Quick probability question for you math braniacs-you have 3 dice,

Stat101 take-home final detected

>>8538325
all dice have a six

1/(6^3)

1/54

>>8538325
3 x 100℅ faggot

>>8538325

You can do this.

>>8538683

Well I have an idea. It's not for a test just my homework.

6 different combinations of 3 / 2 / 1

6 / (6*5*4)?

For the first die, there's a 1/6 chance of having a six. Since the next die must be different, the odds change to 1/5. And for the third die again they change to 1/4. So the final answer is 1/120.

>>8538689
Is the question whether the whole roll will add up to six, or whether at least one die will roll six?

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40KB, 300x400px

>>8538692
>For the first die, there's a 1/6 chance of having a six. Since the next die must be different, the odds change to 1/5.

>>8538693

Sorry it's the sum will be six.

>>8538325
(1/6)^(6)

>>8538692
Another way to think about this is you have 6^3 total possible rolls. 6 of these rolls have all 3 dice being the same (one possible for each number). Then another 5*3*6 ways to get two dice being the same number (three different arrangements for each number, times 5 for the third dice in order to not include the 3 of a kind). So the total number of dice with different rolls is 6^3-16*6=20*6. The number of these with one dice as a 6 is 3. So 3/(20*6)=1/60.

>>8538697
Oh I see. So for all the dice to be different, the possibilities of getting a sum of six are:
1 2 3
1 3 2
2 3 1
2 1 3
3 2 1
3 1 2
Going from >>8538708
that gives 6/(20*6)=1/20

>>8538325
Probability of rolling at least one six is 1/2. Probability of the sum being six is 5/108.

>>8538325
>given that they are all different
Well there wouldn't be three dice if they were the same dice

>>8538723
But it says given that they are all different, so you have to reduce the 6^3

There are only 6 combinations of dice that both add up to six and are all different (the permutations of 1 to 3)

this gives 6/6^3 or 1/36

from there you divide by the probability that they're all different. 1*(5/6)*(4/6)

(1/36)/(20/36)=1/20

1/6 + (5/6 * 1/6) + (5/6 * 5/6 * 1/6)

>>8538697
Expand
[eqn] \left( \frac{1}{6} x + \frac{1}{6} x^2 + \frac{1}{6} x^3 + \frac{1}{6} x^4 + \frac{1}{6} x^5 + \frac{1}{6} x^6 \right)^3 [/eqn]
and look at the coefficient in front of [math] x^6 [/math].

>>8538720
This, wtf are the rest of you talking about

>>8538325
We'll split the cases up and add them. First we count the total amount of die rolls. First we find that, if all three dice are different, there are 6 ways to choose 1 dice, 5 to choose the second, and 4 to choose the third. If two are the same, we have 6 ways to choose 1 dice, and five ways to choose the other two. If all three are the same, we have simply six ways.

Then we could how many there are without any sixes. This is the same thing but we only choose from a pool of five option. Then we divide:


[math]6\cdot5\cdot4+6\cdot5+6[/math] total combinations. [math]5\cdot4\cdot3+5\cdot4+5[/math] total combination without a 6. [math]\frac{5\cdot4\cdot3+5\cdot4+5}{6\cdot5\cdot4+6\cdot5+6}=\frac{85}{156}[/math]