roll.
If you do, you should know the probability, q, of not getting a 6 at each, roll, where q=1-p.

Once you know q, you can get to what you want by think this way.

To hit a 6 at least by 4 rolls means, you will not get it in the first 3 rolls, your probability of not getting a 6 in first three is $q^3$.
Now, you should know the p.
• Oct 4th 2009, 07:23 PM
Flipz4226
Actually I think this may be a geometric distribution P(X=x) = (1-p)^x-1 * p . I got P(X >= 4) = 1 - [f(0) + f(1) + f(2)+ f(3)] = .57 Does this seem reasonable? Thanks
• Oct 4th 2009, 09:38 PM
novice
Quote:

Originally Posted by Flipz4226
Actually I think this may be a geometric distribution??

The process, you are speaking of , is known as Binomial Process. Most commonly known as Binomial Distribution or Bernoulli trial.

Your f(0)+f(1)+f(3) is an expression of the binomial expansion. Look carefully and compare the following:

Binomial expansion:

$\Sigma\left(\begin{array}{c}n\\k\end{array}\right)$ $p^k q^{n-k}$

With b(k;n,p) = b(0;3,p), your binomial expansion will become

$(q+p)^3=q^3 +3q^2p+3qp^2+p^3$

Binomial Theorem:
$\Sigma\left(\begin{array}{c}n\\r\end{array}\right)$ $a^{n-r} b^{r}$ which becomes

$(a+b)^3=a^3 +3a^2b+3ab^2+b^3$

In every lesson in binomial process, an instructor has an obligation to point out this similarity to help you understand how the Binomial Theorem was developed, but it does not mean you are to use it for this process of tossing dice for n trials. With n trials, you perhaps must go back to revisit you tree diagrams and stochastic method to remind yourself of the multiplication rule to convince yourself what I explained in my last post.

In short, the expression you used above will not lend you the correct result.

Try it all different ways from a small experiment to a larger ones and discover it yourself. You will learn ever more.

Good luck.
• Oct 4th 2009, 10:00 PM
Flipz4226
I see it's definitely not a binomial distribution now. But isn't a geometric distribution the number of trials until a success which is when one of the dice is a 6? And also is the probability of getting at least one 6 to appear on the two dice 2/12?
• Oct 5th 2009, 12:53 AM
mr fantastic
Quote:

Originally Posted by Flipz4226
I see it's definitely not a binomial distribution now. But isn't a geometric distribution the number of trials until a success which is when one of the dice is a 6? And also is the probability of getting at least one 6 to appear on the two dice 2/12?

It is geometric and p = 11/36 (number of favourable outcomes divided by total number of outcomes).