Combination dices

[biostud]

I'm doing some calculations involving dice. Lets say I have 6 throws with one die and want to get the row 1-2-3-4. What are my chances of getting that?

If I had 4 throws it would be (1/6)^4

With 6 throws would it be (1/6^4 + 1/6 + 1/6)?

If I throw all 6 dices at the same time could I use the unordered with replacements formula?
n=6 r=4
(n-1+r)!/((n-1)!*r!)

That will give 126 combinations out of which 1-2-3-4 would be in 6*6 of those combinations so 36/126

 

[Soundmanred]

Dice
Die

 

[pete6032]

The plural of die is dice, not dices.

 

[biostud]

The plural of die is dice, not dices.

Thx for clearing up that...

 

[pete6032]

Thx for clearing up that...

Np I got you broseph.

 

[Ken g6]

With 6 throws would it be (1/6^4 + 1/6 + 1/6)?

That doesn't sound right. (1/6^4)*3 sounds closer. You need to get that row, but it can be in any of 3 places.

 

[Schmide]

This is combinations with repetition.

So there are n=6 things to choose from and we choose r=4

(n+r-1)C(r)

(n+r-1)C(r) = (6+4-1)! / 4!(6-1)! = 9! / 4! 5! = 126

 

[biostud]

This is combinations with repetition.

So there are n=6 things to choose from and we choose r=4

(n+r-1)C(r)

(n+r-1)C(r) = (6+4-1)! / 4!(6-1)! = 9! / 4! 5! = 126

So 126 combinations, is the probability the 1/126 or something else?

 

[Jaepheth]

There are 3 ways to get the desired sequence in a string of 6 rolls:
I. x-x-1-2-3-4
II. x-1-2-3-4-x
III. 1-2-3-4-x-x

Each x can be any outcome so there are 6^2 of each I through III. 36*3 = 108
6^6 total combinations (46,656) for a total probability of 108/46656 = 0.0023148148...

The probability of it happening at least once in 4 rounds is the same as 1 - (probability of it never happening) So 1-(1-.0023148148)^4 = approximately 0.0092271586
Six attempts = 1-(1-P)^6 = approx 0.013808761


Throwing 6 dice simultaneously:
The probability of any single number not showing up at all on 6 dice simultaneously thrown is (5/6)^6 = 0.3349
Probability of no 1s showing up = A
2s = B
3s = C
4s = D

For any two events x and y the probability of x OR y = P(x) + P(y) - P(x)*P(y)

A or B or C or D = ((A+B-A*B)+C-C*(A+B-A*B))+D-D*((A+B-A*B)+C-C*(A+B-A*B))

Since they're all equal we can just call them all A though
((A+A-A*A)+A-A*(A+A-A*A))+A-A*((A+A-A*A)+A-A*(A+A-A*A))
=((2A-A^2)+A-A*(2A-A^2))+A-A*((2A-A^2)+A-A*(2A-A^2))
=(2A-A^2+A-2A^2+A^3)+A-A*(2A-A^2+A-2A^2+A^3))
=2A-A^2+A-2A^2+A^3+A-2A^2+A^3-A^2+2A^3-A^4
=4A-6A^2+4A^3-A^4

=4*0.3349-6*0.3349^2+4*0.3349^3-0.3349^4
= 0.80431701
And that's the probability of not getting either a 1,2,3, or 4. So the probability of getting them all is
1 - 0.80431701
= 0.19568299

At least, I think I have that all right.

 

[Rakehellion]

dices

 

[DrPizza]

I do a lot of statistics with students related to dice.

What the heck is a "row"? This doesn't seem too clear to me, though it seems you mean a 1, then a 2, then a 3, then a 4 in that order.

If so, then for 4 rolls in a row, you're correct: 1/6 * 1/6 * 1/6 * 1/6

If you roll one die 6 times in a row, it's possible that have success with 1,2,3,4,x,x, or success with x,1,2,3,4,x, or success with x,x,1,2,3,4,

So, the probability is
1/6 * 1/6 * 1/6 * 1/6 * 6/6 * 6/6 plus
6/6 * 1/6 * 1/6 * 1/6 * 1/6 * 6/6 plus
6/6 * 6/6 * 1/6 * 1/6 * 1/6 * 1/6

= 3* (1/6)^4 Note, this equals 0.0023148148, the same answer Jaepheth has above, but by using the product of individual probabilities instead. Were it 7 rolls in a row, you can rapidly see that the probability would be 4*(1/6)^4. More than 7 rolls and the problem gets a little more complicated, since you get get 1,2,3,4 multiple times in, say, 100 rolls.

If I throw all 6 dices at the same time could I use the unordered with replacements formula?
n=6 r=4
(n-1+r)!/((n-1)!*r!)

Here is where your post confused the above part - this is a completely different problem, since the order doesn't count. I'm assuming that of the 6 dice that are rolled, you want at least one 1, at least one 2, at least one 3, and at least one 4. Is this correct?

 

[biostud]

Here is where your post confused the above part - this is a completely different problem, since the order doesn't count. I'm assuming that of the 6 dice that are rolled, you want at least one 1, at least one 2, at least one 3, and at least one 4. Is this correct?

Exactly

And btw, it was because I'm making some questions for 10th grade mathematics and this is already far too complicated.

That's the reason why I don't like combinatorics and probability. You can ask a very simple question, only to find out the mathematical solutions are extremely complex, and often very difficult to break down into smaller pieces. :/