Probability of sum is 20 (4 dice are rolled)

In summary: Did you read it?Basically, it's the number of ways of arranging the four dice given that two are the same (5, 5) and the other two are different (6, 4).
  • #1
jaus tail
615
48

Homework Statement


4 dice are rolled. Find probability that sum is 20.

Homework Equations


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

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.
 
Physics news on Phys.org
  • #2
jaus tail said:

Homework Statement


4 dice are rolled. Find probability that sum is 20.

Homework Equations


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

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.
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!}##
jaus tail said:
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.
 
  • Like
Likes jaus tail
  • #3
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?
 
  • Like
Likes jaus tail
  • #4
jaus tail said:

Homework Statement


4 dice are rolled. Find probability that sum is 20.

Homework Equations


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

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.

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:
  • Like
Likes jaus tail
  • #5
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.

Mark44 said:
##\binom{4!}{2!, 1!, 1!}##
Could you explain this formula?

.Scott said:
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?

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?

Ray Vickson said:
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.

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:
  • #6
jaus tail said:
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?
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).
 
  • Like
Likes jaus tail
  • #7
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!}##
jaus tail said:
Could you explain this formula?
It's explained in the permutations link I included in my earlier post.
 
  • Like
Likes jaus tail

FAQ: Probability of sum is 20 (4 dice are rolled)

1. What is the probability of getting a sum of 20 when rolling four dice?

The probability of getting a sum of 20 when rolling four dice is approximately 0.0772 or 7.72%. This means that out of every 100 rolls, you can expect to get a sum of 20 about 8 times.

2. How did you calculate the probability of getting a sum of 20 with four dice?

To calculate the probability, we first need to determine all the possible combinations of four dice that can add up to 20. In this case, there are 27 possible combinations. Then, we divide the number of combinations that add up to 20 by the total number of combinations (1296) to get the probability.

3. Is the probability of getting a sum of 20 the same for all four dice?

No, the probability is not the same for all four dice. The first dice has the highest probability of contributing to the sum of 20, followed by the second and third dice, and finally the fourth dice has the lowest probability. This is because the first dice has the most number of possible combinations (6) that can contribute to a sum of 20, while the fourth dice only has 1 possible combination.

4. How does the probability change if we add or remove dice?

If we add more dice, the probability of getting a sum of 20 increases because there are more possible combinations that can contribute to the sum. On the other hand, if we remove dice, the probability decreases because there are fewer possible combinations that can result in a sum of 20.

5. Is there a way to improve the chances of getting a sum of 20 when rolling four dice?

No, as the outcome of each dice roll is completely random, there is no way to guarantee or improve the chances of getting a sum of 20 when rolling four dice. However, rolling the dice multiple times can increase the likelihood of getting a sum of 20, but it does not guarantee it.

Back
Top