Probability sum of 4 dice rolling greater than

[Aaron7]

Homework Statement

What is the probability of 4 independent dice summing to 20 or more?

Homework Equations

N/A

The Attempt at a Solution

I am not too sure how to approach this in an exam.
I have attempted to do it this way but it would be easy to miss out a term:

(I have put together numbers so that a number to the right is less than the number to its left.)

6666
6665 6664 6663 6662
6655 6654 6653
6644
6555 6554
5555

Then working out the combinations for each one:
eg. 6664 => 4C1
or 6654 => 6 first then 3C2 ways, 5 first then 3! way, 4 first then 3 ways => 12 ways
(not too sure if best method to do above calculation)

To get p + 4p + 4p +... + p = 70p where p = 1/6⁴ = 1/1296
so I get the answer to be 35/648.

Is this correct and is there a better method/thought process for when I am in an exam?

Many thanks.

 

[Joffan]

I think you probably do need to identify all the possibilities for achieving a score of 20 or more, so that step is tricky but necessary.

Arrangements of a group size $n$ with $n_a$ repeats of element $a$, $n_b$ repeats of element $b$, etc:
$$\frac{n!}{n_a! n_b! n_c! ...}$$

so there are $\frac{4!}{2!1!1!} = \frac{24}{2} = 12$ arrangements of 6654, for example.

 

[RoshanBBQ]

>> x = ones(1,6)/6

x =

   0.166666666666667   0.166666666666667   0.166666666666667   0.166666666666667   0.166666666666667   0.166666666666667

>> fourSum = conv(x,conv(x,conv(x,x)))

fourSum =

  Columns 1 through 6

   0.000771604938272   0.003086419753086   0.007716049382716   0.015432098765432   0.027006172839506   0.043209876543210

  Columns 7 through 12

   0.061728395061728   0.080246913580247   0.096450617283951   0.108024691358025   0.112654320987654   0.108024691358025

  Columns 13 through 18

   0.096450617283951   0.080246913580247   0.061728395061728   0.043209876543210   0.027006172839506   0.015432098765432

  Columns 19 through 21

   0.007716049382716   0.003086419753086   0.000771604938272

>> sum(fourSum((20-3):end))

ans =

   0.054012345679012

>> 35/648

ans =

   0.054012345679012

I can at least confirm you got the right answer.

 

[Aaron7]

Thanks for both of your help.

 

[Ray Vickson]

Homework Statement

What is the probability of 4 independent dice summing to 20 or more?

Homework Equations

N/A

The Attempt at a Solution

I am not too sure how to approach this in an exam.
I have attempted to do it this way but it would be easy to miss out a term:

(I have put together numbers so that a number to the right is less than the number to its left.)

6666
6665 6664 6663 6662
6655 6654 6653
6644
6555 6554
5555

Then working out the combinations for each one:
eg. 6664 => 4C1
or 6654 => 6 first then 3C2 ways, 5 first then 3! way, 4 first then 3 ways => 12 ways
(not too sure if best method to do above calculation)

To get p + 4p + 4p +... + p = 70p where p = 1/6⁴ = 1/1296
so I get the answer to be 35/648.

Is this correct and is there a better method/thought process for when I am in an exam?

Many thanks.

Homework Statement

Homework Equations

The Attempt at a Solution

In this problem the counting is a bit easier if we start by assuming each die gives 6 minus the number of 'holes', where each die has holes = 0,1,2,3,4,5 and these are independent and equally likely. Thus, we start at 24 (0 holes), then go to 23 (1 hole distributed among 4 dice), then 22 (2 holes distributed among 4 dice), etc, down to 20 (4 holes distributed among 4 dice).

Alternatively, we can consider the mgf M(z) = (1/6^4) * (z+z^2+z^3+z^4+z^5+z^6)^4 = (1/6^4)*z^4(1-z^6)^4/(1-z)^4. When we expand this as sum_{k=4..inf} c[k]*z^k we have c[k] = Pr{total = k}. Using a computer algebra system we can determine c[k] for k = 20,...,24 and sum them up. Again, though, it is easier to consider 'holes' instead of 'spots' and to get the mgf of the number of holes, then sum the results for k = 0 to k = 4 holes.

RGV