# Thread: Probability with a dice

1. ## Probability with a dice

I throw a dice 6 times. What are the odds of getting higher than 4 on at least 5 throws?

Can I write this as: $\displaystyle \left(\frac{2}{6}\right)^5+\frac{2}{6}$ ?

In other words..The odds of getting a 5 or 6, 5 times in a row..Plus the probability of getting it in the 6th throw?

2. ## Re: Probability with a dice

So getting higher than 4 on 5 or 6 throws? Yes, the probability of getting 5 or 6 on one through is 2/6= 1/3 so the probability of throwing 5 or 6 6 consecutive times is $\displaystyle (1/3)^6$. The probabilty of get 5 or 6 exactly 5 times is $\displaystyle 6(1/3)^5(2/3)$. The probabilty of either of those happening is the sum, $\displaystyle (1/3)^6+ 6(1/3)^5$.

Notice the "6" in that? That is because there are 6 different orders: YYYYYN, YYYYNY, YYYNYY, YYNYYY, YNYYYY, and NYYYY where "Y" means a 5 or 6 and "N" means not a 5 or 6.

(And, by the way, the word "dices" is a verb meaning "cuts something into small cubes". The word you want is "dice" which is itself the plural of the word "die".)

3. ## Re: Probability with a dice

Hi:

You want to use the binomial distribution for this problem. Are you familiar with that distribution?

Howard Heller
InteractiveMathTutor.com

4. ## Re: Probability with a dice

Originally Posted by Paze
I throw a dice 6 times. What are the odds of getting higher than 4 on at least 5 throws?
Can I write this as: $\displaystyle \left(\frac{2}{6}\right)^5+\frac{2}{6}$ ?
In other words..The odds of getting a 5 or 6, 5 times in a row..Plus the probability of getting it in the 6th throw?
I think that there is a language(translation) difficulty here.

First I do not like the word odds and never use it. So I change it to probability.

Thus "What are the probability of getting higher than 4 on at least 5 throws?"
That means "getting a five or six at least five times out of six throws".

If you agree that is what it means, then the answer is:
$\displaystyle \binom{6}{5}\left( {\frac{2}{6}} \right)^5}\left( {\frac{4}{6}} \right) + {\left( {\frac{2}{6}} \right)^6$

5. ## Re: Probability with a dice

That is correct now.

Howard Heller
InteractiveMathTutor.com

6. ## Re: Probability with a dice

Originally Posted by IMTinstructor
That is correct now.

Howard Heller
InteractiveMathTutor.com
Someone of your low IQ would not know if that is correct or not.
Anyone stupid enough to fall for you scam deserves the outcome.
I can find no entry for you at MathGenealogy Project

So what equips you to tutor?

7. ## Re: Probability with a dice

Math major, math expert, actuarial science major, MBA

8. ## Re: Probability with a dice

Originally Posted by IMTinstructor
Math major, math expert, actuarial science major, MBA
Oh my goodness. Do you really think that has anything to do with mathematics?
What does "Math major" mean? "actuarial science major" has very little to do with mathematics.

I live in a state in which it is illegal to call oneself a "blank" if one does not have a PhD in "blank".
I can thank my brother for that, "educational psychologist is not a psychologist".

Do you have a PhD?

9. ## Re: Probability with a dice

Thanks guys. This helped me understand my problem! On to the next...

10. ## Re: Probability with a dice

Originally Posted by Plato
I think that there is a language(translation) difficulty here.

First I do not like the word odds and never use it. So I change it to probability.

Thus "What are the probability of getting higher than 4 on at least 5 throws?"
That means "getting a five or six at least five times out of six throws".

If you agree that is what it means, then the answer is:
$\displaystyle \binom{6}{5}\left( {\frac{2}{6}} \right)^5}\left( {\frac{4}{6}} \right) + {\left( {\frac{2}{6}} \right)^6$
Or wait, hold on. Your answers differs from Halls's answer, doesn't it? I got the same answer as HallsOfIvy. Where does the 4/6 come from?