Question

What is the probability that the total after rolling 4 fair dice is 21?
And of 22?
http://projects.iq.harvard.edu/files/stat110/files/strategic_practice_and_homework_1.pdf

Answer 1

I don't understand the answer given

Answer 2

If you roll 4 fair dice, how many total possible outcomes are there?

Answer 3

1296, right? (6*6*6*6)

Answer 4

You have 6 possible outcomes with each roll.
For each of the 6 outcomes of the first die, there are 6 possible outcomes of the second die. Then there are 6 possible outcomes of the third die and another 6 possible outcomes of the 4th die.
The total number of outcomes is 6 * 6 * 6 * 6 = 6^4 = 1296

Answer 5

Ok, you got that correct.
Now we need to see how many of the outcomes add to 21.

Answer 6

Yep...that's where I get confused.

I know we can use (6,6,6,3) (6,5,5,5) and (6,6,5,4). My problem is knowing how many of each one there are

Answer 7

Just using three 6's and one 3, there are 4 ways:
3,6,6,6
6,3,6,6
6,6,3,6
6,6,6,3

Answer 8

5,5,5,6
5,6,5,5
5,5,6,5
5,5,5,6

Answer 9

I don't think we need to count all the ways.
The question is asking what is more likely to happen a sum of 21 or a sum of 22 using 4 rolls of a die. You don't need to know each one. You just need to compare the results.

Answer 10

Yep. But if you check the answer they explain using permutations.
" To get a 21, the outcome must be a permutation of (6, 6, 6, 3) (4 possibilities), (6, 5, 5, 5) (4 possibilities), or (6, 6, 5, 4) (4!/2 = 12 possibilities). To get a 22, the outcome must be a permutation of (6, 6, 6, 4) (4 possibilities), or (6, 6, 5, 5) (4!/22 = 6 possibilities). "

Answer 11

I would like to understand why it is 4!/2...or how to do this in a less exhaustive way (so I can do this with big numbers)

Answer 12

4! = 4 * 3 * 2 * 1

Answer 13

Yep...but from where does the 4! come for this example?

Answer 14

Use the counting principle.
In how many way can you arrange 1,2,3,4?
There are 4 possibilities for the first position, then 3 possibilities for the second position, then 2 possibilities for the third position then 1 for the last position.
That gives 4 * 3 2 * 1
That means the digits 1,2,3,4 (without repetition) can make 4! different combinations.

Now what about the letters A,A,B,C?
Again you can do the counting principle and get 4 * 3 * 2 * 1, but since there are two A's, you must divide by 2 to eliminate groupings that look the same.
If we call one A "A1" and the other A "A2"
A1A2BC is indistinguishable from A2A1BC.
That's why we divide by 2.

Answer 15

Ok...so lets go with the example of 3,6,6,6. There are 4 possibilities for every position. So that would be 4! = 24. Then how did that decrease to 4?

I'm thinking in: (n!)/(r!*(n-r)!)
So that would be 4! / 3! (using r=3), but I'm not sure

Answer 16

Since there are three 6's, you must divide by 3! = 6
4!/3! = 24/6 = 4

Answer 17

Oh so 4 + 4 + 4!/2!. You divide by the repeated ones

Answer 18

So now for the (5,5,6,6) it is the same 4!/(2 * 2!)
Thanks!

Answer 19

When you roll 1 die, there is an equal probability of landing any of the 6 outcomes.

When you roll 2 dice, the minimum you can get is 2 and the maximum is 12.
(2 + 12)/2 = 7
The most likely outcome is 7. The least likely are 2 and 12. As you go down from 7 to 2, each outcome is less likely. As you go up from 8 to 12, each outcome is less likely.

With 3 dice, the outcomes are from 3 to 18.
The average is (3 + 18)/2 = 10.5.
The most likely outcomes are equally 10 and 11.
Outcomes from 9 to 3 and outcomes from 12 to 18 become progressively less likely.

When 4 dice are rolled, the outcomes range from 4 to 24.
The average is (4 + 24)/2 = 14.
Outcomes from 13 to 4 become progressively less likely, as do outcomes from 15 to 24.

Since 22 is more than 21, 22 is less likely than 21.

Answer 20

Take a look at this:
http://alumnus.caltech.edu/~leif/FRP/probability.html

Answer 21

Thanks a lot, much better now

Answer 22

You're welcome.