Dice sum probability

Three dice are rolled. What is the probability their sum is equal to 9? #math #maths #mathematics #probability #shorts

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[dagneytaggart]

I couldn't have answered that, but I understand the explanation of the answer. That's a good day in the world of mathematics for me.

[spiderjerusalem]

envisage these dices as three distinct boxes.
9 items are to be distributed into these, the minimum number of items each must at least hold is 1 (since dice's lowest number is 1, no zero)

so we first put one in each box,
leaving (9–3)=6 items left.

6 items now to be distributed into these,
use star&bars method:
ₙHᵣ = ₃H₆ = (6+3–1 6)
= (8 6) = 8!/(6!2!) = 28

invalid cases = {(7, 1, 1), (1, 7, 1), (1, 1, 7)}

so (28–3)=25 ways

Probability = 25/6³ = 25/216

[ricofilberto]

Or you can use stars and bars, then do some exclusion from the invalid cases that is 8C2 = 8!/(2!*6!) = 28 ways, and the invalid cases is (1, 1, 7); (1, 7, 1); (7, 1, 1); so there will be 28-3 = 25 ways, therefore the probability is 25/216.

[andreaspatounis]

Another way to solve it is by saying that if the sum of two of the three dice is x then:
If 2<x<9 then there is a 1/6 probability that the sum is 9
Else we are sure that the sum is not 9
The probability that x is less or equal than 2 is 1/36
The probability that x is bigger or equal to 9 is (4 + 3 + 2 + 1)/36 = 10/36
So the probability that 2<x<9 is 1 - 1/36 - 10/36 = 25/36
So the probability that the sum of the three dice to be equal to 9 is
(25/36) * (1/6) = 25/216

[CreativeThinker]

Coefficient of x^9 in (x+x^2+x^3+x^4+x^5+x^6)^3 divided by 216

[jamesgregory]

Very nice quick method, I wouldn’t of found that line of reasoning! The questions are getting easier!!

[wiggles]

I did it by thinking of the outcome tree. The root node has 6 branches, then each node on the next layer has 6 branches, and then each node on the next layer has 6 branches. If your first die is 1, then the next die has to be at least 2 so say it is 2, and the last die is then 6. Thus the first way to get 9 is (1, 2, 6). The next way to get 9 is (1, 3, 5). The next way to get 9 is (1, 4, 4). The next way is (1, 5, 3). The next way is (1, 6, 2). That's 5 ways so far. Then when your first die is 2, there are 6 ways to get a sum of 9. When the first die is 3, there are 5 ways. When the first die is 4, there are 4 ways. When the first die is 5, there are 3 ways. When the first die is 6, there are 2 ways. In total there are 5+6+5+4+3+2=25 ways to get a sum of 9. There are 216 ways to roll 3 dice, so the answer is 25/216. This is close to the same amount of work as the solution in the video.

[TheEulerID]

I just found it simpler to count the possibilities with the first die being 1, 2, 3 ... 6. Maybe not as mathematically sophisticated, but quick and easy, especially as a clear pattern emerges so you can short cut it.

[trumpetbob]

I didn't do the proper way to solve this, but I got the right answer anyway. In Microsoft Excel (to make it easier on myself), I created a 2x2 chart of all 36 combinations for two dice. Then, I counted how many results were between 3 and 8 (knowing we need to add one more die). I then took this result (25/36 total combinations) and multiplied it by the 1/6 chance the third die will land on the "needed" number.

[anonymouskitten]

My brain saw dice and thought “which ones”

[fluffytheostrich]

This makes an assumption that the six sided die is used. As it is not clarified, you could instead assume any number of alternative die types. Maybe you need to factor in whether dice are randomly pulled from a nerds dice bag? Maybe it's Yahtzee? Simply why the three dice are rolled together changes which ones night be used!!!

As such, I assume caltrops. 4*4*4= 64, and posit 4 / 64 = 1 / 16 instead

[Boonetube]

We don’t know, it never said how many sides.

[alxjones]

First die shows n, second shows m. We must have m between 1 and 6 but also between 3-n and 8-n. The third die is determined by these two. Count the cases:

n = 1; m is between 2 and 6, so 5 possibilities
n = 2; m is between 1 and 6, so 6 possibilities
n = 3; m is between 1 and 5, so 5 possibilities
n = 4; m is between 1 and 4, so 4 possibilities
n = 5; m is between 1 and 3, so 3 possibilities
n = 6; m is between 1 and 2, so 2 possibilities

Add them all up, (5+5) + (6+4) + (2+3) = 25

Then, just count the number of total die rolls, obviously 6^3.

Probability is 25/6^3

[stratonov]

Below C++ program worked for me like a charm:
int a[6] = {1, 2, 3, 4, 5, 6};
int count = 0;
for(int i = 0;i < 6; ++i)
for(int j = 0; j < 6; ++j)
for(int k = 0; k < 6; ++k)
if(a[i] + a[j] + a[k] == 9)
count++;

cout << count <<endl;

[jdhouse]

To count this you can also look at the polynomial
((x-x^7)/(1-x))^3 and look at the coefficient of x^9

This is because the thing under the ^3 is equal to (x+x^2+...+x^6), so raising it to the third power counts all the ways to sum up each exponent using 3d6

[GourangaPL]

or expand (1 + z + z^2 + z^3 + z^4 + z^5 + z^6)^3 and the answer will be the number next to z^9

[VivBrodock]

this is exceptionally easy, the odds of getting 3-8 on two dice is 25/36 multiply that by the odds of getting the number you need on the other die 1/6 and you get 25/216

[brookestephen]

the order does not matter. You also missed 1, 1, 7. This is a combination problem, not a permutation problem.

[jpcurley]

This is a method that doesn't extend well to large numbers. Using the product of polynomial combinations solves large problems like rolling 10 dice to get a sum of 27

[Mystery]

Just added the possibilities for 3-8 with 2 dice (2/36, 3/36...) and multiplied with 1/6.