Probability of sum is 20 (4 dice are rolled)

1. Oct 20, 2016

jaus tail

1. The problem statement, all variables and given/known data
4 dice are rolled. Find probability that sum is 20.

2. Relevant equations
If a dice is rolled the outcome can be 1, 2, 3, 4, 5, 6

3. The attempt at a solution
Well the combinations for sum to be 20 are:
5, 5, 5, 5 = 20
5, 5, 6, 4 = 20
6, 6, 5, 3 = 20
6, 6, 4, 4 = 20
6, 6, 6, 2 = 20

Total outcomes is 6 * 6 * 6 * 6
I don't know how to calculate number of favorable outcomes.

I mean with combination 5, 5, 5, 5 only 1 arrangement is possible that is all dice is 5.
How to proceed with 5, 5, 6, 4 and other combinations?

Thanks in advance.

2. Oct 20, 2016

Staff: Mentor

Look at the various possible dice combinations for each of the five you listed above. For four 5's, there's only one way it can occur.
For the last combination, three 6's and one 2, you could have any of these:
2, 6, 6, 6
6, 2, 6, 6
6, 6, 2, 6
6, 6, 6, 2
IOW, the 2 could appear on the first die, the second die, the third die, or the fourth die.

The second and third combinations are slightly harder, as you have to look at the possible permutations of three symbols. That is, you need to count the ways that, for example, the numbers 5, 5, 6, 4, can be rearranged.

As it turns out, this number is called the multinomial coefficient, which is also discussed in the wiki article in the link above.
$\binom{4!}{2!, 1!, 1!}$

3. Oct 20, 2016

.Scott

5, 5, 5, 5 = 20
5, 5, 6, 4 = 20
6, 6, 5, 3 = 20
6, 6, 4, 4 = 20
6, 6, 6, 2 = 20

Not all combinations are created equal.
5,5,5,5 can only happen when all are 5.
6,6,6,2 can happen 4 ways (4!/3!)
5,5,6,4 can happen 12 ways (4!/2!)

Do you need more help or can you figure what 6,6,4,4 would be?

4. Oct 20, 2016

Ray Vickson

Have you looked at probability generating functions? They make such problems surprisingly easy.

Alternatively: since you are looking at an event {total = 20} near the maximum total of 24, it seems easier to look down from 24 instead of up from 4. So, each die can show 6-h, where h = 0,1,2,3,4,5 is the "hole". You want the probability that four independent holes add up to 4. That sounds a lot easier than the original problem.

Last edited: Oct 20, 2016
5. Oct 24, 2016

jaus tail

Thanks for the replies but I didn't understand how to solve the problem. Till now we've done sums like pick up balls from 1 bag to other. So you have 8C2, etc. Though I don't know why we use C and why not P at times.

Could you explain this formula?

How did you get the underlined part? The 4!/3!. I guess it's 4P3. Where did 3 come from?
And in next why is it 4P2?

Like P(24), then P(23), then P(22), then P(21), then P(20) like try to find a sequence?

Is it like a mirror image? P(24) = 1 = P(4).
P(23) = 4 = P(5)
Likewise P(22) = P(6)
P(21) = P(7)
P(20) = P(8) = 35/(6*6*6*^6)

Like combinations for sum = 8 are:
2, 2, 2, 2

2, 2, 3, 1
2, 2, 1, 3
2, 3, 1, 2
2, 1, 3, 2
2, 1, 2, 3
2, 3, 2, 1
1, 2, 3, 2
3, 2, 1, 2
1, 3, 2, 2
3, 1, 2, 2
1, 2, 2, 3
3, 2, 2, 1

4, 2, 1, 1
2, 4, 1, 1
4, 1, 1, 2
2, 1, 1, 4
1, 1, 4, 2
1, 1, 2, 4
4, 1, 2, 1
2, 1, 4, 1
1, 4, 1, 2
1, 2, 1, 4
1, 2, 4, 1
1, 4, 2, 1

3, 3, 1, 1
3, 1, 1, 3
1, 1, 3, 3
1, 3, 3, 1
3, 1, 3, 1
1, 3, 1, 3

1, 1, 1, 5
1, 1, 5, 1
1, 5, 1, 1
5, 1, 1, 1

Last edited: Oct 24, 2016
6. Oct 24, 2016

.Scott

The number of ways to order n unique items is n!.
If some of the n are not unique, then you need to start dividing.

Say you have a group that totals "n"; perhaps some are unique, but there are three subgroups numbering "p", "q", and "r" each with identical elements within its number.
In each of those three subgroups, the order of the elements does not matter. So instead of p! possible arrangements of the "p" subgroup, there is only one arrangement of those elements. So you will need to divide by p!. And so on for "r" and "s". The final result will be n!/(p!q!r!). For example, if we ignore suits, the number of ways or ordering a regular deck of 52 cards would be 52!/((4!)^13).

7. Oct 24, 2016

Staff: Mentor

It's explained in the permutations link I included in my earlier post.

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