# Thread: Roll a Pair of six-sided dice 5 times and getting four 4's?

1. ## Roll a Pair of six-sided dice 5 times and getting four 4's?

What is the probability of rolling a pair of six-sided dice five times and getting four 4's?

@Plato this was the other questions I am really struggling with.

This is what I came up with, but I'm not sure if it is correct.

p(x=4) = C(5,4)
× (1/36)^4 × (35/36)^5-4

which equals 2.894180045
×10^-0.6 ....That looks wrong to me..?

2. ## Re: Roll a Pair of six-sided dice 5 times and getting four 4's?

P(4) = 7/36

5choose 4 ((7/36)^4)(29/36) = .0057

so about one in 17543 attempts

3. ## Re: Roll a Pair of six-sided dice 5 times and getting four 4's?

Originally Posted by tdotodot
What is the probability of rolling a pair of six-sided dice five times and getting four 4's?

When you say 'getting four 4's' do you mean a sum of four or do you mean four spots on one die?

4. ## Re: Roll a Pair of six-sided dice 5 times and getting four 4's?

Originally Posted by Plato
When you say 'getting four 4's' do you mean a sum of four or do you mean four spots on one die?
^That's how to question was worded. I think it means sum of four. (eg: 2 dots on one die and 2 dots on the other, or 3 dots and 1dot) So the Question is asking the probability of getting the sum of four, 4 times.

5. ## Re: Roll a Pair of six-sided dice 5 times and getting four 4's?

Originally Posted by tdotodot
^That's how to question was worded. I think it means sum of four. (eg: 2 dots on one die and 2 dots on the other, or 3 dots and 1dot) So the Question is asking the probability of getting the sum of four, 4 times.

That would be the way I would expect to read it.
If you roll a pair of dice there are three ways to get a sum of four.

So in five rolls getting a sum of four four times:
$\displaystyle \binom{5}{4}\left(\frac{3}{36}\right)^4\left(\frac {33}{36}\right)$.

6. ## Re: Roll a Pair of six-sided dice 5 times and getting four 4's?

Originally Posted by Plato
That would be the way I would expect to read it.
If you roll a pair of dice there are three ways to get a sum of four.

So in five rolls getting a sum of four four times:
$\displaystyle \binom{5}{4}\left(\frac{3}{36}\right)^4\left(\frac {33}{36}\right)$.
Could I rewrite that as:

p(x=4) = C(5,4) × (3/36)^4 × (33/36)
p(x=4) = 2.210326646×10^-04

My calculator is outputting a weird number?

OR is it NOT C(5,4) And just (5/4), which gives me 5.525816615×10^-05

___Compared to what my original post shows, I was only wrong with the probability of getting a 4. I had 1/36 and 35/36 WhereAs you have 3/36 and 33/36.

Once again -your a life saver man!

7. ## Re: Roll a Pair of six-sided dice 5 times and getting four 4's?

Originally Posted by tdotodot
Could I rewrite that as:

p(x=4) = C(5,4) × (3/36)^4 × (33/36)
p(x=4) = 2.210326646×10^-04

My calculator is outputting a weird number?

$\displaystyle \binom{5}{4}\left(\frac{3}{36}\right)^4\left(\frac {33}{36}\right)=0.000221032664609$

8. ## Re: Roll a Pair of six-sided dice 5 times and getting four 4's?

Thanks, that actually looks right, unlike the number I keep getting on my calculator?!?!?! *Don't Get WHY* I tried using an online calculator, input everything exactly the same and got 0.000221032664609.... Guess there's something wrong with my calculator! JUST VERY GLAD I finished those two questions!

0.000221032664609. ONLINE CALCULATOR
2.210326646×10^-04 MY CALCULATOR
^It seems as though it is just multiplying it a few thousand times? Why?

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### ehat is the probability of getting four one in five time rollung a die

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